Show that .
Solution. For every positive integer we have by definition .

It means that for we have .

Note now that if a prime then ; but since then there exists a positive integer such that .

Also this trivial inequality holds: : so .

Summing up, .

Extendig this product, we have
.

For every integer it’s also true that .

Then we can say that there exists a positive constant such that .

Turning back to our problem, we proved that .

Now since for every the following inequality holds definitively, then we finished, indeed:

. []

In Fisher Black and Myron Scholes published their pathbreaking paper [Ref: The pricing of options and corporate liabilities, Journaly of Political Economy] on option pricing, where the main idea (attribuited to Rober Merton) is the use of trading in continous time and the notion of arbitrage. The simple and economically very convincing “principle of no-arbitrage” allows one to derive, in certain mathematical model of financial markets (such as Samuelson model, the so-called “Black-Scholes model” based on geometric Brownian motion), unique prices for options and other contingent claims. The main role of no-arbitrage has been started to be studied later, in the late seventies (see Harrison, Kreps and Pliska), and it can be summarized in the “Fundamental theorem of asset pricing”(the name was given by Ph.Dybvig and S.Ross), which is not just a theorem but rather a general principle deeply related with martingale theory. We’ll begin this article with the basic proof of the theorem, as in its first formultion by Harrison and Pliska, under contraints of discrete time and finite (this implies that all the function spaces are all finite dimensional and isomorphic to ). Clearly, these assumption are very restrictive and do not even apply to the very first example of the theory, such as Black-Scholes model ot the much older model considered by L.Bachelier. So we’ll establish some more general version of this theorem, first in a continous time model, and then assuming infinite dimensional spaces: these version are result of a long line of increasingly general research achieved in the past two decades up to nowadays. Loosely speaking, it states that a mathematic model of financial market is free of arbitrage if and only if it is a martingale under an equivalent probability measure (i.e. it has same sets of null measure); it plays a truly fundamental role in the theory of pricing and hedging of derivatives securities (or, synonymously, contingent claim, i.e., element of ) by no-arbitrage arguments; in this way the relation to martingale theory and stochastic integration opens the gates to the application of a powerful mathematical theory.

The whole essay can now be downloaded here: LF1302108

“The unique solution of with   is .”

Since the whole proof uses cylotomic fields, Galois theory, and other non elementary facts, we’ll present here some cases that can be completely solved with basic knowledge:

1- .

Consider the equation .

We have to show that, for all integer , is not a perfect non trivial power of another integer, i.e. has no integer solutions in except the trivial one .

It’s clear that has not non-trivial solution, so from now on we can assume that .
     
If then and in particular implies , that is , contradiction. We showed that is even, and is odd.

Working now in we have: . Since since is a unit. It means also that is divisible by , and it cannot be since we showed that has to be odd. In other words, and are both power of some complex number in .

So there exist such that .

Taking only imaginary parts, in the summation we have to consider only even , that implies: , a integer number divisible by : then , and in particular .

Then, it’s equivalent to say that .

Since and must be odd, then is a even integer. Looking the previous equation in , we must have , that is equivalent to

Taking in consideration the identity , we can say that for all integers the following chain of inequality holds: .

It’s enough to conclude that the equation has no non-trivial solution in integers greater than 1.

2- with .

If is a power of , then there are no solutions, directly by point 1.

Otherwise there exist a prime and a integer such that ; note now that has a solution if and only if for some integer . Then : it means that both factors are powers of . But , and we get a contradiction since:
. []

3- with for all .

Since , it’s clear that . Directly from point 1, we have that if than there are no solutions, so we’ll assume that is odd. Since , we have that if is a solution then , directly from point 2. It means that if is a (odd) solution, then .

Assume that there exists a prime such that . Then, we get a contradiction since:
.

Now ; moreover is a concave function (it means, with decreasing derivative function), so the previous inequality will not hold for all integer . And since, for there are no solutions, then has to be a power of .

Let a integer such that . Then will be even, and in particular the following inequality holds:

.

It’s a contradiction, since the previous inequality is false for all integers . []

Corollary: for all integer define the set   has no solutions in integers >1 }. Then .

Indeed for the inclusion-exclusion principle, it’s true that .

4- .

(Solution by dario2994).  Assume that for some there exists a solution ; as before must be odd and even. For every odd prime that divides  we have: . It implies , so it’s true also that , that’s a contradiction for every . So is exactly . []

5- and .

(Under construction)

6- .

(Under construction)

For each define the binary entropy , and let a fixed integer. Show that the following inequality holds for all .

Solution. Substitute the explicit value of in the inequality, and just regroup to obtain the inequality . It means that we can reformulate the problem in the equivalent form: “Let be positive integers with sum , then ”.  So, for every fixed we obtain that left hand side if this inequality is a function of the unique variable , since . We can show that attains its minimum at ,so that it will be enough to show that . Observe also that the function is symmetric with respect to .

Case1: for some . Fix , then , that is true if and only if : it can be rewritten as . Now it is well-known that the function is strictly increasing in , so that this inequality remains true if and only if , and in fact by construction . So, for the case even, we have to show that , that is for all positive integer .

Case2: for some . Just repeat the same ideas of previous lines: fix , then , that is true if and only if : it can be rewritten as . And, as before, it is enough that , and in fact by construction . So, for the case odd, we have to show that . Now multiply both members to , and we obtain: . Set and observe the following chain of inequalities: . We have shown that for the case odd, it is enough to prove that for all positive integer .

First, let’s verify manually that is greater than , defined as the maximum between and  for all :  (C, k(C),  h(C)): { (1, 2,  1.3), (2, 6,  3.2), (3, 20, 9.1), (4,70 , 28.4), (5,  252, 93), (6,  924, 315), (7, 3432, 1092), (8, 12870, 3855), (9, 48620, 13797),(10,  184756, 49932),(11, 705432, 182361)}.

Note now that for all real the inequality holds, so that the whole problem can be summarized in for all positive integer .

Define now the sequence of of positive reals by . It is clear that such sequence is (strictly) decreasing, with . Integrating two times by part, also : we can conclude that:

, that implies .  So, for the proof of our problem, it is enough that , that is ; using the weak inequalities and , it is enough that , that is . The proof is complete since we assumed . []

Problem 17- A sum of prime powers is not a power of 2

Find all such that ,where .

Solution. We will prove that if is a solution that .

1. If then , so that .

2.If and then . If then in we have , so that , but is absurd. Otherwise, , and in we have : this is enough to find the other trivial solution .

3. If and then . Since we have that , and taking we obtain , absurd.

4. From now, we can suppose , so that implies that .

5. In we have that , so and .

6. In we have that , so .

7. In we have that but residues of are exactly , so that we can conclude that .

8. To sum up all observations in points 5,6,7 , we can reformulate the problem as “find all such that for some .

9. In we have that ; and since then also . Now residues of are exacty , and we are searching , that is equivalent to .

10. To sum up all observations in points 8,9 , we can reformulate the problem as “find all such that . We divide the rest of the solution in two cases:

PART 1.

11. Let’s find all solutions in the case , that is in non-negative integers.

12. The equation in point 11 is equivalent to , and we can see the trivial solution with that is . Let’s show that no other integer solutions exist.

13. Consider the 2-adic valuations: it is true that . And by the lifting lemma it is equal to , that implies , so .

14. Consider the equation in (after observing that ), and we obtain: that is equivalent to

15. To sum up points 13,14, the equation becomes equivalent to for some positive integers.

16. After observing that (that is equivalent to ), consider the equation of point 15 in : we have that and . This is a contradiction.

PART 2.

17. Now let’s show that in the case , the equation has no solutions.

18. In (the information in this point makes the difference from the previous part) we have . Now, since , we conclude that , that is equivalent to and .

19. Observe that and take the equation in , we have: . We can see directly by hand that residues of are exactly and residues of are . And we can see that is a solution if and only if .

20. Since we know that for point 18, then it is also true that . In particular if is a divisor of then is constant in .

21. In we have . Now and : complete residues of LHS are , and complete residues of RHS are .

22. Looking previous residues we can see that if is a solution modulo ( we can say that because ) then .

23. In noone of previous couples modulo we verify that , as requested in point 18. This is a contradiction. []

Problem 18. A equation on 4 non negative integers.

Solve the equation in nonnegative integers.

Solution. Since the equation is simmetric in , we can assume without loss of generality that .

Moreover, we must have : in fact, if then there exists a integer such that is a perfect non-trivial power. And it’s well known (working in ) that the unique solution is , that is , but or do not work.

If then , that is a contradiction. And, if then that implies , so if is a solution then and are coprime positive integers.

Also, looking the equation in and we deduce also that and .

If then and, as before, is never a non trivial perfect power.

If then, since , we have to verify if , and we have the solutions (and the simmetrical ).

If then, since , we have to verify only two cases: and , and both have no solutions. From now, we can assume .

Since then we can examine first cases (when is “big”): if for some then (this congruence is implies , so there exist such that and ). It means that :

- if then , but we already examined this case;

- if then , and . But and we have the solution and its simmetrical one.

- if then , and since we have to check only cases . But in both cases is not a integer, since is not a quadratic residue in . From now, we can assume that if then : it means that .

To sum, other than two previous solutions, if verify the original equation, then and , and .

If then and if and only if , that is a contradiction.

If then (that is implies ), or (that is , contradiction).

If then , that is a contradiction.

We proved that if is a solution (with ) then it belongs in . []

Problem 19. Simmetric divisibilty

Find all such that .

Solution.  It’s clear that is a infinite class of solutions and that otherwise we must have . It holds also . Since then it holds also : divides .  But , that’s a contradiction. []

Problem 20. Even implies divisible by 8

If are fixed such that is a perfect square, then is idivisible by .

Solution. Define the integer , then is a perfect square. Since is even, then for all . If in then . Otherwise , that’s our aim. []

Problem 21. Sum of powers of 2 is a fixed integer

Find all non negative integers such that .

Solution. It’s well-known that every positive integer has a unique binary representation. If is a fixed positive integer, then define the number of 1 that are in his binary representation.  Since then . Now, if then and it means that .  Instead, if and then : it means that if or if . In all previous cases . It means that if a triple solution exists then wlog . But since the binary representation is unique, then we must have and all his permutations. []

Problem 22.  Divisibility again

Define . How many positive integers satisfy ?

Solution. Since and is or (but in all cases coprime with ), then by chinese remainder theorem we’ll have exactly a solution in the interval to the system of congruences: , where and . The number of way to choose and is evidently , and noone solution of them is exactly . Otherwise we have the cases in which for some , but we have to exclude the case that leads to the solution .  So, we can conclude that there are exactly distinct positive integers such that . []

Problem 23.  A Easy equation in positive integers

Find all such that .

Solution. Since then it implies that , and since then also , that’s . Now the following inequality holds: , that’s equivalent to : it means that , but also this couple does not satisfy the original equation, so there are no solutions. []

Probem 24. Exponential equation with integers

Find all such that .

Solution. We’ll show that if is a solution then .

- If then .

- If then .

- If then is a integer, since it’s difference of two integers. It means that a positive integer exists such that  . The equation becomes , that’s . If then , and , implying that . Otherwise : then .

- If , then is a integer, since it’s difference of two integers. It means that a positive integers exists such that . The equation becomes . Since then implies that .  The following chain of inequalities holds: , that’s a contradiction.

- If we have no solution since the equation becomes and is not defined in for all integer .

- If , then define the positive integers ; the equation becomes and it implies and . This system has no solution in positive integers . []

Proposition 1.  The statement is true if . (TST Germany 2006)

Solution. We’ll show that it is true for infinitely many values of , where . We can note that if  then . At the same way if then . Assume that , then and : so if (where is the Mangoldt function) then for some ; but it has only finitely many solutions, since by the Lifting Lemma , so , that is false definitively.It means that , so  is a square. Assuming that  by contradiction, we should have . But we can easily show a more general fact: is never a non-trivial power of some positive integers. In fact, if for some positive integer , then is even (otherwise modulo we should have ). But working in the extension  we have , so they are both -power. So it must exist such that , but seeing the -coefficient we must have , from which the unique trivial solution follows. It means that : in other words, if then is a power of . 

By proving the original statement, it is enough to assume that it holds for only finitely many values of , and we’ll obtain a contradiction. So, in particular, assume that if is not a power of then  definitively. It means that if is enough large to avoid all finitely many solutions of above cases then , that is definitively. And it implies that , where and is the -th prime of . Taking the natural logaritm, it is true that  (where is a Chebychev function). But the Prime Number Theorem is equivalent to , so we have that is asymptotically greater than . And PNT is again equivalent to , so we should have that is definitively greater than , where represents a positive fixed constant. And it is clearly false. []

Proposition 2. The statement is true if . (Brazilian Olympiad 1995 and Bulgaria 1995)

Solution. It is enough to fix for some .

Infact , and assume that :it means that . Assume again that then we’ll have, after continuing this algorithm infinitely many times, for all . Note that cannot be definitively a power of since it is in the form (see proposition 1). So, if for some then , so definitively, that is clearly a contradiction. []

Proposition 3. The statement is true if . (Unknown source)

(Under construction)

Problem 2 Let be a set of boys: they play with each other in a tournament of Pro Evolution Soccer 2009, in respect of the following rules:
i) every boy play one and only one time against each other boy (so we can assume that every match has the form for some );
ii) if the match , with , ends with the win of the boy , then gains point, and doesn’t gain any point;
iii) if the match , with , ends with the parity of the two boys, then point is assigned to both boys.
(We assume for simplicity that in the imaginary match the boy doesn’t gain any point).
Show that for some positive integer there exist a set of boys such that, for all choice of the positive integer , the boy gains always a integer number of points in the total of the matches .

Problem 3 Define for all , where .
Let fixed; show that exist infinitely many such that .

Problem 4 Let be a permutation of .
If we know that is not monic, show that if exist such that then .

Problem 5 Let fixed. Show that exist infinitely many such that .

Problem 6. Let be the Lucas sequence defined by for all . Show that every odd divisor of has a even tens digit. (by Gebegb)

 Problem 7. Let fixed, and define , and let a real fixed. Show that holds definitively.

Solution. We can begin noting that the infinite sequence of positive integers defined by for all verify the divisibility relation . Easily by PMI, if for all , then . On the other hand it is also true that the infinite sequence of positive integer defined by verify the divisibility relation too, infact , and again by PMI, if for all then . So it is enough to show that , that is true since in we have .

We can show now that only the sequences and have only a finite number of common elements: in fact if for some then , but it is well known the equation has only finitely many solution in , since by the Lifting Lemma , but , that is false definitively (it can be seen also as a corollary of Mihailescu Theorem).

Obviusly both sequences are strictly increasing, so it is true that each sequences have at least respectively and elements that are not greater than , so definitively.

About the upper bound, it is equivalent to show that .

Note that for all , so that . Now if then for all sufficiently large. It means that , and also that if then . So, in the set there are at most numbers multiple of .

Let fixed and define . Clearly we have that for some positive real fixed.

It is enough to conclude since , but if is sufficiently large, then both limits can be arbitrary small. []

Problem 8. Let a convex quadrilateral fixed such that , . Define the midpoint of ; show that

Solution . X in BE such that CXE=CDE so DECX cyclic so DXE=DCE=180-CDA-CAD=180-CAB-CAD=180-BAD, so ABXD cyclic so BDA=AXB=CXE=CDE.

Problem 9. Find all such that .

Solution. 2 doesn’t divide f(x):= x³ – 5x + 17  that is integer, so min{y,z}>-1 and y = 0. If   there are no solutions, so Now in so we have the only case . So must be a cube, but for all .

Problem 10. Define the function such that if , and otherwise. Define the function such that for all . Define also the sequence of integers such that e , where if , and otherwise. How many distinct elements are in the set ?

Solution. Consider the obvious bijection between {1,…,2048} and the strings of 11 binary digits. The problem states that we begin with , and that if or (where is a string of 10 binary digits) then if we can find i<n+1 such that , otherwise . Define and . Suppose that for some e in {0,1} and a string y of 10 binary digits, then by definition of k, so define the other i such that 1<i<k+1. And, if then define j that integer such that 0<j<k+1 and (it must exist since ). If j>1 then exist and is , then , contradiction, so j=1. It means that $latex y=0 \in S$, and if then , contradiction, so . Clearly ; suppose that . A number and a string of 10 binary digits exist such that , and suppose (clearly ).Define for some 1<w<k+1, then there exist 0<z<k+1 such that $x_z = b1 \in S$. As before if z>1 then exist and is , contradiction; and if z=1 then b=0, again contradiction. So a number does not belong to S only if . But in this way we can define a sequence of number that do not belong to S, and that is eventually null, but .
It shows that .[]

Problem 11.  Let be two positive integer fixed such that . Evaluate .

Solution. It is equivalent to search the minimum positive integer integer n such that and have the same number of digit in base . Working in base , if then , so the max “report” when we compute in the leftmost place is 1.And trivially need to have at least digits. So the minimum integer is .[]

Problem 12. (a) Let positive integer such that . Show that for infinitely many positive integer . (Laurentiu Panaitopol,TST Romania1987)

(b) Show that there exist infinitely many triple of positive integers such that and and for infinitely many positive integer . (Me)

 Solution. (a) Since is a symmetric polynomial, it can be written in terms of , and (see  the fundamental theorem of symmetric function). Now since by assumption, if is made only by monomial of the form , then it has a degree multiple of 3. But .

(b) There exist infinitely many primes such that (in fact, if this set is finite and P is the product of its elements, then should have only prime divisor of the form but it is false since for all prime such that ; in all way, it is a corollary of Dirichlet theorem). Set and then ; it means that if and then for some integer (in fact takes a complete residue system in and  does not work). Define this integer, so and for infinitely many positive integer . This is true, since it is enough to take and by Euler criterion we have in . But a sum of odd number of odd addends is odd, and in particulat it is non null.[]

Problema 13. Fix a strictly increasing sequence of such that ten n-th terms is not bigger than nx+y definitively for some . Show that .

Solution. By contradiction if is set of prime divisor of and    then we’ll have that   definitively, that is clearly a contradiction. []

Problem 14. Prove that in the set there are elements such that any three of them are not in arithmetic progression.

Problem 15. Let be integers such that ; is $500 < a_1$ (always) true ?

Solution.  It is enough to show that if we have distinct positive integers s.t. noone divides another one, so the smallest element is at least . Let be the adic valuation of , so naturally there exist a bijective function from to , where . Consider the numbers s.t. : if and we have contradiction: it follows that , so the only number with has the property that . Now it is enough to prove that , but it is quite easy considering the number in the form and using similar tecnique as above.

Problem 16. Let fixed and let  be three polynomial such that . Show that the equation has only a finite number of solutions, where denotes the sum of digit of a positive integer .

Problem 17.  Let fixed such that and fix also non constant polynomials for each . Now define the function and such that if appears in the decimal representation of , otherwise .
Show that it does not exist a non-zero polynomial such that for all the equation :
holds.

From Cono Sur Olympiad: “Show that for all there exist such that the number has all digits in the set for all “.

Solution. It is enough to set .

Let’s verify our claim. If then , that clearly works. Otherwise, suppose that . We have by construction that for all choice of , so it is enough to concentrate our attention on the digits of the set . Since in decimal representation, for all there are exactly digits (such that the first one is strictly positive) then it is easy to see that the number of digits of is less than the number of digits of for all . In fact the inequality holds always. It means that it is enough to verify that the set of numbers has in total  all digits in the set for all fixed. And this is true, let see why; if is a power of then our claim is trivial, since it is true for . Otherwise, define . Since and , then in every set there is at least one element of for all fixed. But it means that the leftmost digit of such element of is exactly , for all fixed. And it is enough to deduce that our claim is true. []

Problem 2- Extended sum of digits.

Let such that and define the sum of digit function in base . Show that if for all integer and , then . (Gabriel Dospinescu)

Solution. For  sufficiently large choose , so . But for all , so and for all . Choosing we obtain , so that is divisible by if , otherwise divisible by (the case is trivial since and would be a power of 2), that implies . Now if then so otherwise we can assume , so choosing enough large and enough large such that we obtain  then .[]

Problem 3- Number of coprime subset of {1,2,..,n}

Show that is not a perfect square for all integer .

Solution. Define for all ; here denotes the convolution product between two arithmetical function; define also for all ; for all ; for all , and  for all . Clearly , and , and . By Moebius inversion and operating in we have : if is odd then ; if for some odd then , otherwise for some and odd, then . We have concluded that for all , but , so cannot be a perfect square.[]

Problem 4. A special case of Dirichlet theorem

Dirichlet theorem states that if is fixed such that and and defined , then diverges. The problem is: show the statement for .

Solution. Suppose that the statement is false, so we can define the number . Let fixed and define also sets:

;

;

;

.

Obviusly for all t in A(k)  and and . Furthermore, if and are fixed, then in the set there are at most  solution of . And, again, we have that . It is enough to deduce that .

Now, we can show by PMI that if for some enough large then . If it is clear, then suppose that it is true for . For we’ll have that if then can be choosen in ways, where and ; and is clearly a square and can be choosen in ways by assumption. It means that , that is our aim.

It means that if then |D(k)| is not greater than the value  , but , that is a contradiction. []

Problem 5- A inequality with Euler function and gcd(.)

Show that for all the inequality holds. (Nanang Susyanto)

Solution. Let such that . Define such that and .So there exist such that and . So we must have and since by construction then , and it is equivalent to say that , and , that is . Let be its canonical factorization, then by Lifting lemma we know that and for all . But by construction , so . It is sufficient to deduce that . So, it is enough to show that to conclude the problem.

We claim that , and it is enough to show that for all s.t. the relation   holds. In fact if for some then so we can assume wlog . Define and in , so . We have by construction that , so the smallest such that is . Now we have , so . If then in we have so , that is impossible. So we have . Obviusly , so . And because we conclude that , that is our aim. []

Problem 6- A functonal with cubes

Find all function such that for all .

Solution. Define then is identically null. Now implies that and for all implies that is a odd function. Define , then implies that . Now and respectively imply that and . We claim now that all and only functions that we are searching are for all for some fixed. Since is odd, it is enough to check that for all . We have that so , so , so and so also . Suppose now the we have proved the statement for all for some t in .Let’s prove the statement for by PMI: in fact since such that and , so that means and finally so also , and the induction is complete. []

Problem 7- A strange divisibility

Let fixed such that ; show that .

Solution.  Working in we have that and by Euler criterion we have that the constant term of the polynomial is , so we can conclude . []

Problem 8- Minimal sum and difference with primes

Let the -th prime of and   a fixed integer. Find, if it exists, the smallest absolute constant such that for at least one choose of .

Solution. We claim that exists and it is . Define , and obviusly will be for all . It is enough to prove that  is if is even, otherwise is (note that if then since the required sum is invariant modulo ). Let’s begin with some cases and complete the proof by PMI showing that if is fixed, then we can obtain (by a suitable choice of ) all integer such that . If then trivially since we can choose . If then since we can choose . If then  since we can choose . If then since we can choose . With we can obtain all odd integers in the interval (in truth, if we can find always counterxamples), in fact it is enough to see that: , , , , , , , , , , , , . Now, remembering that (Bertand postulate), we can obtain all integers with the parity of  of the intervals (choosing ) and of (choosing ). But the interval is a subset of their union, so the induction is complete. [] 

Problem 9- When n^2 +1 divides n!

Show that for some if and only if the proposition i) and ii) are both false:

i) ;

ii) .

Solution. Part 1: “only if”. If for some then clearly and in particular , that contradicts the proposition i). On the other hand, if then it should be and in particular , that is false for all . Part 2: “if”. If  and then we want to show that . Remember that is never a non-trivial power of some positive integers. In fact, if for some positive integer , then is even (otherwise modulo we should have ). But working in the extension  we have , so they are both -power. So it must exist such that , but seeing the -coefficient we must have , from which the unique trivial solution follows. In particular since then . Part2/case1: . We have . Let fixed: if then ; if then there exist such that and : in particular we have since iff , and it shows that ; otherwise then we have and so . It is enough to show that . But if a prime divides then and then . And so the inequality holds for all integer since and if it is true for then , that is true. Part2/case2: . We have now , so the condition on proposition ii) is never satisfied. Let fixed:  if then ; if then there exist such that and : in particular we have and the conclusion is the same as in part2/case1; otherwise then we have and the conclusion is the same as in part2/case1 again. []

Problem 10- A condition for being a covering system

If  and are fixed integers such that for all there exist that satisfy , then find the smallest possible value of the positive integer . (Turkey National Math Olympiad 2009)

Solution. We claim that the required number is .
Assume by contradiction that for we can find integers and s.t. is a covering system. Obviusly we must have that the system covers all residue classes in , and if then we cannot have a covering system since each congruence can cover at most one of the class residues. At the same way if then there exists a residue class in that is not covered. Now since if then , that is absurd. It means that is exactly . Now we must have that the system convers all and clearly we need have and . Now we must have and since otherwise , that is absurd. So there exist such that , but we have always , so if exists then it is : but in we have that the system covers at most residue class, and the last one is not enough to close such bug. On the other hand and is a covering system. []

Problem 11- Only a few element in harmonic sum determine its divergence

Let fixed and be the set of numbers such that doesn’t occur in their decimal representation, then .

Solution. Define the reciprocal of number of digits of and for all then    . Corollary: since is not convergent,then there exist infinitely many primes that contains in its decimal representation. []

Problem 12- A equation involving Euler Phi-part 1

Find all such that .

Solution. Since for all we have that , so the chain of inequalities holds, so if then . Since is not a power of then cannot be a power of , so . Now if then that is a contradiction, so , i.e. is odd. Suppose that for some then that is ; and it is easy to see that since otherwise that is false. If then i.e. , so is a solution. From now suppose that is not prime. If then since otherwise also i.e. if divides then and so that is again a contradiction. So  and . So there exist primes such that and , i.e. that is absurd. It proves if is a solution then . Define  : if it prime then , i.e. , contradiction. If then , contradiction; case 1: if and then and (as above), so we are lead to with , and they are not solution. Case 2: with , and . If then that is false, so there exist primes such that . It leads to and the unique solution is clearly . We have shown that for some positive integer if and only if . []

Problem 13- A equation involving Euler Phi-part 2

Find all such that .

Solution. Since for all we have that , so the chain of inequalities holds, so if then . Since is not a power of then cannot be a power of , so . Note also that that is and for all . If then ,contradiction and if then  again, since implies ,contradiction. Case 1. If then and . Now implies that : if then , i.e. , that is a solution, otherwise it can only be a square of a prime and , contradiction. Case 2. If and then . Now since otherwise , that is false. In particular it implies that . If and then , i.e. , otherwise, as before, it can only be a square of a prime and , contradiction. If and for some primes then , i.e. and it has no solution; otherwise for some distinct odd primes such that , so we have , i.e. . Now in we have , so cannot lead to some solution. We can also easily see that lead both to solutions . It remains only the last case , so for some primes . So the equation becomes , but , that is false for all ; otherwise we must have and so , but , contradiction. We have shown that for some positive integer if and only if . []

Problem 14- A strictly increasing sequence

Let a strictlyincreasing sequence of positive integers such that for all . Then . (Carlo Sanna)

Solution. We assume to demonstrate it for all . Let a strictly increasing sequence of positive integers such that and define for all . Let the (numerable) set , then by assumption . It means that in worst case we would have . It means that . So it is enough to show that , that is , but it is trivial since is a strictly increasing sequence.

Remark: The inequality is true for . In fact, as before, , so the last inequality can be replaced by . It means that for all . 

Corollary:Define the sequence and for all . Then for all positive integers . For it is trivially true. Suppose it is true for all . Then, by PMI, we want to show that . Summing all identities we have that . So it is enough to show that . Since , we have that it is enough that . If then rhs is at least . On the other hand, if then we have prove that ; the right hand side is a quadratic function in and with negative coefficients of degree 1 and 2 (taking in mind that ): it means that it is a decresing function in , so it is enough to prove the inequality for the worst case, . So we are lead to prove that , that is obvious. []

Problem 15- A generalization from IMO99

Find all such that .

Solution. Trivially . If then implies that also . If then we should have that . Obviusly so that ; and since we have that we have that . Now since we have that so we have . Define , and if it is not defined then . As before . In particular implies that . So if another solution with exists then for some but , that is a contradiction. Now we can suppose that . As before, define , then , and on the other side, since obviusly ,then . It means that . If then that is absurd since . Otherwisev and we have at the same time , that is that is clearly absurd. []

Solution. Define the aritmetical function . Note that the statement is equivalent to . Let’s begin with some estimates about .

We have that for all since is a multiplicative aritmetical function and is a lower bound for the number of distinct odd prime factor of . And define the set of functions  where . Let fixed and define also the set .

Clearly . On the other hand, because of the observation above, we have that . But it means that , or equivalently .(*)

Let show now by PMI that for some constants fixed. For the statement is surely true (see for example well-known Chebychev’s result about ) and suppose that it is true for .

(Under construction)

Solution. There are exactly numbers multiple of and strictly less of in the set . So exist such that ,  since for all . But it implies that : that is, if a prime divides then , and at least one of the inequalities is strict. It mean that for all . Now let the number of quadratic residues in the set , where also is considered a quadratic residue. Since there are exactly even integers in the set , then . It is easy to see by Chinese Therem Remainder that if for some integer then . But  , and . Then the inequality directly implies .[]

Additional observations. If a such permutation exists (see below for examples) and there exist always a even number of such ones since can be exchanged. Suppose that and ; then for all and . It means that if a permutation of exists, then the set of (where is considered a quadratic residue in ) is sent to the same set of   where, by force, we must have for some integer  (in the case of it happens essentially the same thing, but with invariant parity of each set, so that we have in total disjoint cycles). Now for all , so if then   need to be : it means that if   and then is the unique odd in the set of . It means also that if for some and then such permutation does not exist (since and cannot be both odd).

Something else, in the case we must have , otherwise since exponent can be taken modulo then  is a permutation of . If is one of possible primitive root in , then define such that for all . Also, define such that for all . It means that is a permutation of , or equivalently that is a permutation of , where and are both permutation of itself. But , then , that is a contradiction for all odd primes .

Again in the case , since then   that holds iff . Now by Euler criterion we have that in  for all integer such that : so if  then   belongs to , and if then . And for obvious reason in every case holds.

Note that it works for all , infact: works, since we can choose , works, since we can choose , works, since we can choose , works, since we can choose , works, since we can choose , works, since we can choose ,  and so they don’t work, , but and so it does not work.

A note by Tho Tung Nguyen(first comment). Case . Let one of possibile primitive root in and define a integer such that . The number of distinct integers such that there exist z in {1,2,..,p} that verifies is exactly . In fact, is always a -th power and if and for some then if and only if for some . But , so it holds if only if . Now, we return on the problem: assume that , then there exist such that . Define the set of all -th power in . Since , also must have exactly number that are -th power. But , otherwise there will be a number greater of of q-th power. But it means that if and then . It means that , that is but implies .  Because by assumption we have , then , that is . And in fact if then is the unique solutions.

Additional observations (continue). Again about the case . We’ll show that for some . Note that for all prime , so it is enough to suppose by absurd that is greater than . By the previous note we know that , so it implies that for all such that . Define the set of all x-th power in , as before, and .Define also the subset of all multiple of in . It is clear that if a prime divides  then we have that  is a subset of . But , so once is fixed, there must exist a integer such that  and it is one of two smallest divisor with factors, and . It means that  is a permutation of the set  in . It is true if and only if is a permutation of in . And this is true if and only if is a permutation of in . Now and , and if it is really a permutation, then also the product of all elements must be the same. But by force and the product of the first set is that is in and the product of the last set that is in , contradiction.

By finishing the case we’ll show that the necessary condition that is a Sophie Germain prime is also sufficient. In fact, it is enough to construct such a permutation . Set and define the set of non quadratic residues without and the set of quadratic residues without and . Define also the subset of all odd numbers in without and , and the subset of all even numbers in without and and adding $2q-1$. We need to search two permutation of sets and such that every exponent of is chosen in and it is again a permutation of , and every exponent of is chosen in and it is again a permutation of . Evidently, if   is a odd generators of then the set can be represented as where . And the set can be represented as where . It is obvious that we need to work in : now if a element that belongs to can be represented by for some then set . Otherwise (i.e. it can be represented as for some ) set . It can be easily seen that the image of is really a permutation of . The same work can be done with the set . In fact can be reprented as where is a even generator in and and the set as where . And set and if , and if . Also this time it can be easily seen that the image of is really a permutation of in .

We are lead now to the only remaining case for some . The note of Tho Tung Nguyen can be directly extended to this case, and working in the set we have that if for some prime  then , that is , as before. So, if then .

But next observation about cannot be applied since the presence of a double residue system modulo will square the residue from which we have obtained the contradiction. Any help would be appreciated (it can also be that the condition is necessary and sufficient, but I don’t have a proof of it..)

This time we’ll prove that for every integer there exist infinitely many positive integer which can be written in more than ways as sum of two odd primes.

As almost proofs about prime numbers, we’ll show assuming the contrary, i.e. the sequence is bounded by some real constant , where is defined as the number of ways that the number is representable as sum of two odd primes.
Obviusly represents the coefficient of in the infinite serie , where .

Assuming that is a random variable in , we can say that the inequality holds, where we have used the well-known identity for every .

It follows that holds. So it is also true that .

It means that , so it is enough to prove that the well-known serie diverges to get a contradiction (note that we can substitute in the text of the problem the sequence of primes with a general sequence of positive integers such that diverges). (*)

We’ll use now the well-known Euler product: it states that if is a multiplicative aritmetical function such that is absolutely convergent, then the identity holds. We have since for all , and is not null everywhere by the definition of aritmetical function. Define and and , for every real fixed.

It is clear that if a positive integer then , and it means that .

The statement follows taking the limit , in fact the product converges absolutely since , that converges by assumption. So we have shown that , taking in mind the additive stucture of . In particular if is a completely multiplicative function then .

Assuming the above notation, it is clear that . (**)

It can be useful also to know the Abel partial summation. Let a divergent and strictly increasing sequence of positive real numbers and let a sequence of complex numbers. Define a arbitrary function and , for every real fixed. Then .

In particular, if (so it is continous and derivable), and and , then is constant in and in , so we can say that:
.

Now, if for all and for all , then , where is the well-known Eulero Mascheroni constant.

Turning at (*) and noting that for all , we can conclude with:
, that is clearly unbounded.[]

We can note finally that we have shown a strong quantitative lower bound on , in fact the famous Mertens formula for prime numbers states that exist a fixed such that for all sufficiently large we have .

We remember that first, second and third Mertens formula states respectively that: (with the usual notation) and the formula of primes, useful in the proof of Prime Number Theorem, can be obtained with the Abel partial summation directly on the second one, but it can be found in all number theory book.