## IMO 2010 Problem 3

Find all functions such that

is perfect square for all

*IMO 2010 Problem 3*

**Solutio:** First of all, see that for any we have . The reason for this is rather simple, since if we suppose the opposite and replace with or we see that the expression is exactly between two consecutive perfect squares and therefore cannot be a perfect square.

Secondly, we prove that . For this, suppose the opposite and take a prime divisor of . Then and have the same residue modulo . Take such that and have odd power order in their factorization. If then if is odd, set to be an odd power of greater than and if is even, then pick such that is a large power of plus . Then one of the numbers and is not divisible with and therefore, the corresponding product is not a perfect square.

This leads to forall and therefore where .

Hi Beni, excellent solution.

I got one question. How did you prove that there exist an integer m such that “m+g(n) and m+g(n+1) have odd power order p in their factorization”?

I proved my self, but I had to split the proof into cases: when p is odd and when is not, and other sub cases.

Thank you for posting your solution.

Regards

My proof is similar to yours. Some cases, subcases, but it is not a hard argument. Thank you for your comment.

please solve it if you can:

find all functions Z to Z such that we have :

f(mn+1) = f(n)f(m) + f(m+n)

for all m,n in Z

please!!!!!