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IMO 2010 Problem 3

Find all functions g:\mathbb{N}\rightarrow\mathbb{N} such that
\left(g(m)+n\right)\left(g(n)+m\right) is perfect square for all m,n\in\mathbb{N}.
IMO 2010 Problem 3

Solutio: First of all, see that for any n \in \Bbb{N} we have g(n)\neq g(n+1)\neq g(n+2)\neq g(n). The reason for this is rather simple, since if we suppose the opposite and replace m with n+1 or n+2 we see that the expression \left(g(m)+n\right)\left(g(n)+m\right) is exactly between two consecutive perfect squares and therefore cannot be a perfect square.

Secondly, we prove that |g(n+1)-g(n)|\leq 1. For this, suppose the opposite and take p a prime divisor of a=g(n+1)-g(n). Then g(n+1) and g(n) have the same residue modulo p. Take m such that m+g(n) and m+g(n+1) have odd power order p in their factorization. If |g(n)-g(n+1)|=p^k q then if k is odd, set m+g(n) to be an odd power of p greater than k and if k is even, then pick m such that m+g(n) is a large power of p plus p. Then one of the numbers g(m)+n and g(m)+n+1 is not divisible with p and therefore, the corresponding product is not a perfect square.

This leads to g(n+1)=g(n)+1 forall n\in \Bbb{N} and therefore f(n)=n+k,\ \forall n \in \Bbb{N} where k \in \Bbb{N}.

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  1. July 18, 2010 at 7:16 am

    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.


    • February 19, 2012 at 6:13 pm

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

  2. function
    February 19, 2012 at 5:14 pm

    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


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