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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.

Regards

• 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. 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