11. find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(m)^{2}+f(n)$ is a divisor of $\left(m^{2}+n\right)^{2}$ for all $m, n \in \mathbb{N}$.

## 1st solution
With $m=n=1$ it follows that $f(1)^{2}+f(1)$ is a divisor of 4, therefore $f(1)=1$. For $m=1$ we get

$$
f(n)+1 \mid(n+1)^{2}
$$

For $n=1$ we get

$$
f(m)^{2}+1 \mid\left(m^{2}+1\right)^{2}
$$

Let $p$ be a prime number. With $n=p-1$ it follows from (1) that $f(p-1)+1$ is a divisor of $p^{2}$. Therefore, $f(p-1)=p-1$ or $f(p-1)=p^{2}-1$. Assume the latter is the case. From (2) then follows $\left(p^{2}-1\right)^{2}-1 \leq\left((p-1)^{2}+1\right)^{2}$. However, for $p>1$ the following applies

$$
\left((p-1)^{2}+1\right)^{2}=\left(p^{2}-2 p+2\right)^{2} \leq\left(p^{2}-2\right)^{2}2 n>n$. We claim that there are natural numbers $a, b \leq n-1$ so that $f(n)+a^{2}$ and $f(n)+b^{2}$ are divisors. If $f(n)$ is even, then choose $a=1$ and $b=3$, if $f(n)$ is odd, then choose $a=2$ and $b=6$. This is possible because of $n \geq 7$. With the divisor strangeness and (3), the following follows

$$
\left(a^{2}+f(n)\right)\left(b^{2}+f(n)\right) \mid(f(n)-n)^{2}
$$

On the other hand, the left-hand side is greater than $f(n)^{2}$, but the right-hand side is less than $f(n)^{2}$ because $f(n)>n$. This is a contradiction, so $f(n)=n$ applies.