APMO

First we perform the following substitutions on the original relation:

1. With $a=b=1$, we find that $f(1)+1\mid f(1{)}^2+1$, which implies $f(1)=1$.

2. With $a=1$, we find that $b+1\mid f(b)+1$. In particular, $b\le f(b)$ for all $b\in {\mathbb{Z}}^{+}$.

3. With $b=1$, we find that $f(a)+1\mid a^2+f(a)$, and thus $f(a)+1\mid a^2-1$. In particular, $f(a)\le a^2-2$ for all $a\ge 2$.

Now, let $p$ be any odd prime. Substituting $a=p$ and $b=f(p)$ in the original relation, we find that $2f(p)\mid p^2+f(p)f(f(p))$. Therefore, $f(p)\mid p^2$. Hence the possible values of $f(p)$ are $1$, $p$ and $p^2$. By (2) above, $f(p)\ge p$ and by (3) above $f(p)\le p^2-2$. So $f(p)=p$ for all primes $p$.

Substituting $a=p$ into the original relation, we find that $b+p\mid p^2+pf(b)$. However, since $(b+p)(f(b)+p-b)=p^2-b^2+bf(b)+pf(b)$, we have $b+p\mid bf(b)-b^2$. Thus, for any fixed $b$ this holds for arbitrarily large primes $p$ and therefore we must have $bf(b)-b^2=0$, or $f(b)=b$, as desired.

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Alternative solution.

As above, we have relations (1)-(3). In (2) and (3), for $b=2$ we have $3\mid f(2)+1$ and $f(2)+1\mid 3$. These imply $f(2)=2$.

Now, using $a=2$ we get $2+b\mid 4+2f(b)$. Let $f(b)=x$. We have $$\begin{array}{r}1+x\equiv 0(\mathrm{mod}b+1)\\ 4+2x\equiv 0(\mathrm{mod}b+2)\end{array}$$ From the first equation $x\equiv b(\mathrm{mod}b+1)$ so $x=b+(b+1)t$ for some integer $t\ge 0$. Then $$0\equiv 4+2x\equiv 4+2(b+(b+1)t)\equiv 4+2(-2-t)\equiv -2t(\mathrm{mod}b+2)$$ Also $t\le b-2$ because $1+x\mid b^2-1$ by (3).

If $b+2$ is odd, then $t\equiv 0(\mathrm{mod}b+2)$. Then $t=0$, which implies $f(b)=b$. If $b+2$ is even, then $t\equiv 0(\mathrm{mod}(b+2)/2)$. Then $t=0$ or $t=(b+2)/2$. But if $t\ne 0$, then by definition $(b+4)/2=(1+t)=(x+1)/(b+1)$ and since $x+1\mid b^2-1$, then $(b+4)/2$ divides $b-1$. Therefore $b+4\mid 10$ and the only possibility is $b=6$. So for even $b$, $b\ne 6$ we have $f(b)=b$.

Finally, by (2) and (3), for $b=6$ we have $7\mid f(6)+1$ and $f(6)+1\mid 35$. This means $f(6)=6$ or $f(6)=34$. The latter is discarded as, for $a=5$, $b=6$, we have by the original equation that $11\mid 5(5+f(6))$. Therefore $f(n)=n$ for every positive integer $n$.

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Alternative solution.

We proceed by induction. As in Solution 1, we have $f(1)=1$. Suppose that $f(n-1)=n-1$ for some integer $n\ge 2$.

With the substitution $a=n$ and $b=n-1$ in the original relation we obtain that $f(n)+n-1\mid n^2+f(n)(n-1)$. Since $f(n)+n-1\mid (n-1)(f(n)+n-1)$, then $f(n)+n-1\mid 2n-1$.

With the substitution $a=n-1$ and $b=n$ in the original relation we obtain that $2n-1\mid (n-1{)}^2+(n-1)f(n)=(n-1)(n-1+f(n))$. Since $(2n-1,n-1)=1$, we deduce that $2n-1\mid f(n)+n-1$.

Therefore, $f(n)+n-1=2n-1$, which implies the desired $f(n)=n$.