Tintarn wrote:

I think you missed the condition that $m,n$ are integers, not necessary positive. Solilin wrote: Find all pairs of integers $(m,n)$ such that $\frac{m^5+n}{m^2+n^2}$ and $\frac{m+n^5}{m^2+n^2}$ are integers. I will analyze only case $mn \not = 0$ Let $d=gcd(m,n)$. From first we have $d^2|n$, from second - $d^2|m$, so $d^2|d \Rightarrow d=1$. Then $gcd(mn, m^2+n^2)=1$. We have (by $mod(m^2+n^2)$) $0=m^5+n=-m^3n^2+n=-n(m^3n-1) \Rightarrow 0=m^3n-1$. From second we get $0=n^3m-1$. So, $0=mn(n^2-m^2) \Rightarrow 0=n^2-m^2$. So, $m=\pm n$. Case 1) $m=n$. Then since $gcd(m,n)=1$ we have $m=n=\pm 1$. They both works. Case 2) $m=-n$. Since $gcd(m,n)=1$ we have $(m,n)=(1,-1); (-1,1)$. This works. If $mn=0$, wlog $n=0$. Then $m^2|m \Rightarrow m = \pm 1$. So, answer is $(m,n)=(1,1),(-1,-1),(1,-1),(-1,1),(0,1),(0,-1),(1,0),(-1,0)$.

OK, but my solution is still correct until the conclusion $m^2+n^2 \mid 2m^2,2n^2$ and the rest is a trivial casework (if $mn \ne 0$, we still get $|m|=|n|$ and then $m,n=\pm 1$, otherwise if $m=0$, we get $n=\pm 1$).

Actually, the official exam paper only asked for positive integers pairs. I think the OP did a typo or misread the problem.

My bad. I fixed it.

A slightly different solution which is probably much less efficient but I found pretty motivated. We claim that the only pair of positive integers $(m,n)$ for which both the given fractions are positive integers is $(1,1)$. It is easy to see that this pair of solutions in fact works. Now, we shall show that there are no other solutions. We first prove the following key claim. Claim : If $(m,n)$ is a pair of solutions, then $m$ and $n$ must be relatively prime. Proof : Consider $m,n$ for which $\text{gcd}(m,n) >1$. Then, consider some prime $p \mid \text{gcd}(m,n)$. Then, WLOG assume that $\nu_p(m) \geq \nu_p(n)=c\geq 1$. Note that if we let $m=p^cm_1$ and $n=p^cn_1$ we have \[\frac{m^5+n}{m^2+n^2}=\frac{(p^cm_1)^5 + p^cn_1}{(p^cm_1)^2 + (p^cn_1)^2} = \frac{p^c (p^{4c}m_1+n_1)}{p^{2c}(m_1^2+n_1^2)}=\frac{p^{4c}m_1+n_1}{p^{c}(m_1^2+n_1^2)}\]But, since $p\nmid n_1$ (we considered $c$ as the maximum power of $p$ which divides $n$), the numerator is not divisible by $p$ but the denominator clearly is. This is a clear contradiction and thus, such prime $p$ cannot exist which proves the claim. Now, note that in order for both the given fractions to be integers we must have the divisibility conditions, \[m^2+n^2 \mid m^5+n \text{ and } m^2 + n^2 \mid m+n^5\]By taking the sum of these two we have $m^2+n^2 \mid m^5+n^5+m+n$. Further, we know that, \[m^2+n^2 \mid (m^2+n^2)(m^3+n^3)=m^5 + n^5 + m^2n^3 + m^3n^2\]Taking the difference of these two we have that \[m^2+n^2 \mid m^2n^3 + m^3n^2 - m - n = (m^2n^2 -1)(m+n)\] Taking the difference of the two initial divisibility conditions and following a similar process we also have \[m^2+n^2 \mid (m^2n^2+1)(m-n)\] Now, we observe that $\text{gcd}(m^2+n^2,m+n)= \text{gcd}(2n^2,m+n)$. So, if there exists a prime $p>2$ which divides both $2n^2$ and $m+n$, $p \mid 2n^2 \implies p\mid n^2 \implies p\mid n$. Thus, since $p\mid m+n $ we must also have that $p\mid m$. Thus, $p\mid \text{gcd}(m,n)$ which is a contradiction to the first claim. Thus, $\text{gcd}(m^2+n^2,m+n)\leq 2$. Similarly, we can also show that $\text{gcd}(m^2+n^2,m-n)\leq 2$. We know have two separate cases. Case 1 : If $m+n$ is odd, then $\text{gcd}(m^2+n^2,m+n) = \text{gcd}(m^2+n^2,m-n) = 1$. Thus, \[m^2+n^2 \mid (m^2n^2-1)(m+n) \text{ and } m^2+n^2 \mid (m^2n^2+1)(m-n)\]infact implies that \[m^2+n^2 \mid m^2n^2-1 \text{ and } m^2+n^2 \mid m^2n^2+1\]But this implies that $m^2+n^2 \mid 2$ which implies $m^2+n^2 =1$ since $m+n$ is odd (this implies $m^2+n^2$ is odd quite easily). This infact has no solutions for positive integers $(m,n)$. Case 2 : If $m+n$ is even, then $\text{gcd}(m^2+n^2,m+n) = \text{gcd}(m^2+n^2,m-n) = 2$. Thus, \[m^2+n^2 \mid (m^2n^2-1)(m+n) \text{ and } m^2+n^2 \mid (m^2n^2+1)(m-n)\]in fact implies that \[m^2+n^2 \mid 2(m^2n^2-1) \text{ and } m^2+n^2 \mid 2(m^2n^2+1)\]But this implies that $m^2+n^2 \mid 4$, so $m^2+n^2 \in \{1,2,4\}$. Since $m,n$ are positive integers where $m+n$ is odd, $1$ and $4$ are not possibilities which implies $m^2+n^2 =2 $ and thus, $m=n=1$ which is indeed the solution we claimed at the start. Thus, we are done and all solutions are indeed of the claimed forms.

The only solution is $\boxed{(m,n)=(1,1)}$. Notice that the following expression must be an integer. $$\frac{n(m^5+n)+m(m+n^5)}{m^2+n^2}-1=\frac{mn(m^4+n^4)}{m^2+n^2}$$Let $d=\gcd(m,n)$ and let $m=ad$ and $n=bd$. We get that $a^2+b^2|a^4+b^4\Rightarrow $ $a^2+b^2|2a^2b^2 \Rightarrow a^2+b^2|2$. So we must have $a=b=1$ or $m=n$. So $\frac{m^5+m}{m^2+m^2}=\frac{m^4+1}{2m}$ is an integer so we must have $m=n=1$.

The answer is $\boxed{(m,n)=(1,1)}$. Rewrite the problem in divisibility form, $$m^2+n^2\mid m^5+n\text{ and }m^2+n^2\mid m+n^5$$It is easy to see that $$m^2+n^2\mid m^6+mn\text{ and }m^2+n^2\mid mn+n^6$$Which implies $$m^2+n^2\mid m^6+n^6+2mn$$and since $m^2+n^2\mid (m^2+n^2)(m^4-m^2n^2+n^4)=m^6+n^6$, now, $$m^2+n^2\mid 2mn$$By AM-GM inequality, $m^2+n^2\geqslant 2mn$ yields $m=n$. Plugging back into the problem and getting that $m=n=1$ is the only solution. Done!

Claim: $\gcd(m,n)=1$ Proof: Let $\gcd(m,n)=d$. Then $m=d m_1$ and $n=d n_1$. Then \[ d^5 m_1^5 +d n_1 = k (d^2 m_1^2 +d^2 n_1^2 ) \implies d^3 m_1^5 + \frac{n_1}{d} = k ( m_1^2 + n_1^2) \]Thus, $d \mid n_1$. Similarly, from the second condition, $d \mid m_1$, contradiction. \[ m^2 + n^2 \mid (m^5 +n) - (m^3-mn^2)(m^2 +n^2)= n+mn^4 \implies m^2+n^2 \mid 1+mn^3\]Similarly, from the second condition, $m^2+n^2 \mid 1+m^3 n$. Hence, \[m^2 + n^2 \mid 2+mn^3+m^3 n =2+mn(m^2+n^2) \implies m^2+n^2 \mid 2 \]Hence, $(m,n)=(1,1)$.

Let $gcd(m,n) = d, d^2|m^2+n^2 \Rightarrow d^2|m^5 +n \Rightarrow d^2|n$ Similarly $d^2|m$ $\therefore d^2 = d, \Rightarrow d = 1$, Let $m^2+n^2 = p$, $p\mid m^5+n\text{ and }p\mid m+n^5 \Rightarrow p\mid m^6+n^6+2mn$ $m^2+n^2\mid (m^2+n^2)(m^4-m^2n^2+n^4)+2mn \Rightarrow m^2+n^2\mid 2mn \Rightarrow m^2+n^2 \le 2mn$ But from A.M $\geq$ G.M it says otherwise. $\therefore m^2+n^2= 2mn \Rightarrow m=n$, now from the original equation we get that $m=n=1$.