fattypiggy123 wrote: Find all pairs of integers $(a,b)$ such that $n|( a^n + b^{n+1})$ for all positive integer $n$ If $a=0$ or $b=0$, the only solution is $a=b=0$ If $a,b\ne 0$, let $n$ coprime with $b$ $n|a^n+b^{n+1}$ and so $n|a^{2n}+a^nb^{n+1}$ $n|2n|a^{2n}+b^{2n+1}$ So $n|a^nb^{n+1}-b^{2n+1}=b^{n+1}(a^n-b^n)$ and so $n|a^n-b^n$ So $n|(a^n+b^{n+1})-(a^n-b^n)=b^n(b+1)$ and so $n|b+1$ and so $b=-1$ So $n|a^n-(-1)^n$. Choosing then $n=|a|$, we get $|a|=1$ It's then easy to see that $a=1$ does not fit and that $a=-1$ fits. Hence the answer : $\boxed{(a,b)\in\{(0,0),(-1,-1)\}}$

fattypiggy123 wrote: Find all pairs of integers $(a,b)$ such that $n|( a^n + b^{n+1})$ for all positive integer $n$ Consider an odd prime $p\ge5$ that does not divide $ab$. For $n=p$ we get (applying Fermat) that $a^p+b^{p+1}\equiv a+b^2\equiv0\bmod{p}$. For $n=2p$ we get that $2p|a^{2p}+b^{2p+1}$, and hence (applying Fermat) $a^{2p}+b^{2p+1}\equiv a^2+b^3\equiv0\bmod{p}$. This means that $a+b^2$ and $a^2+b^3$ both are divisible by infinitely many odd primes, which implies $a+b^2=0$ and $a^2+b^3=0$. Hence $(a,b)=(0,0)$ and $(a,b)=(-1,-1)$ are the only solutions.

Solution: First we take $n$ to be an arbitrary odd prime say $p$.Then $p \mid a^p+b^{p+1} \implies p \mid a+b^2$ where e have used FLT.This means that $a+b^2$ is divisible by infinitely many odd primes $p$ which means that $a+b^2=0$. Thus $n \mid (-b^2)^n+b^{n+1}$.Taking $n$ to be an even integer,say $2k$ we get $2k \mid b^{2k+1}(b^{2k-1}+1)$.The integer $b$ has finitely many primes and we take $k$ to be an arbirary prime power of a prime divisor $q$ of $b+1$ which does not divide $b+1$.Say $q^{\alpha} \nmid b+1$.Then $k=q^{\alpha} \implies 2q^{\alpha} \mid {b^{2q^{\alpha}-1}+1 \implies q^{\alpha} \mid b+1}$ where the last line follows from LTE.This a contradiction to our assumption which arises on assuming that $b+1$ has a prime divisor and every prime divisor of $b+1$ has a maximum power which divides $b+1$.So $b+1=1,0$ or $-1$ ,i.e, $b=0,-1,-2$..The corresponding values of $a$ are $0,-1,-4$.Clearly $(a,b)=(0,0),(-1,-1)$ satisfy our problem statement.The pair $(-4,-2)$ does not satisfy the statement since $n$ then cannot be odd.So $(a,b)=(0,0),(-1,-1)$ are the only solutions.