Anyone has a solution？

Answer: all positive integers that differ by 1. Solution: Denote $d=\mid s-t \mid$. If $d \ge 2$, take an odd prime divisor $p$ of $7^d+1$ (when $d>=2$ this number is not divisible by $16$, larger than it and so not the power of $2$ , so $p$ exists). Set $a=7, b=p+7$, we have $a^s+b^t \equiv 7^s+7^t \equiv 0 (mod p)$, and so $p \mid a^n+b^{n+1} \equiv 7^n*8 (mod p)$, which is clearly impossible (clearly $p \neq 7$). If $d=0$, take coprime positive integers $a,b$, greater than $1$. We have $a^s+b^s \mid a^n+b^{n+1}$. Let $n=ks+r, 0\le r \le s-1$, so $a^{ks+r} \equiv (a^s)^k*a^r \equiv (-b^s)^k * a^r+b^{ks+r+1} (mod a^s+b^s)$. Since $a$ and $b$ are coprime, we can cancel out $b^{ks}$ and get $a^s+b^s \mid (-1)^k*a^r+b^{r+1} < (a^s+b^s)$ and $(-1)^k*a^r+b^{r+1}>-(a^s+b^s)$, and it is not zero because $a,b$ are coprime and $b>1$, so again we arrive at a contradiction. Finally, we show that $d=1$ works. If $t=s+1$ just take $n=s$. If it is not the case, we need $a^{t+1}+b^t \mid a^n+b^{n+1}$. We inspect all the prime divisors of $a^{t+1}+b^t$. Let $p^r \mid \mid a^{t+1}+b^t$ . If $p \mid a,b$, for all $n \ge r, p^r \mid a^n+b^{n+1}$. So suppose that $p$ is coprime to $a$ and $b$. Since $1 \equiv a^{\varphi{(p^r)}} \equiv b^{\varphi{(p^r)}} (mod p^r)$, we get for all integers $l, a^{l \varphi{(p^r)}-t-1} \equiv -b^{l \varphi{(p^r)}-t}$, so for that particular prime divisor all $n$ such that $\varphi{(p^r)} \mid n+t+1$ satisfy the condition. It means we can choose $n=u\varphi(a^{t+1}+b^t)-t-1$ (since $\varphi$ is multiplicative for coprimes) for $u$ sufficiently large (to make $n$ larger than all the exponents of prime divisors of $a^{t+1}+b^{t}$ because of those divisors that divide $a,b$), and the proof is completed.

Select $gcd(a,b)=1$, $a^s+b^t \mid a^n+b^{n+1}$, then $a^s+b^t \mid b^{n+1}(b^{t-s-1}-1)$. Clearly, $gcd(a^s+b^t, b)=1$, then $a^s+b^t \mid b^{t-s-1}-1$, $a^s+b^t > b^{t-s-1}-1$, then $t-s=1$.

@WypHxr I think it is quite wrong - as shown in #3, also the pairs $s-t=1$ work...

Wyphxr wrong solution Aldagansan Laqqmisanyo

IMO2022Goldinshallah wrote: Wyphxr wrong solution Aldagansan ++++