It's clear that for n=k there's infinitely many solutions. so just gotta consider the case n>k right? (im new to these problems so don't mind, my approaches might be really stupid)

It's easy to show that for any prime $p \mid a^k+b^k-1$, we have either $p \mid a, p \mid b-1$, or $p\mid b, p \mid a-1$. This follows by using Euler's totient function and by using the fact that $\gcd(t^k-1, t^l-1)=t^{\gcd(k, l)}-1$. Any idea how to finish?

a_507_bc wrote: It's easy to show that for any prime $p \mid a^k+b^k-1$, we have either $p \mid a, p \mid b-1$, or $p\mid b, p \mid a-1$. This follows by using Euler's totient function and by using the fact that $\gcd(t^k-1, t^l-1)=t^{\gcd(k, l)}-1$. Any idea how to finish? I think you can finish with LTE with a lot of case bash (as you already has $p|b-1$ or $p|a-b$) the writeup is a bit tedious though

After geting p| a or b If p divide b it easy to claim that the valuation p of L.H.S is smaller then a-1 and if p divide a we get smaller than b-1 so if a and b are different than 1 we get L.H.S smaller than (a-1)(b-1) impossible so a=1 and b free or b=1 and a free

Hamzaachak wrote: so if a and b are different than 1 we get L.H.S smaller than (a-1)(b-1) impossible so a=1 and b free or b=1 and a free In fact there is some $(a,b,k)$ satisfying $|a^k+b^k-1| \leq |(a-1)(b-1)|$ and $k>1$ (such as $a+b=0$ and $k$ is an odd number)

jungle_wang wrote: Hamzaachak wrote: so if a and b are different than 1 we get L.H.S smaller than (a-1)(b-1) impossible so a=1 and b free or b=1 and a free In fact there is some $(a,b,k)$ satisfying $|a^k+b^k-1| \leq |(a-1)(b-1)|$ and $k>1$ (such as $a+b=0$ and $k$ is an odd number) I don't read the question well i think that a and b are positif so it's my bad

jungle_wang wrote: Hamzaachak wrote: so if a and b are different than 1 we get L.H.S smaller than (a-1)(b-1) impossible so a=1 and b free or b=1 and a free In fact there is some $(a,b,k)$ satisfying $|a^k+b^k-1| \leq |(a-1)(b-1)|$ and $k>1$ (such as $a+b=0$ and $k$ is an odd number) But i think that this inequalitie juste add some cases

Hamzaachak wrote: jungle_wang wrote: Hamzaachak wrote: so if a and b are different than 1 we get L.H.S smaller than (a-1)(b-1) impossible so a=1 and b free or b=1 and a free In fact there is some $(a,b,k)$ satisfying $|a^k+b^k-1| \leq |(a-1)(b-1)|$ and $k>1$ (such as $a+b=0$ and $k$ is an odd number) But i think that this inequalitie juste add some cases In fact there's still a lot of details to go after knowing that $|a^k+b^k-1| \leq |(a-1)(b-1)|$(when $a,b\neq 1$), good luck :⁠-⁠)