I'm interesting to look the solution of this equation :$$(p^y+1 )\bmod (p^x+1)=0 $$ at a least to see an example of the two coprime $y, x$ for which $(p^y+1 )\bmod (p^x+1)=0 $ but i don't succed , I tried to check wolfram alpha for $y=3$ and $x=2$ for the fisrt list of prime numbers but i don't accross any example of that .
My question :Is there any pair of the two coprime $(x,y)$ satisfy the Equation:
$$(p^y+1 )\bmod (p^x+1)=0 $$ ? Note:I have tried to use some definition of diaphontine equation (elliptic curve or any similar of that ) but no result
Thank you for any help
No such examples exist. \begin{align*} p^y+1=0(\text{mod } p^x+1) &\iff p^x+1|p^y+1\\ &\iff p^x+1|p^y-p^x\\ &\iff p^x+1|p^x(p^{y-x}-1)\\ &\iff p^x+1|p^{y-x}-1\\ &\iff p^x+1|p^{y-x}+p^x\\ &\iff p^x+1|p^x(p^{y-2x}+1)\\ &\iff p^x+1|p^{y-2x}+1\\ &\cdots\\ &\iff p^x+1|p^{y\text{ mod } x}\pm 1 \end{align*}
And the last line doesn't hold since the RHS is smaller than the LHS. In the circumstances where the equation holds, this last line has $y=0\text{ (mod }x)$ and the $\pm 1$ is a $-1$ so that the RHS is $0$. The statement $(x,y)=1$ rules this out. This proof also gives a necessary and sufficient condition for the equation, as the RHS being $0$ is the only way this division can work.