Find all $(x,y) \in \mathbb{Z}×\mathbb{Z}$ such that $$690x+1230y=210$$ $$23|(x+y)^{7777} $$
I solved the diophantine equation and got $$x=-112+41n ~\text{ and }~ y=63-23n$$ I simplified the congruence, and ended up with $$(-49+18n)^{11}\equiv 0 \ (\text{mod }23)$$
And I'm stuck. I have no clue how to follow, any ideas?
Rewrite as $(23n-46-5n-3)^{11}\equiv -(5n+3)^{11} \pmod{23}$. Now it's equivalent to $(5n+3)^{11}=0\pmod {23}$. Now if $gcd(5n+3,23)=1$ then this contradicts Fermat's theorem since $11=\frac{23-1}{2}$. Thus $gcd(5n+3,23)$ is not $1$ and thus $23|5n+3$, this is necessary.