I was wondering if there was a way to reduce the exponents in this expression, so that it can be calculated for very large value of exponent, in particular for large $y$ (i.e. $3000$).
$$(x^y+1)\bmod (y^x+1) = 0 $$
I want to find for which positive integers x and y this is true. For this reason, it is necessary to test very large numbers for $y$ and because of hardware limitations I need to find a way to reduce it beforehand, if it exists.
Using properties of the modulo operator does not bring me anywhere, and I do not see how I can reduce at least the first exponent. I could do something if I could prove that:
$$x^y\bmod a = x\bmod \sqrt[y]{a}$$
But I did not find anything similar so far.
EDIT: Since I need the modulo to be equal to zero, I can say that the first expression is equal to: $$x^y-1=n \times (y^x+1)$$ where $n$ is a positive integer, and so: $$x=\sqrt[y](n \times (y^x+1)-1)$$ I got rid of the large exponent $y$ but I have now to check for any value of $n$ (possibly extremely large) to evaluate the equation. If I could go back to the modulo operator from here it would be perfect.