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Your problem amounts to solving $2^{128} c - a^y z = 1$. I assume you are given $a,y$ and must find $c,z$ that satisfy the equation. I suggest you first solve $$a^y z \equiv 1 \pmod{2^{128}}.$$ This has as solution $$z \equiv (a^{-1})^y \pmod{2^{128}},$$ so you can find a solution for $z$ by computing the inverse of $a$ modulo $2^{128}$ (using one ...


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So I have to find a manipulation of $p$ such that I only have even numbers as solution. Not so. You have to find a manipulation of $p$ such that if you can find a solution / all the solutions of the manipulated $p'$ in even numbers, then you can find a solution / all the solutions of $p$. That's a very different problem.


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