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Prove, z is odd if both x and y are odd and $3^x + 2^y ≡ 15^z \pmod {20}$ where $y > 1$.

Taking modulo 4 gives $ - 1 + 0 ≡ 3^z \pmod {4}$, then what?

There are several such problems, what is the general method? Please refer me a online document/book chapter if possible.

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A "general rule" is that problems like this simplify when using CRT to split them into congruences modulo smaller moduli. Doing that here reduces to simple parity arguments, namely

$\qquad\begin{align} &\bmod 20\!:\,\ 3^{\large 1+2j}+ 2^{\large 1+2k}\equiv 15^{\large z}\\[.2em] \iff &\bmod 20\!:\,\ 3\cdot 9^{\large j}\ + 2\cdot 4^{\large k}\ \equiv 15^{\large z}\\[.2em] \Rightarrow\ &\bmod 4\!:\,\ {-}1\equiv (-1)^{\large z} \!\iff z\equiv 1\!\!\pmod{\!2}\\[.2em] \Rightarrow\ &\bmod 5\!:\,\ 3(-1)^{\large j}-3(-1)^{\large k}\equiv 0\iff (-1)^{\large j}\equiv (-1)^{\large k}\iff j\equiv k\!\!\pmod{\!2} \end{align}$

Bill Dubuque
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