Existence of solution of Congruence based on Exponent

Question

What is the general approach of proving existence of integer solution of congruence based on its exponent? For example, if $x, y, z$ are all arbitrary odd number, then
$43^x + 102^y \equiv 35^z \pmod {120}$ has no integer solution, how can I prove this? what are the approaches?

Answer 1

Note:

$43^{2k+1}\equiv 43$ or $67 \mod 120$

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added in response to inquiry in comment:

because $43^1=43, $

$43^2=1829=15\times120+49\equiv49,$

$43^3\equiv43\times49=2107=17\times120+67\equiv67, $

and $43^4\equiv49^2=2401=20\times120+1\equiv1\pmod{120}$

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$102^{2l+1}\equiv 102$ or $48$ or $72 \mod 120$

$35^{2m+1}\equiv 35\mod120$

Answer 2

A first step would be to apply the Chinese remainder theorem. Because $120=2^3\times3\times5$ the problem reduces to solving the system of congquences \begin{eqnarray*} 43^x + 102^y \equiv 35^z \pmod {2^3},\\ 43^x + 102^y \equiv 35^z \pmod {3\hphantom{^3}},\\ 43^x + 102^y \equiv 35^z \pmod {5\hphantom{^3}}. \end{eqnarray*} Of course then we can reduce the bases to get the equivalent system \begin{eqnarray*} 3^x + 6^y \equiv 3^z \pmod {2^3},\\ 1^x + 0^y \equiv 2^z \pmod {3\hphantom{^3}},\\ 3^x + 2^y \equiv 0^z \pmod {5\hphantom{^3}}. \end{eqnarray*} Next you can list the different values of these powers, where the powers mod $p^k$ cycle with a period dividing $\varphi(p^k)=p^{k-1}(p-1)$ or eventually become $0$. For example:

• The first congruence shows that if $y\geq3$ then $x\equiv z\pmod{2}$.
• The second congruence shows that $z\equiv0\pmod{2}$.
• The third congruence shows that $x\equiv y\pmod{2}$.

As you require $x$, $y$ and $z$ to be all odd, the second observation shows that there are no such solutions.