<s>[INST] Suppose that $m \ge 0$ show that $49 \mid 5\cdot3^{4m + 2} + 53\cdot2^{5m}$ I've re-written the equation in a few different ways hoping for a few different approaches:
$$49y = 5 \cdot 3^{4m + 2} + 53 \cdot 2^{5m} $$
I think the first equation has more potential, since it looks like I should be able to just solve for $m$, or get to a point where I can say that any positive $m$ will work, but there doesn't seem to be an easy way to proceed to solve for $m$.
or 
$5\cdot3^{4m + 2} + 53 \cdot 2^{5m} \equiv 0 \mod 49$
I'm not fabulous at congruence equations, but I was thinking I could reduce this down to something more manageable to solve the problem, but none of my congruence equation tricks really help.
Anyways I'm at a loss as to where to proceed, if you all have any tips on ways to proceed, that'd be great :D
 [/INST] \begin{align*}
5\cdot 3^{4(m + 1) + 2} + 53 \cdot 2^{5(m + 1)} &= 5 \cdot 3^{4m + 2} \cdot 3^4 + 53 \cdot 2^{5m} \cdot 2^5 \\
&= 81 \cdot \Big(5\cdot 3^{4m + 2}\Big) + 32 \cdot \Big(53 \cdot 2^{5m}\Big) \\
&= (49 + 32) \cdot\Big(5\cdot 3^{4m + 2}\Big) + 32 \cdot \Big(53 \cdot 2^m\Big) \\
&= 49 \cdot \Big(5 \cdot 3^{4m + 2}\Big) + 32 \cdot\Big(5 \cdot 3^{4m + 2} + 53 \cdot 2^m\Big)
\end{align*}
Can you finish from here?
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