<s>[INST] Find $a, b, c, d \in \mathbb{Z}$ such that $2^a=3^b5^c+7^d$ Solve $2^a=3^b5^c+7^d$ over the positive integer.
I know $a$ is even because: $(-1)^a \equiv2^a = 3^b5^c+7^d \equiv1 \ (mod\  3)$
 [/INST] One solution to your equation  is $a = 2$, $b = 0$, $c = 0$ and $d = 0$. Another one would be $a = 3$, $b = 0$, $c = 0$, $d = 1$. Yet another one would be $a = 4$, $b  =1$, $c = 1$, $d = 0$. And $a = 5$, $b = 0$, $c = 2$, $d  =1$.
One is clever to recognize that $a + b + c+d = 2(a - 1)$ for all $a,b,c,d \in \mathbb{N}$ that satisfy your equation. Since we want positive integer solutions, it turns out that there is only one tuple which satisfies the above: $(6,1,1,2)$.
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