<s>[INST] Solutions of $2^a+5^b=c^2$ I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
\begin{align*}
2^{2n}+5^b &= c^2 \\
(2^n)^2+5^b &= c^2 \\
5^b &= c^2-(2^n)^2 \\
5^b &= (c-2^n)(c+2^n)
\end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided by 5, $c-2^n \neq c+2^n$, so $c-2^n=1$ and $c+2^n=5^b$.
From this equations, I got $1+2^{n+1}=5^b$.
But I don't know how to continue.
 [/INST] Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} \equiv -1 \pmod{5^b}$$
implies $n+1$ is divisible by $\varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 \geq 2(5^{b-1})$.
Hence $$5^b \geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.

Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
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