bumpbump it's related to catalan's conjecture

3, 2 works

I claim that the only solutions for $(a,b)$ are $\boxed{(1,0), (3,2)}.$ Not hard to confirm that they work. I'm thinking back to... Fermat's Sandwich ... where the equation $y^2 = x^3 - 2$ has the solutions $y = 5, x = 3.$ Rearranging, $2^b + 2 \equiv 0 \pmod{3} \implies 2 | b$. Using the above observation, it's sufficient to show that $3 | a.$ Suppose $a > 2$ (not hard to argue that $(1, 0)$ is the only solution for $a \leq 2$), then since $9 | 5^b + 2$, we have $b = 6k + 2 = 3m + 2$. But then $$25 (125)^m + 2 = 3^a \implies 3^a \equiv 27 \pmod{31} \implies a \equiv 3 \pmod{30}$$... yea and in particular, $3 | a$. Eh?

Firstly it is not hard to observe by modulo 3 that $2|b$ and by modulo 5 that $3|b.$ Now the equation become with integers $a'$ and $b'$: $$27^{a'} - 25^{b'} = 2$$. Suppose that $a';b' >1$. Note that $\phi_25(27) = 20$. So for every $b'>1$ we get: $\phi_{25^{b'}}(27) = 20.25^{b' -1}$ Then $27^ {20.25^{b' -1} +1} \equiv 2 \pmod (25^{b'})$ And $27^ {20.25^{b' -1} +1} > 27^{b'} > 25{b'}$. Then if $a';b' >1$, it is impossible to get $27^{a'} - 25^{b'} = 2$. --> $b' = 0 \implies b=0 \implies a=1$ --> $b' = 1 \implies b=2 \implies a=3$. And we are done.

find all integers a,b>=0 such as 3a²=5^b +2