Circumcircle wrote: Find all natural numbers $x$, $y$ such that $2^x+5^y+2$ is a perfect square. Assuming that $0$ is a natural number, here is my solution:

Just noticed the edit. So the first case can be ignored.

Let $$2^x+5^y+2=a^2$$If $x\geq 2, LHS=RHS=a^2\equiv 3\pmod 4$, which is impossible. So, $x=1$ $5^y=a^2-4=(a-2)(a+2)$ $gcd(a-2,a+2)=4$, so both cannot be divisible by $5$. Hence, $a-2=1\implies a=3\implies y=1$ Hence, the only solution is $x=y=1$

Circumcircle wrote: Find all positive integers $x$, $y$ such that $2^x+5^y+2$ is a perfect square. For all $x\geq 2$ $2^x+5^y+2\equiv 3\mod 4$ which can never be perfect square. Hence $x=1$ $5^y+4=z^2\implies 5^y=(z-2)(z+2)\implies z\equiv 2\mod 5, z\equiv -2\mod 5$ Hence $z-2=1, z+2=z^2-4\implies z=3$ is only solution.

Mm very clever...how did you know to take it mod 4 though?

samrocksnature wrote: Mm very clever...how did you know to take it mod 4 though? Because $5^n \equiv 1 \pmod 4$ and $2^n \equiv 0 \pmod 4$ for $n \ge 2$, the expression for $n \ge 2$ will be congruent to $3 \pmod 4$, which can never be a perfect square. This puts a very nice bound for $n$, and it's easy to check the case of $n = 1$. I'm kind of just restating what people before me have said, to be honest.

Circumcircle wrote: Find all positive integers $x$, $y$ such that $2^x+5^y+2$ is a perfect square. First use mod 4 then you will get x can't be greater than 1. Then find out (x,y) if x=1