a_507_bc wrote: Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$ If $y=0$ then $x=1, p=2$. If $y=1$ then by mod 3 $x=0$ and there are no solutions. Now let $y>1$, so $4 | RHS$. Note that since $4|7-3$ we have $2 \not | x$. After this look at mod 8 to get $y<3$ ($p \not = 2$).

$p=2 \to y=0,x=1$ $p$ is odd $\to 3^x+p^2 \equiv 2,4 \pmod {8} \to y<3$ for $y=1$ there are no solutions For $y=2: x=1,p=5$

We only need that $p$ is odd if $p \geq 3$. Indeed, mod 8 we have $3^x \equiv 3,1$ and $p^2\equiv 1$ and so $3^x + p^2 \equiv 2, 4$ while $7 \cdot 2^y \equiv 0$ for $y\geq 3$, contradiction. Hence either $p=2$ or $y\leq 2$. If $p=2$, then parity insists on $y=0$, so $x=1$. If $y=2$, then only $p=5$ and $x=1$ works. If $y=1$, then there are no solutions. If $y=0$, then $p=2$ and $x=1$ works. Hence all solutions are $(x,y,p) = (1,0,2), (1,2,5)$.

Simply by applying mod 8 which restricts or bounds the value of y i.e y<3 By checking manually which gives us 2 solutions i.e (1,2,0);(1,2,5)