2021/062) Solve in integers x,y such that $2^x-3^y=7$

x and y have to be non -ve integers because otherwise we shall get a fraction

We have $3^y \equiv 0 \pmod 3$ for  y positive and and 1 for y zero

So let us take the 2 cases separately

$y = 0 => 2^x = 8$ or x = 3 so a solutions (x=3,y=0)

Now let us take y positive so we get

$2^x \equiv 1 \pmod 3$ so x has to be even as $2^x \equiv 1 \pmod 3$ for even x and -1 for odd x

so x = 2n.

Working in mod 4 we get $3^y \equiv 1 \mod 4$ so y has to be even as  $3^x \equiv -1 \pmod 4$ for even x and -1 for odd x

So we get y = 2m

Now $2^(2n) - 3^(3m) = 7$ or factoring we get

$(2^n+3^m)(2^n- 3^m) 7$ and as 7 is prime we have

$2^n+3^m = 7$ and $2^n-3^m = 1$

or solving we get $2^{nd}$ solution (n= 2, m = 1) or (x-4, y=2)

So 2 solutions are $(x,y) = \{(3,0), (4,2)\}$