Number Theory or Algebra? Part 10 - Finale

We can solve this problem by the properties of odd and even numbers. Let's begin.

$2^{x +3}$ is always an even number.

$3\cdot4^{y+5}$ is always an even number.

$\therefore 2^{x +3} - 3\cdot4^{y+5} + 5$ is always an odd number.

For the solution to exist, $z$ must be an odd integer.

So,

$z^{2} \equiv 1 (mod 8)$ (This can be proved when $z$ is an odd integer.)

$2^{x +3} - 3\cdot4^{y+5} + 5 \equiv 5 (mod 8)$

$\therefore 2^{x +3} - 3\cdot4^{y+5} + 5 \not\equiv z^{2}$

$\therefore 2^{x +3} - 3\cdot4^{y+5} + 5 \neq z^{2}$

Let's solve this looking at congruence $\bmod{8}$ . $n^2 \equiv \{ 0, 1, 2, 4 \} \bmod{8}$ While $2^{x+3} -3 \cdot 4^{y+5} + 5 \equiv 5 \bmod{8}$ Thus $5 \neq n^2 \;\forall n$ , there are no solutions.