How can be proved $\sqrt{n^8+2\cdot 7^{m}+4}$ is irrational?

Question

How can be proved $\sqrt{n^8+2\cdot 7^{m}+4}$ is irrational, where $n,m \in \mathbb{N}$. I tried to write: $$a^2-n^8=2(7^m+2)$$ and to try finding the last digit in the left and in the right side.

So, the last digit for the RHS can be $0,2,6$. The problem is when the last digit might be $0$.

have you any other idea?

Answer 1

$a^2-n^8$ is a difference of two squares.

A number is a difference of two squares iff it is odd or a multiple of $4$.

$2(7^m+2)$ is even but not a multiple of $4$ and so cannot be a difference of two squares.

Answer 2

$\sqrt{n^8+2\cdot 7^{m}+4}$ is rational iff $a^2={n^8+2\cdot 7^{m}+4}, a\in \mathbb Z$

The square of any odd number $\equiv 1 \bmod 4$

The square of any even number $\equiv 0 \bmod 4$

Analyzing the equation $\bmod 4$, we see that it requires $(0,1)\equiv (0,1)+2(-1)^m+0 \bmod 4$ or $(0,1)\equiv (0,1)\pm 2 \bmod 4$

The latter is impossible, so $\sqrt{n^8+2\cdot 7^{m}+4}$ is irrational.