Search naturals as powers

[MM]

Find all natural numbers m and n such that [tex]2^{m}7^{n}+1[/tex] is a perfect square.

[martin123456]

Find all natural numbers m and n such that [tex]2^{m}7^{n}+1[/tex] is a perfect square.

[tex]2^m7^n = (r+1)(r-1)[/tex]. If r is even then impossible
let [tex]r=2^kt+1[/tex], where k is natural, t is natural or zero.
If r=1 - impossible
so t is natural
[tex]2^m7^n=(2^{k-1}t+1)2*2^kt[/tex] => m = k+1.
[tex]7^n=(2^{k-1}t+1)t[/tex] It is not posible both multiples to be divisible by 7, so one of them equals 1. the first one greater 1. so t=1.
[tex]7^n=2^{k-1}+1[/tex]. mod 3 we get that [tex]1=2^{k-1}+1[/tex], so [tex]3/2^{k-1}[/tex] which is impossible
so no solutions