Perfect square for n in N

Question

a and b in N , Such a.2^n+b is a perfect square for every n in N.

    Prove that a=0

Can anyone help please ?

Answer 1

Let $n\in \Bbb N$. We are given that there are natural numbers $u, v, w$ with $u^2=2^na+b$, $v^2=2^{n+1}a+b$, and $w^2=2^{n+2}a+b$. There are two cases.

1. Assume $b=0$. Then $v^2=2u^2$, which is only possible for $v=u=0$ because $\sqrt 2$ is irrational. Hence $a=0$ as desired.

2. Assume $b\ne 0$. Then $3b=4u^2-w^2$ and so $2u\pm w\ne 0$. More precisely, $$m:=\left|2u-w\right|=\left|\frac{4u^2-w^2}{2u+w}\right|=\left|\frac{3b}{2u+w}\right|$$ is a positive integer. Hence $m\ge 1$. It follows that $2u+w= \frac{3b}m\le 3b$ and from this, $$ 2^na+b=u^2\le 9b^2.$$As this holds for all $n$, we conlcude $a=0$.