Let $a,b\ge 0$ be integers such that $2^na+b$ is a perfect square $\forall n>0$

Question

Let $a,b\ge 0$ be integers such that $2^na+b$ is a perfect square $\forall n>0$. Show that $a=0$.

I have no idea how to start this problem since I don't see any arithmetic proprety that I can use in such a problem.

Thanks !

Answer 1

Let $2^n a+ b =a_n^2$ for some sequence $(a_n)_{n=1}^\infty$. We shall prove that the sequence $(a_n)$ is constant, which will establish that $a=0$.

Suppose that $a\neq 0$. Then, $a_n$ is a strictly increasing sequence. Observe that, $$ a_n^2 = 2^n a+b\Rightarrow (2a_n)^2 = 2^{n+2}a+4b=a_{n+2}^2 +3b \Rightarrow 3 b=(2a_n-a_{n+2})(2a_n+a_{n+2}). $$ Now, this immediately implies, for all $n$, $2a_n +a_{n+2}$ divides $3b$, which in turn yields $2a_n+a_{n+2}\leqslant 3b$ for all $n$. However, $b$ is fixed; and thus, if $a\neq 0$, the sequence $(a_n)$ is unbounded, yielding a contradiction.

Edit To address John's comment below: This all hold of course provided $2a_n-a_{n+2}\neq 0$ for all $n$. Say, there is an $n$ making this object $0$, which would yield $b=0$. Now, we have $2^n a$ is a perfect square for all $n$, which is a contradiction: If $2^n a =x^2$ then $2^{n+1}a =2x^2$, which cannot be a perfect square (easy to see, let $y^2=2x^2$, then $2\mid y$, letting $y=2y_1$, we get $x^2=2y_1^2$, yielding $2\mid x$, which, in turn yielding $y_1^2=2x_1^2$, an infinite descent is occurring).