Track / Stage 6 / 99 of 400 #1099 of 2000

Problem 1099

National olympiad, first round

Number theory Difficulty 6.1 Prove it

LII OM - III - Task 4

Given such integers $a$ and $b$ that for every non-negative integer $n$ the number $2^na + b$ is a square of an integer. Prove that $a = 0$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

If $b = 0$ , then $a = 0$ , because for $a \ne 0$ , the numbers $a$ and $2a$ cannot both be squares of integers.
If the number $a$ were negative, then for some large natural number $n$ , the number $2^n a + b$ would also be negative, and thus could not be a square of an integer.
The only case left to consider is when $a \geq 0$ and $b \neq 0$ .
For every positive integer $k$ , the numbers

are squares of different non-negative integers, say

Then $x_{k}+y_{k}\leq(x_{k}+y_{k})|x_{k}-y_{k}|=|x_{k}^{2}-y_{k}^{2}|=|3b|$ , hence

Thus the sequence $(x_k)$ is bounded, which is only possible if $a = 0$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.