Let $A\subset\Bbb N$ such that for all $a\in A,b\in\Bbb N$, there exists some $c$ such that $a+bc$ is a square. What is $A\cap\{1,2,\dots,2018\}$?

Question

This is the first problem of a local Math Olympiad in Spain. I'll translate it, avoiding some funny definitions:

Let $\Bbb N^*$ be the set of positive integers, and $A$ the subset of numbers $a$ in $\Bbb N^*$ such that for all $b\in \Bbb N^*$ there exists some $c\in\Bbb N^*$ such that $a+bc$ is a square. What is $B\equiv A\cap\{1,2,\ldots,2018\}$?

I have found that suitable (that is, from $1^2$ to $44^2$) perfect squares are in $B$, but I have stumbled upon the equation

$$a\equiv m^2\pmod b$$

My question: is it possible for a non square number to be a square in $\Bbb Z_b$ for every positive integer $b$?

Secondary questions:

• Is there any simpler solution for this problem?
• Are the Chinese Remainder Theorem and Quadratic Reciprocity Theorem necessary to solve this? (I find these theorems too strong for a local Olympiad).

Answer 1

As requested in the comments:

If $a$ is not a square then $v_p(a)=k$ is odd for some prime $p$. Letting $b=p^{k+1}$ we see that $v_p(a+bc)=k$ for all $c$, hence $a+bc$ can never be a square.