# Find all $k$ such that there are infinitely many positive integers $a$ such that $a(a+k) + k$ is a perfect square

Problem: For a given positive integer $$k$$, we call an integer $$n$$ a $$k$$-number if both of the following conditions are satisfied: (i) The integer $$n$$ is the product of two positive integers which differ by $$k$$. (ii) The integer $$n$$ is $$k$$ less than a square number. Find all $$k$$ such that there are infinitely many 𝑘-numbers. (BMO2 2022)

Solution: Note that $$n$$ is a $$k$$-number if and only if the equation $$n = m^2 − k = r(r + k)$$ has solutions in integers $$m, r$$ with $$k \geq 0$$. The right-hand equality can be rewritten as $$k^2 − 4k = (2r + k)^2 − (2m)^2$$ , so $$k$$-numbers correspond to ways of writing $$k^2 − 4k$$ as a difference of two squares, $$N^2 − M^2$$ with $$N > r$$ and $$M$$ even (which forces $$N$$ to have the same parity as $$k$$). Any non-zero integer can only be written as a difference of two squares in finitely many ways (because each gives a factorisation, and a number has only finitely many factors). If $$k \neq 4$$ then $$k^2 − 4k \neq 0$$, and as a result, if $$k \neq 4$$ then there are only finitely many $$k$$-numbers. Conversely, if $$k = 4$$ then setting $$m = r + 2$$ for $$r \geq 0$$ shows that there are infinitely many $$4$$-numbers.

"so $$k$$-numbers correspond to ways of writing $$k^2 − 4k$$ as a difference of two squares" - how does this work? Also, is there any motivation for doing this, or is the difference of squares just a very common method to prove there does not exist an infinite amount of something?

Just as it says, $$m^2 - k = r(r+k)$$ is equivalent to $$k^2 - 4 k = k^2 - 4 (m^2 - r(r+k)) = k^2 + 4rk + 4 r^2 - 4 m^2 = (2r+k)^2 - (2m)^2$$ which says that $$k^2 - 4 k$$ is the difference of two squares. The point is that writing something as the difference of two squares, say $$x = a^2 - b^2$$, tells you that $$x = (a-b)(a+b)$$, and a nonzero integer has only a finite set of factors.