Show that the sum of the first $n$ natural numbers is a perfect square for infinitely many $n$.

Question

Show that the sum of the first $n$ natural numbers is a perfect square for infinitely many $n$.

The question doesn't make any sense to me. Any help is appreciated.

Answer 1

$1+2+\cdots + n = \dfrac{n(n+1)}{2}=k^2 \to n^2+n - 2k^2 = 0 \to \triangle = 1-4(1)(-2k^2) = 1+8k^2= m^2 \to m^2-8k^2 = 1$. This is a Pell equation with initial solution $(m,k) = (3,1)$, then it has infinitely many solutions, proving the claim.

Answer 2

Using the formula for the sum of the first $n$ natural numbers, you want to find infinitely many natural numbers $n$ and $y$ such that $$\frac{n(n+1)}{2} = y^2,$$ which you can rearrange to $$(2n+1)^2 - 8y^2 = 1.$$ That is Pell's equation, which has infinitely many integer solutions for $2n+1$ and $y$.

We still need to prove that $2n+1$ is odd for infinitely many of those solutions so that there will be infinitely many solutions for which $n$ is a natural number. But $(2n+1)^2$ must be odd because $8y^2$ is even and $1$ is odd, so $2n+1$ is also odd.

Thus, the sum of the first $n$ natural numbers is a perfect square for infinitely many $n$.