Exercise 7.8.5* ($\sum_{i=1}^n i = m^2$ has infinitely many solutions)

Proof. The sum $S_n$ of the first $n$ natural numbers is $S_n={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{n}i=(n(n+1))∕2$.

For all positive integers $n,m$, we have the following equivalence:

Let $(x,y)$ be any positive solution of $x^2-8y^2=1$. Then $x^2=8y^2+1$ is odd, so $x$ is odd. There exists some integer $n$ such that $x=2n+1$. Put $m=y$. Then ${(2n+1)}^2-8m^2=1$, so $\frac{n(n+1)}{2}=m^2$.

By Theorem 7.25, there are infinitely many positive solutions of $x^2-8y^2=1$. For each such solution, $(n,m)=((x-1)∕2,y)$ is a solution of $\frac{n(n+1)}{2}=m^2$.

Therefore the sum of the first $n$ natural numbers is a perfect square for infinitely many values of $n$. □