Square Triangles

Here is my proof of @Jon Sy challenge:

We are trying to prove that the equation $\frac { n(n+1) }{ 2 } =m^{ 2 }$ , such that $m,n\in \aleph$ . Note that $n$ and $n+1$ have distinct parity, therefore one of them is even and the other is odd; also, note that $gcd((n+1)/2,n)=1$ , thus, the product can be expressed as $n\cdot \frac { n+1 }{ 2 }$ or $(n+1)\cdot \frac { n }{ 2 }$ , where the two factors are coprime integers. As the product is a perfect square, each factor is also a square. Therefore, there exist some integers $x,y$ such that $n=x^2$ and $n+1=2y^2$ , then $x^2-2y^2=-1$ , in the first case. In the second case, $n+1=x^2$ and $n=2y^2$ , so $x^2-2y^2=1$ .

Hence, we're looking for the solutions to the equation ${ x }^{ 2 }-2{ y }^{ 2 }=\pm 1$ . Note that once we find a pair $(x,y)$ that satisfies the equation, then $(w,z)$ , with $w=x+2y$ and $z=x+y$ , also satisfies the equation, cause

${ w }^{ 2 }-2{ z }^{ 2 }={ (x+2y) }^{ 2 }-2{ (x+y) }^{ 2 }={ x }^{ 2 }+4xy+4y^{ 2 }-2{ x }^{ 2 }-4xy-2{ y }^{ 2 }=-({ x }^{ 2 }-2{ y }^{ 2 })=\mp 1$

We can find easily a solution: $(1,1)$ , which generates the infinite sequence of solutions $(x,y)$ : $(3,2),(7,5),(17,12),(41,29),(99,70),...$

Hello By guess and check, we find $\boxed{36}$ is the smallest that works. Extension problem: What is the next smallest triangular number that is also a perfect square, or prove none exists. Extension to extension problem: Prove or disprove: There are infinitely many triangular numbers that are also perfect squares. How about perfect cubes? Or higher perfect powers? Thank