Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.

Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a^2$), $n \in \mathbb{N}.$

I have proved the above question which appeared in one of the Math-Olympiad. And I do know the solution. Sharing the question only because the question has a cute solution.

• Let $n = 4x^4$, we have: $$n(n+1) = (4x^4)^2 + (2x^2)^2 = (4x^4-2x^2)^2 + (4x^3)^2$$
• Let $n = (u^2 + v^2)^2$, we have $$\begin{align} n(n+1) = & ((u^2 + v^2)^2)^2 + (u^2+v^2)^2\\ = & (u^4 - 2uv - v^4)^2 + (2uv^3-v^2+2u^3v+u^2)^2\\ = & (u^4 + 2uv - v^4)^2 + (2uv^3+v^2+2u^3v-u^2)^2 \end{align}$$
• Let $n = (x+y)^2 + (2xy)^2$, we finally have an example that $n$ is not a square: $$\begin{align} n(n+1) = & (4x^2y^2+2xy+y^2-x^2)^2 + (4x^2y+y+x)^2\\ = & (4x^2y^2+2xy-y^2+x^2)^2 + (4y^2x+y+x)^2 \end{align}$$

Let $n=t^2$
$n(n+1)=(t^2)^2+t^2$

Also $t^4+t^2=(t^2-1)^2+3t^2-1$ and $3t^2-1=k^2$ for infinitely many k . (It is a Pell like equation and 3 is odd.)

Pretty happy with approaches. Considering $n$ as a square gives one fine way.

Consider $n=m^2=p^2+q^2$

Now, $n(n +1)= (p^2+q^2) (m^2+1)$

$=(pm+q)^2+(qm-p)^2$

Note that they are two distinct ways.

Thus, for example, $m=5k, p=4k,q=3k$

$n(n +1)= (25k^2)^2+ (5k)^2=(15k^2+4k^2)^2+(20k^2-3k^2)^2$

And we know that there are infinite numbers of the form $n=p^2+q^2$ (Pythagorean Triplets)