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. 
 A: *

*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:
$$\begin{align}
n(n+1) = & (4x^2y^2+2xy+y^2-x^2)^2 + (4x^2y+y+x)^2\\
       = & (4x^2y^2+2xy-y^2+x^2)^2 + (4y^2x+y+x)^2
\end{align}$$

A: 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.)
A: 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)
A: If $n$ is a square then $n(n+1)$ is a sum of two squares, $n(n+1)=n^2+n$. If $n$ is a square of the form $n=(k+1)^2$ with $2k+2$ a square then $n(n+1)$ can be written as a sum of two squares in another way: $$n(n+1)=(k+1)^2(k^2+2k+2)=(k+1)^2k^2+(k+1)^2(2k+2).$$
Of course there are infinitely many $k\in\mathbb N$ such that $2k+2$ is a square.
A: If $n=t^2$ is a square, all odd prime factors of $n+1$ are equal to $1 \mod 4$, and by using the Chinese remainder theorem for equations $t^2+1=0 \mod p_i$ it can be arranged that $n+1$ is divisible by many different such primes.
The conclusion is that for any $k$, there is an arithmetic progression of values of $t$ such that $n(n+1)$ has more than $k$ different representations as sum of two squares, when $n=t^2$.  This is a denser set of $n$ than the other solutions, and it would be interesting to see if there is an explicit construction that has higher density, possibly as high as linear (ignoring logarithmic factors).
