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.

  • $\begingroup$ RMO or INMO by any chance? $\endgroup$ – Soham Chowdhury May 17 '13 at 13:46
  • $\begingroup$ @SohamChowdhury: Yes. Its from RMO. $\endgroup$ – Inceptio May 17 '13 at 14:59
  • 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: $$\begin{align} n(n+1) = & (4x^2y^2+2xy+y^2-x^2)^2 + (4x^2y+y+x)^2\\ = & (4x^2y^2+2xy-y^2+x^2)^2 + (4y^2x+y+x)^2 \end{align}$$
  • $\begingroup$ (+1) This is very close to my answer: $4x^4=(k+1)^2\Rightarrow k=2x^2-1\ldots$ $\endgroup$ – P.. May 12 '13 at 7:07
  • $\begingroup$ Very nice (+1). $\endgroup$ – Inceptio May 14 '13 at 6:55

Let $n=t^2$

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


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)


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.


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).

  • $\begingroup$ I don't believe an explicit construction is known, but a result of Hooley (1974) shows that $n$ and $n+1$ are both sums of two squares with frequency $\Theta(x/\log x)$ for $n \le x$. I believe it can be strengthened to arithmetic progressions, so that one can build up arbitrarily many distinct representations. $\endgroup$ – Erick Wong May 12 '13 at 7:42
  • $\begingroup$ Actually I suppose the density here is so high that one cannot help but have many representations in most values of $n$ and $n+1$. For instance, only $O(\sqrt{x/\log x})$ numbers up to $x$ admit a unique representation as a sum of two squares. $\endgroup$ – Erick Wong May 12 '13 at 8:03
  • $\begingroup$ Hooley's result looks optimal up to a bounded positive factor, and hints that the frequency for the $n(n+1)$ form of the problem ought to be $Cx/\sqrt{\log x}$. @ErickWong $\endgroup$ – zyx May 12 '13 at 20:04
  • 1
    $\begingroup$ That $n$ and $n+1$ both have to meet the condition also explains why it is hard to do better than $x^{1/2}$ with an explicit construction. $\endgroup$ – zyx May 12 '13 at 20:15
  • 1
    $\begingroup$ I had the unfair advantage of reading Hooley's paper which explicitly uses bounds on $\sum r_2(n(n+1))$... :). $\endgroup$ – Erick Wong May 12 '13 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.