Find smallest integer $n$ so that $\sqrt{\frac{1^2+2^2+3^2+...+n^2}n}$ is an integer

Question

Find smallest positive integer $n$ (not equal to $1$) so that $\frac{1^2+2^2+3^2+...+n^2}n$ is a perfect square.

I tried to use : $1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}6$

That gives us $\frac{1^2+2^2+3^2+...+n^2}n=\frac{(n+1)(2n+1)}6$

I tried to brute-force it using python, and I got $n=337$, but I should probably use modular arithmetic (or something else).

According to PARI/GP, $337$ is the smallest solution beyond $1$ — Peter, May 02 '18 at 12:54
@Landuros Usually, perfect powers do not include trivial powers like $1^n$ or $n^1$ — Peter, May 02 '18 at 12:54
If you're asking for an easier way to come up with 337, I think only modular arithmetic would help to figure out which things you could skip in your tests. — , May 02 '18 at 13:08

score 9 · Answer 1 · answered May 02 '18 at 13:08

9

You want to solve $$(n+1)(2n+1)=6m^2.$$ This is $$16n^2+24n+8=48m^2$$ or $$(4n+3)^2-48m^2=1.$$ This is basically Pell's equation $x^2-48y^2=1$, but with the extra stipulation $x\equiv3\pmod 4$. The positive solutions of this Pell equation are given by $$x_k+y_k\sqrt{48}=(7+\sqrt{48})^k.$$ The solutions with $x_k\equiv 3\pmod 4$ are those with $k$ odd. The first is $x_1=7$ giving $n=1$. The next is $x_3=1351$ giving $n=337$. The next is $x_5=262087$ giving $n=65521$, etc.

answered May 02 '18 at 13:08

Angina Seng

158,341

Can you explain Pell's equation a little more? – Bilal HIMITE May 02 '18 at 13:14
1

https://en.wikipedia.org/wiki/Pell%27s_equation – Rhys Hughes May 02 '18 at 13:20
1

@BilalHIMITE See Pell's equation I and II on Keith Conrad's page. – Trevor Gunn May 02 '18 at 13:26
Desperately trying to see anyhow to unstuck this problem except from using pell the diophantine savior. no luck. – Abr001am May 02 '18 at 13:38

Find smallest integer $n$ so that $\sqrt{\frac{1^2+2^2+3^2+...+n^2}n}$ is an integer

1 Answers1