# When is $2^2+...+n^2=p^k$.

Find all $$n>1$$ such that $$\sum_{i=2}^ni^2=p^k$$ where $$p$$ is a prime and $$k\ge 1$$.

I tried to use the formula for the sum of the first $$n$$ perfect squares, $$\frac{n(n+1)(2n+1)-6}{6}=p^k$$ $$(n-1)(2n^2+5n+6)=6p^k$$ Furthermore $$d=\gcd(n-1,2n^2+5n+6)=13$$ or $$1$$.

Let’s see when is equal $$1$$, we get these subcases $$\cases{n-1=1 \\ 2n^2+5n+6=6p^k}, \cases{n-1=2\\ 2n^2+5n+6=3p^k}, \cases{n-1=3\\ 2n^2+5n+6=2p^k}$$ $${\cases{n-1=6\\ 2n^2+5n+6=p^k}}$$ There are some restrictions on $$p$$ in the last $$3$$ subcases, but they’re obvious. from these cases we get these solutions $$(p,q)\in\{(2,2),(13,1),(29,1),(139,1)\}$$ But the problem lies in the case when $$d=13$$, I can’t get a handle on the subcases.

• I think it would be more profitable to expand and refactor the numerator of the original term with $-6$ in it, since being equal to $p^k$, you can start to look at their gcd which puts a strong restriction if they can't share prime factors. Jan 18 at 10:46
• Indeed, $n(n+1)(2n+1)-6$ has a factorization. Can you find it? Look for a root. Then, you won't need to push the $6$ to the other side. Jan 18 at 11:14
– PNT
Jan 18 at 11:26
• artofproblemsolving.com/community/c6h1639347p10323485 Might be helpful Jan 18 at 13:39

If $$d=1$$, then you've done the computation correctly.

If $$d = 13$$, then note that $$p=13$$ is forced, so dividing by $$13^2$$ on both sides, $$\frac{(n-1)}{13}\frac{(2n^2+5n+6)}{13} = 6 \times 13^{k-2}$$

Since the quantities on the left are co-prime, we get that one of $$\frac{n-1}{13}$$ or $$\frac{2n^2+5n+6}{13}$$ is a divisor of $$6$$, and the other is a multiple of $$13^{k-2}$$. However, for $$n \geq 5$$, $$2n^2+5n+6 \geq 2\times 5^2+5 \times 5 +6 \geq 81 > 78 = 13 \times 6$$

Therefore, $$\frac{2n^2+5n+6}{13}$$ can be a divisor of $$6$$ only if $$n<5$$. In this case, however, $$\frac{n-1}{13}$$ isn't even an integer (we want $$n \geq 2$$ so $$n=1$$ is ruled out).

It follows that $$\frac{n-1}{13}$$ must be a divisor of $$6$$ i.e. that $$n-1 = 13,26,39,78$$ i.e. that $$n= 14,27,40,79$$.

We will diverge from the approach in the AoPS post here, to show a different method. Indeed, one can substitute the values above and check that no more solutions are found , but we can do something that minimizes numerical computation.

We've already seen that $$\frac{2n^2+5n+6}{13}>6$$ for $$n>5$$. So if this expression is to be a multiple of a power of $$13$$ times a divisor of $$6$$ ,that power of $$13$$ must be at least $$13$$. Therefore, $$2n^2+5n+6$$ must be a multiple of $$169$$.

The idea now is to use the fact that we only need to test numbers belonging to the arithmetic progression $$13k+1$$. So we compute the following polynomial modulo $$169$$: $$2(1+13k)^2 + 5(1+13k)+6 \pmod{169} = 2(26k+1)+65k+5+6 = 13(9k+1)$$

Therefore, we need $$9k+1$$ to be a multiple of $$13$$ : this doesn't occur for $$k=1,2,3,6$$. So no more solutions exist.