Sum of squares of 3 consecutive numbers is not a perfect square

I'm trying to show that $$(n-1)^2+n^2+(n+1)^2=a^2$$ does not have a solution for $$n,a\in \Bbb N$$. I've written $$(n-1)^2+n^2+(n+1)^2=3n^2+2$$, so what I need to show is that $$3n^2+2$$ cannot be a perfect square. From here and here I understand that there is some relationship between perfect squares and modulo, but I fail to see which is it or how should apply it in my case. I do understand, however, that, in the case of 5 consecutive numbers, $$n^2+2$$ needs to be a multiple of 5.

• Your title says something completely different from your actual question. Of course the sum of 3 consecutive natural numbers can be a perfect square, e.g. 2+3+4=9. – bof Sep 22 '19 at 6:28

You are nearly there. Now just use the fact that $$a^2 \ne 2 \pmod 3$$.

The proof is here, which relies on the fact that you can just check $$n=0,1,2$$ (or even better, $$n=-1,0,1$$) for a result that holds for all natural $$n$$.

• Why "even better" $n=-1,0,1$? – Patricio Sep 23 '19 at 11:05
• Out of any three consecutive numbers, there will always be one that is $0 \pmod 3, 1 \pmod 3$ and $2 \pmod 3$. $n=-1,0,1$ is a better choice because $(-1)^2 = 1^2 = 1$ and $0^2 = 0$ which are even smaller numbers than $2^2$. – Toby Mak Sep 24 '19 at 2:02

Toby Mak's hint settles the question but I would like to rephrase the answer without using congruences to make it comprehensible to those not-so-much into math, as this result comes up on the first page in Google search.

In order to show that

$$(n - 1)^2 + n^2 + (n + 1)^2 \neq a^2$$ $$n, a \in \mathbb N$$

We will try to perform reductio ad absurdum by assuming that there is $$a \in \mathbb N$$ that satisfies the equation: $$(n - 1)^2 + n^2 + (n + 1)^2 = a^2$$ We expand the left-hand side of the equation and thus get: $$3n^2 + 2 = a^2$$

Notice that the left-hand side of the equation divided by $$3$$ gives a remainder of $$2$$. That implies a question - what is the remainder of the right-hand side when divided by $$3$$?

We've got three cases to consider $$a = 3k + 0 \lor a = 3k + 1 \lor a = 3k + 2, k \in \mathbb N$$ It's clear that if $$a = 3k$$ then the remainder of $$a^2$$ is $$0$$.

If $$a = 3k + 1$$ , then $$a^2 = 3(3k^2 + 2k) + 1$$ and thus the remainder is $$1$$.

If $$a = 3k + 2$$ , then $$a^2 = 3(3k^2 + 4k + 1) + 1$$, so the remainder is again $$1$$.

Hence, we arrived to a contradiction as it is imposible for a number which divided by 3 gives the remainder of $$2$$ to be equal to a number which divided by $$3$$ gives a remainder of $$0$$ or $$1$$.

Q.E.D.