Elementary Number Theory – Why (x-1)^2 + x^2 + (x+1)^2 Cannot Be a Perfect Square

Question

Prove that it is impossible for three consecutive squares to sum to another perfect square.
I have tried for the three numbers $x-1$, $x$, and $x+1$.

Answer 1

As you stated, you let the $3$ consecutive numbers be $x-1$, $x$, and $x+1$. This will give you a sum of their squares to be $3x^2 + 2$. Consider any integer $n$ and $r = 0,1$ or $2$. Then $(3n+r)^2 = 9n^2 + 6nr + r^2$, so the possible remainders when divided by $3$ are just $r^2$, i.e., $0$, $1$, plus $4$ which has a remainder of $1$ also. Thus, all perfect squares have a remainder of either $0$ or $1$ when divided by $3$, but this sum has a remainder of $2$. Thus, it cannot be a perfect square.

In general, you should try to handle these types of questions by checking the remainders (sometimes called congruences in higher math) of various small integers to see if you can find any particular pattern, such as determine anything which doesn't fit.