arithmetic problem: Prove that if you add the squares of three consecutive integer numbers and then subtract two, you always get a multiple of 3.

we get $(x-1)^2+x^2+(x+1)^2-2=3x^2$

Hint: Consider the numbers mod $3$: they will be $0,1,2$ or $1,2,0$ or $2,0,1$. Square and add and subtract $2$.

Note that

$$n^2 + (n+1)^2 + (n+2)^2 - 2 = 3n^2+6n+3.$$

What do you notice?

Because all of the reasonable solutions have been taken, here is my overkill solution.

Suppose that $3\mid x^2+(x+1)^2+(x+2)^2-2$, then $$3\mid x^2+(x+1)^2+(x+2)^2 -2 + 3(2x+3)$$ $$3\mid x^2+(x+1)^2+(x+2)^2 + (x+3-x)(x+3+x)-2$$ $$3\mid x^2+(x+1)^2+(x+2)^2 + (x+3)^2-x^2-2$$ $$3\mid (x+1)^2+(x+2)^2 + (x+3)^2-2$$ also note that that $(-1)^2+0^2+1^2-2 = 0 = 3\cdot 0$

Similarly, one can also show that $$3\mid x^2+(x+1)^2+(x+2)^2-2\implies 3\mid (x-1)^2+x^2+(x+1)^2-2$$

HINT

We want to prove that:

$$x^2 + (x+1)^2 + (x+2)^2 - 2$$

is a multiple of $3$. We know that if a number is a multiple of three then it must have a form like $3$ times some number. Let the number be $n$ so we have $3n$. If our expression is of that form then we know that it's a multiple of three. Expand and simplify.

HINT $2$

$$x^2 + x^2 + 2x + 1 + x^2 + 4x +4 - 2$$ Simplify and see if it's of the form $3n$.