It is impossible for $(x-1)^2+x^2+(x+1)^2$ to be a perfect square 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 $x+1$.
 A: 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.
A: I will attempt a proof by contradiction.
Assume that the given sum of three consecutive squares sums to the square $(x+n)^2$, $n$ being a positive integer like $x$. Thus the statement or proposition we conjecture to be true is
$$(x-1)^2+x^2+(x+1)^2=(x+n)^2\tag{1}$$
which simplifies to
$$2x^2+2=2nx+n^2\tag{2}$$
Since the left hand side of equation 2 is even, it immediately follows that n must be even. Thus let $n=2m$ giving
$$x^2+1=2mx+2m^2\tag{3}$$
Now from (3) we can immediately say that $x$ must be odd. Thus let $x=2u-1$, where u is any non-zero positive integer. Substituting for $x$ on the left hand side, (3) then simplifies to
$$2u^2-2u+1=mx+m^2\tag{4}$$
The left hand side of (4) is always odd, therefore $m$ cannot be even. However if $m$ is odd with $x$ odd as required above, $mx+m^2$ will always be even. [The result of multiplying two odd whole numbers together is always odd, as proved by for example by algebraicly multiplying $(2u-1)$ and $(2v-1)$.] 
Therefore we have reached a contradiction, as $x$ cannot be both odd and even at the same time, and so the original conjecture (1) must be false.
A: The smallest solution you may think of:
Note that $(x-1)^2+x^2+(x+1)^2=3x^2+2\equiv 2\pmod{3}$, but $$y^2\equiv 0,1\pmod{3}$$for all integers $y$. Therefore, $(x-1)^2+x^2+(x+1)^2$ can never be a perfect square.
