Show that the sum of squares of four consecutive natural numbers may never be a square. 
Show that the sum of squares of four consecutive natural numbers may never be a square.

I know (and I have the proof) a theorem that says that every perfect square is congruent to $0, 1$ or $4$ $\pmod8$ and wanted to demonstrate this using it.
 A: $0^2 + 1^2 + 2^2 + 3^2 \equiv 0 + 1 + 0 + 1 \equiv 2 \pmod 4$
$x^2 \equiv 2 \pmod 4$ doesn't have a solution.
A: Let $n$, $n+1$, $n+2$, and $n+3$ be any four consecutive natural numbers. Then we have $$ n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = 4n^2 + 12n + 14 = (2n + 3)^2 + 5. $$ Now if this sum were to be the square of a natural number $N$, say, then we must have $$ N^2 - (2n + 3)^2 = 5.$$ However we know that for any two distinct natural numbers $r$ and $k$, whenever $k < r$, we have $k^2 < r^2$ and conversely; in fact, whenever $k < r$, we also have $$ r^2 -k^2 \geq (k+1)^2 - k^2 = 2k+1. $$ That is, the difference between the squres of two natural numbers is at least $1$ more than twice the smaller. 
Now $(2n + 3)^2$ is a natural number whenever $n$ is a natural number. And since $$N^2 - (2n + 3)^2 = 5 > 0,$$ we can conclude that $$2n+3 < N.$$ So we must have $$ N^2 - (2n + 3)^2 = 5 \geq 2(2n+3) + 1 = 4n+7.$$ or $$ 5 \geq 4n+7.$$ So $$ -2 \geq 4n,$$ which implies that $$-\frac{1}{2} \geq n,$$ which is clearly impossible because $n$, being a natural number is at least $1$. 
A: Take sum of squares of four consecutive natural numbers starting with k. 
$$k^2 + (k+1)^2 + (k+2)^2 + (k+3)^2 $$
$$= k^2 + k^2+2k+1 + k^2 + 4k + 4 + k^2 + 6k + 9 $$
$$= 4k^2 + 12k + 14$$
Take this expression in $(mod\,8)$, you have $4k^2 + 4k + 6 (mod\, 8) = 4(k^2 + k) + 6 (mod\,8) = 6(mod 8)$ as no matter k is odd or even, $k^2+k$ is even and then $8|4(k^2+k)$, that is $4(k^2+k) = 0 (mod\,8)$. 
As sum of squares of four consecutive natural numbers is always $6 (mod\, 8)$, it cannot be a square because it can never be $0,1$ or $4 (mod \, 8)$. 
A: $$n=2k+r\\r=\{ 1,2\;\text{or}\;3 \}\\n\equiv1\pmod4\Longrightarrow n^2\equiv1\pmod4\\n\equiv2\pmod4\Longrightarrow n^2\equiv0\pmod4\\n\equiv3\pmod4\Longrightarrow n^2\equiv9\equiv1\pmod4$$Ie, every number is congruent to the square $1$ or $0$ $\pmod4$

Then, $$n^2+(n+1)^2+(n+2)^2+(n+3)^2=\\4n^2+12n+14\equiv0+0+2\equiv2\pmod4$$Therefore, there can be a square, it leaves the rest $2$ and not $0$ or $1$.

A: Assume, contrary to the statement, that $m$ and $n$ are natural numbers such that
\begin{align}
  n^2 &= m^2+(m+1)^2+(m+2)^2+(m+3)^2  \\
    &= 4m^2+12m+14  \\
    &= (2m+3)^2+5.
\end{align}
Hence $5 = n^2-(2m+3)^2 = (n-2m-3)(n+2m+3)$, contradicting $n+2m+3 \ge 6$.
