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.

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 $1$. 
