Counting Solutions

$n^2 \equiv 2\text{ (mod 8)}$ $=> 8k + 2 = n^2$ where k is a non-negative integer.

(8k+2)=2(4k+1) is not a perfect square for any value of k.

Hence there are no solutions.

It is impossible for the square of an integer to have a remainder of $2$ when divided by $8$ . Let us show this by examining all possible cases where $k$ is an even whole number $\le 8$ , and $n \pmod 8 \equiv k$ .

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$n \pmod 8 \equiv 0$

$0^2=0$

$0 \pmod 8 \equiv 0$

$n^2 \pmod 8 \equiv 0$

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$n \pmod 8 \equiv 2$

$2^2=4$

$4 \pmod 8 \equiv 4$

$n^2 \pmod 8 \equiv 4$

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$n \pmod 8 \equiv 4$

$4^2=16$

$16 \pmod 8 \equiv 0$

$n^2 \pmod 8 \equiv 0$

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$n \pmod 8 \equiv 6$

$6^2=36$

$36 \pmod 8 \equiv 4$ .

$n^2 \pmod 8 \equiv 4$

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Since no cases show that $n^2 \pmod 8 \equiv 2$ the answer must be $\boxed{0}$ .

There are lots of easy ways to do this problem, but I just felt like showing all cases :P

If $n^2 \equiv 2 \: (mod \: 8)$ , then $n$ must be even, since odd $n$ would yield an odd $mod \: 8$ . An even $n$ would be of the form $2k$ , where $k$ is an integer. Then $n^2$ would be $4k^2$ , and it would be $0 \: mod \: 4$ , so it would be either $0 \: mod \:8$ or $4 \: mod \: 8$ . Since $n^2$ cannot be $2 \: mod \: 8$ for either even or odd integers, we conclude that the answer is $\fbox 0$ .

$n^{2} \left( \mod 8 \right)$ generates the series 1, 4, 1, 0, 1, 4, 1, 0, ... What more need be said?