Perfect Squares!!

A number of form $4t-1$ or $4t+3$ can never be a perfect square.

$\textbf{Case I :}$ when $N$ is even-

$N$ = 2k for some integer k.

Then $X$ can re-written as $4a-1$ (where $a = 4k$ ) which is not a perfect square.

$\textbf{Case 2 :}$ when $N$ is odd-

$N$ = $2j+1$ for some integer j.

Then $X$ can re-written as $4b+3$ (where $b = 4j+1$ ) which is not a perfect square.

Hence for any positive integer $N$ , $X$ cannot be a perfect square.

If 8N-1 is to be a perfect square, then either (8N-1)≡ 0 mod 4 or (8N-1)≡ 1 mod 4.

So either 8N≡ 1 mod 4 or 8N≡ 2 mod 4.

But the former is not possible since it implies that 8N is odd.

The implication of the latter is that 8N leaves a remainder of 2 when divided by 4, which is not possible since any multiple of 8 is a multiple of 4.

Thus, since 8N is congruent to neither 1 mod 4 nor 2 mod 4, there are no integer values for N such that 8N-1 is a perfect square.

The result that is needed is that the square of an odd integer is always of the firm 8j+1, where j is an integer. To prove this let X be an odd integer. Set X=2k+1 where k is an integer Squaring X we obtain after factoring the first two terms 4k(k+1)+1 Since k and k+1 are consecutive integers, their product is divisible by 2 and hence the form of the square of an odd integer is 8j+1. Since the difference of 8j+1 and 8s-1 is never 0 for any choice of integers j and s, these integers are not equivalent module 8. Thus an integer cannot simultaneously have both forms. Thus the answer is 0.