number of solutions of $ x^2$ $\equiv$ $-1$ mod $2^k$

Question

How can i prove that the congruence $x^2$ $\equiv$ $-1$ has no solutions in modulo $2^k$ where $k$ is greater than or equal to $2$?

Answer 1

It can be proved like this:

$x^2 \equiv -1 \mod 2^k \Longleftrightarrow x^2 + 1 \equiv 0 \mod 2^k \Longleftrightarrow 2^k \mid x^2 + 1; \tag 1$

now suppose $x$ is even; let

$x = 2n; \tag 2$

then

$x^2 + 1 = 4n^2 + 1, \tag 3$

which is odd; thus for no $k \in \Bbb N$ does

$2^k \mid 4n^2 + 1; \tag 4$

thus $x$ cannot be even; if $x$ is odd, we write

$x = 2n + 1, \tag 5$

$x^2 + 1 = 4n^2 + 4n + 2 = 2(2n^2 + 2n + 1); \tag 6$

but

$2^2 = 4 \not \mid 2(2n^2 + 2n + 1), \tag 7$

thus

$2^k \not \mid 2(2n^2 + 2n + 1), \; k \ge 2; \tag 8$

thus, $x$ can be neither even nor odd, so in fact can be no integer whatsoever. $OE\Delta$.