1
$\begingroup$

Show that $-1\not\equiv x^2\mod 4$, i.e. $-1$ is a quadratic nonresidue.

If I consider first $x$, s.t. $1\equiv x^2\mod 4$ then this is fulfilled, if I take any odd number since then;

$1\equiv x^2\mod 4\iff x^2-1\equiv0\mod 4\iff (x+1)(x-1)\equiv0\mod 4$

because if $x$ is odd, then $x\pm1$ are even, and one of them has to be divisible by $4$

Now consider $-1\equiv x^2\mod 4$, then $x$ should be again odd, but any odd number squared is $1\mod 4$ by the first case.

Is this OK ?

  • 4
    Explicitly square every element mod 4. Does $-1$ show up as a square? – 2017-01-06

1 Answers 1

1

Your proof is correct but is a little bit long. Since $4$ is not too big, you can list the squares mod $4$ : $$0^2=0,\; 1^2=1,\; 2^2 \equiv 0,\; 3^2 \equiv 1,\; \dots$$ You immediately see that only $0$ and $1$ are quadratic residues mod $4$, but $-1 \equiv 3$ is not.

  • 1
    Another way is to say: if $x=2n$ is even then $$x^2=4n^2 \equiv 0 \neq -1 \pmod 4,$$ and if $x=2n+1$ is odd, then $$x^2=4n^2+4n+1 \equiv 1 \neq -1 \pmod 4.$$ – 2017-01-06