True or false: There is no square $6$ mod $7$. If you find an example, then you are finish. If you cannot find an example, then prove that the below statement is not true.
$$ x^2 \equiv 6\mod 7$$
When I try some examples I get $0,1,2,4 \mod 7$ so I would have to prove that there is no square $6 \mod 7$ but I am having a hard time. Any ideas?
On
Just try all integers from 0 to 6. Square them and check the remainder modulo 7. If there's no such number among them (with a remainder of 6), then there's no such number at all.
On
For any real $x$, we have
$$x \equiv 0,1,2,3,4,5,6 \pmod 7$$ and $$x^2 \equiv 0,1,2,4 \pmod 7$$
Hence the statement is true.
On
According to the first supplement to quadratic reciprocity https://en.wikipedia.org/wiki/Quadratic_reciprocity#.C2.B11_and_the_first_supplement
$x^2\equiv-1\pmod{7}$ has no solution since $7\not\equiv1\pmod{4}$
Suppose $x^2\equiv 6$ then $x^4\equiv 36 \equiv 1$, and clearly $x$ is not a multiple of $7$, so little Fermat tells us that $x^6\equiv 1$ but then $x^6=x^2\cdot x^4\equiv 6\times 1\equiv 6$ is a contradiction.
This also shows by easy generalisation that the congruence $x^2\equiv -1 \bmod p$ cannot be solved for any prime $p\equiv 3 \bmod 4$