How to solve $x^2+1\equiv0~\mod 7$

Question

How to solve $x^2+1\equiv0~\mod 7$ .

Initial thoughts: We want to look for $x$ such that $x^2\equiv-1~\mod 7$.

By Lagrange's we have that $x^2+1$ has at most $2$ solutions but I have no idea if they are actually met or not.

Let $x\in\Bbb{Z}$ then $x \equiv0,1,2,3,4,5,6 \mod7$ and $x^2 \equiv0,1,4,2,2,4,1\mod7$ so I would guess that the two solutions are met and are $x\equiv1\mod7$ and $x\equiv6\mod7$ yet take $1\equiv1\mod7$ but $1^2=1\neq-1 \mod7$.

So obviously I am making some serious errors/misunderstanding. Could anyone explain why I am wrong and how to solve equations of this type (i.e simple non linear congruences modulo a prime).

Thanks!

http://math.stackexchange.com/questions/122048/1-is-a-quadratic-residue-modulo-p-if-and-only-if-p-equiv-1-pmod4 — lab bhattacharjee, Nov 21 '16 at 18:04
Why do you "guess that the two solutions are met"? You just listed all of the values of $x^2$ (modulo $7$), and none of them was $6$ (which is $\equiv -1)$. Doesn't that demonstrate that the equation has no solutions? — Brian Tung, Nov 21 '16 at 18:06
Comment: When $p$ is prime, only for half the numbers $a=0,1,\dots,p-1$ can you solve $x^2\equiv a\pmod p$. — Ted Shifrin, Nov 21 '16 at 18:12
@TedShifrin: For odd primes $p$, exactly half of the nonzero residues are quadratic (zero always being a square). — hardmath, Nov 21 '16 at 18:34
If $p$ is prime and $p+1$ is divisible by $4$ then $x^2\equiv -1 \pmod p $ has no solution. — DanielWainfleet, Nov 21 '16 at 20:43

score 8 · Answer 1 · answered Nov 21 '16 at 18:27

8

Hint $\,\ {\rm mod}\ 7\!:\ \left[a^{\large 2}\equiv -1\right]^{\large 3}\!\Rightarrow\, a^{\large 6}\equiv -1\,$ contra little Fermat.

answered Nov 21 '16 at 18:27

Bill Dubuque

272,048

Above is the easy necessity direction of Euler's Criterion $\ \ $ – Bill Dubuque Jun 20 '21 at 09:11

score 3 · Answer 2 · answered Nov 21 '16 at 18:17

3

The following theorem might be of some help:

Theorem: The quadratic congruence $x^2 + 1\equiv 0 \pmod p$, where $p$ is an odd prime, has a solution if and only if $p\equiv 1 \pmod 4$. The solution being $\{(\frac{p-1}{2})!\}^2+1\equiv 0\pmod p.$

answered Nov 21 '16 at 18:17

Shrey Aryan

393

In your case $7\not\equiv 1\pmod 4.$ Therefore no solution exits. – Shrey Aryan Nov 21 '16 at 18:18
+1, for the useful theorem you have mentioned,it is helpful. – Vidyanshu Mishra Nov 22 '16 at 07:46

How to solve $x^2+1\equiv0~\mod 7$

2 Answers2