Equation $x^2 + y^2 + 1 = 0$ (mod $p$)

Question

How to prove that equation $x^2 + y^2 + 1 = 0$ (mod $p$) has roots?

Hints are acceptable.

Hint: If you take $y=0$, then all $p$ with $(\frac{-1}{p})=1$ are solved. — vadim123, Oct 31 '14 at 17:01
You can show that for $p\equiv 3 \pmod 4$, $y=0, x=\pm\sqrt{-1}$ is a solution where $\sqrt{-1}$ exists because of the requirement $p\equiv 3\pmod 4$... — AlexR, Oct 31 '14 at 17:01
Hint: Every element of the ring $\mathbb{Z}/p\mathbb{Z}$ is a sum of two squares in that ring. — Geoff Robinson, Oct 31 '14 at 17:03
You can directly write the solution of the equation: $x^2+y^2=pz-1$ It's easier. — individ, Oct 31 '14 at 17:04
@GeoffRobinson it seems pretty equal to the whole statement. — Jihad, Oct 31 '14 at 17:07
@Jihad: It is essentially equivalent, I agree, but it is not so hard to prove directly. — Geoff Robinson, Oct 31 '14 at 17:12

score 9 · Accepted Answer · answered Oct 31 '14 at 17:16

9

Hint: Let $p$ be an odd prime. Note that $x^2$ takes on $\frac{p+1}{2}$ distinct values modulo $p$, and so does $-(y^2+1)$. Now use the Pigeonhole Principle.

answered Oct 31 '14 at 17:16

André Nicolas

514,336

How about if we impose $x,y\neq 0$? See this question. – Darth Geek Feb 29 '20 at 13:38
Why does $x^{2}$ take on $\frac{p+1}{2}$ distinct values modulo $p$? – sanguine Dec 08 '21 at 14:18

Equation $x^2 + y^2 + 1 = 0$ (mod $p$)

1 Answers1