Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I am reading "A Classical Introduction to Modern Number Theory", and I need help for one question:

Show that $$x^2 \equiv y^2-d \pmod p$$ has $p-1$ solutions for $\gcd(p,d)=1$ and $2p-1$ for $\gcd(p,d)>1$, where $p$ is a prime number greater than 3.

I am a little confused, should the answer both be $2p-1$?

share|cite|improve this question
What is $gcd(p,d)$ in the second instance? 2? >1? – Ross Millikan Nov 1 '11 at 3:07
Oh, sorry, it should be >1 – Rob Nov 1 '11 at 3:14
up vote 5 down vote accepted

We are looking at the congruence $(x-y)(x+y)\equiv d\pmod{p}$. If $d$ is divisible by $p$, the solutions are $(a,a)$ and $(a,-a)$, where $a$ travels from $0$ to $p-1$. Since $p$ is odd, these are all distinct modulo $p$, except when $a=0$. So there are $2(p-1)+1=2p-1$ solutions.

Suppose now that $d$ is not divisible by $p$. Let $x-y = a$, where $a$ travels from $1$ to $p-1$ (clearly $y-x$ cannot be congruent to $0$). For any such $a$, there is a unique $b$ such that $ab\equiv -d\pmod{p}$. Then $x^2-y^2\equiv d\pmod{p}$ if and only if $x+y\equiv b\pmod{p}$.

Since $p$ is odd, $2$ is invertible modulo $p$, and therefore the system $x-y\equiv a\pmod p$, $x+y \equiv b\pmod{p}$ has a unique solution $(x,y)$ modulo $p$. It follows that there are as many solutions of the original congruence as there are choices for $a$, namely $p-1$. The case $p=3$ is not special, we can take $p \ge 3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.