If prime $p \equiv 1\pmod 4$ and $b = ((p-1)/2)!$ then show that $b^2 \equiv -1\pmod p$.

Question

This question is from Victor Shoup's book on number theory chapter 2. The problem statement is as mentioned in the title of the question. I haven't been able to crack this one till now. I focused on solving this using the following information:

If p is an odd prime then $(p-1)! = -1 \pmod p$, otherwise called Wilson's theorem.
When $p \equiv 1 \pmod 4$ then any non square in $Z_p^*$ yields a square root of -1 modulo p.

I think we have to prove here that $b=((p-1)/2)!$ doesn't belong to $(Z_p^*)^2$ but I am unable to use Wilson's theorem or any other result to prove it.
PS: Not a homework for me but might be for someone else one day so tagging it as the same.

Something slightly more general was proved in an answer on MathSE. The answer has been written several times, including at least once by me, but of course I can't find that one. — André Nicolas, Aug 30 '12 at 04:24
I don't get your second point. For $p=13$ the squares are $1$,$3$, $4$, $9$, $10$, $12\equiv -1$, and the square roots of $-1$ are $5$, $8$. How exactly does every non-square (take $2$ for instance) "yield" a square root of $-1$? — Marc van Leeuwen, Aug 30 '12 at 04:58
Marc, any non-square in $Z_p^*$ when raised to power $(p-1)/4$ will give you a square root of -1 modulo p. Here $p=13$ so $(p-1)/4 = 3$ and $2^3 = 8$ which is a square root of -1 as you have pointed out. — Abhishek Anand, Aug 30 '12 at 05:06

score 1 · Accepted Answer · answered Aug 30 '12 at 04:26

1

HINT: Since $p\equiv 1\pmod4$, we can write $p=4n+1$ for some integer $n$. Then $\frac12(p-1)=2n$. Suppose that $1\le k\le 2n$; then $-k\equiv p-k\pmod p$, and $2n+1\le p-k\le p-1$. Now consider

$$\prod_{k=2n+1}^{p-1}k\equiv\prod_{k=1}^{2n}(-k)\pmod p\;;$$

what is this in terms of $b$? And then what is $b^2$ in terms of $(p-1)!$?

answered Aug 30 '12 at 04:26

Brian M. Scott

616,228

Thanks Brian, worked it out. Though Bill's answer is also acceptable and lets us work out some more on our own. – Abhishek Anand Aug 30 '12 at 04:47
@Abhishek: Yes, Bill’s answer and mine are basically the same thing, just with more detail on my part. – Brian M. Scott Aug 30 '12 at 04:50

Bill Dubuque · Answer 2 · 2023-08-31T17:45:47.007

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Hint $ $ compute $\rm\:(p\!-\!1)!\:$ in two ways: first by Wilson's theorem, and second by pairing up each factor with its negation (i.e. exploit reflection (involution) symmetry $\rm\:x\to -x\:$ on $\rm\:\Bbb Z_p^*)$ as here.

edited Aug 31 '23 at 17:45

answered Aug 30 '12 at 04:34

Bill Dubuque

272,048

Bill, your answer is also acceptable but I saw Brian's first :) – Abhishek Anand Aug 30 '12 at 04:48

score 1 · Answer 3 · answered Aug 30 '12 at 15:44

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By taking the product $(p-1)!$ and pairing each $x$ with $p-x\equiv -x \pmod p$ we get

$$ (p-1)! \equiv \prod_{x=1}^{(p-1)/2} -x^2 = (-1)^{(p-1)/2}b^2=b^2 $$

since $p\equiv 1 \pmod 4$ gives that $(p-1)/2$ is even. So by Wilson's theorem $$b^2\equiv (p-1)! \equiv -1 \pmod p$$

answered Aug 30 '12 at 15:44

Richard Evans

46

A very well written and concise answer. – Jakub Konieczny Sep 22 '12 at 08:34

If prime $p \equiv 1\pmod 4$ and $b = ((p-1)/2)!$ then show that $b^2 \equiv -1\pmod p$.

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