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About LinkBacksIf p is an odd prime, prove that 1^2 * 3^2 * 5^2 * ...(p-2)^2 = (-1)^(p+1)/2 (mod p)
and
2^2 * 4^2 * 6^2 * ... (p-1)^2 = (-1)^(p+1)/2 (mod p)
I know that (p-1)! = -1 mod p using Wilson's theorem, but then I am stuck. Help?
If p is an odd prime, prove that 1^2 * 3^2 * 5^2 * ...(p-2)^2 = (-1)^(p+1)/2 (mod p)
and
2^2 * 4^2 * 6^2 * ... (p-1)^2 = (-1)^(p+1)/2 (mod p)
I know that (p-1)! = -1 mod p using Wilson's theorem, but then I am stuck. Help?
For the first one : first, notice thatare all the squares in
. The squares of
, by Euler's criterion, are the roots of the polynomial
. Therefore, since
is a unique factorization domain, we can factor this polynomial as :
Equating the constant terms, you get the theorem.
The second one is exactly the same.
Very interesting information. The problem is that I haven't learned Euler's criterion in class yet, so I don't really understand your solution. I do know how to manipulate mods, use Wilson's Theorem, and Euler's generalization of Fermat's little theorem. Is there another way using only basic modular arithmetic and congruence to prove that 1^2 * 3^2 * 5^2 * ...(p-2)^2 = (-1)^(p+1)/2 (mod p) and 2^2 * 4^2 * 6^2 * ... (p-1)^2 = (-1)^(p+1)/2 (mod p) are both true? Thanks to all.
Also, you can take a primitive root module, call it
.
Since you have all the quadratic residues (*) there -in both cases-, and they are congruent to:we find that both products are congruent to
But-since
and it's primitive- and so we find that the products are congruent to:
(*)is a quadratic residue module p if and only if there exist an integer
such that
EDIT: What you have there is congruent tonow note that
...
but then we have:
^2\equiv 1\cdot(-(p-1))\cdot ... \cdot \left(\tfrac{p-1}{2}\right)\cdot \left(p-\tfrac{p-1}{2}\right)\equiv (-1)^{\tfrac{p-1}{2}}\cdot (p-1)!)
^{\tfrac{p+1}{2}}}(\bmod.p))
-just by applying Wilson's Theorem-
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