Congruence Proof using Wilson's Theorem
Dear all,
I'm having trouble completing the proof for the following result, that requires Wilson's Theorem. I've shown this, along with my work, below.
Thank you!
scherz0
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For any odd prime $\displaystyle p $, show that:
$\displaystyle 1^2 \cdot 3^2 \cdot ... \cdot (p - 2)^2 \equiv 2^2 \cdot 4^2 \cdot ... \cdot (p - 1)^2 \equiv (-1)^{\frac{p+1}{2}}$
Work: By Wilson's Theorem, I know that: $\displaystyle p$ prime $\displaystyle \Longleftrightarrow (p - 1)! \equiv -1\pmod {p} \Longleftrightarrow (p - 2)! \equiv 1\pmod {p} $.
Also, for any $\displaystyle k$, I know that $\displaystyle p - k \equiv -k\pmod {p}$. This means that:
$\displaystyle p - 1 \equiv -1\pmod {p}, p - 2 \equiv -2\pmod {p}, p - 3 \equiv -3\pmod {p} \implies (p - 1)(p - 2)(p - 3)... \equiv (-1)(-2)(-3)... \pmod{p}$
But I can't see how to apply the results above to prove the congruence relation?