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Problem from a Theorem

If \(p\) is an odd prime number, then prove that \( \left( \left( \dfrac{p-1}{2} \right)! \right)^2 + (-1)^\frac{p-1}{2}\) is divisible by \(p\).

Note by Fahim Shahriar Shakkhor
2 years, 10 months ago

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From Wilson's Theorem, \((p-1)! \equiv -1 (\mod p)\)

\(\Rightarrow 1\times2\times........\times \frac{p-1}{2}\times\frac{p+1}{2} \times ..........\times (p-2)\times (p-1) \equiv -1 (\mod p) \)

\(\Rightarrow 1\times2\times........\times \frac{p-1}{2}\times(-\frac{p-1}{2}) \times ..........\times (-2)\times (-1) \equiv -1 (\mod p) \)

\(\Rightarrow ((\frac{p-1}{2})!)^2 \times (-1)^{\frac{p-1}{2}} \equiv -1 (\mod p) \)

\(\Rightarrow ((\frac{p-1}{2})!)^2 \equiv (-1)\times(-1)^{\frac{p-1}{2}} (\mod p)\)

\(\Rightarrow ((\frac{p-1}{2})!)^2 + (-1)^{\frac{p-1}{2}} \equiv 0 (\mod p) \) Fahim Shahriar Shakkhor · 2 years, 10 months ago

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Nice and tricky!

For convenience, we let \(p=2n+1\) and so we wanna prove that \((n!)^2=n!\cdot n!\equiv (-1)^{n+1}\pmod p\). Now note that \(n\equiv n-p=-(p-n)\pmod p\). So

\[n!=\prod_{k=1}^n k \equiv \prod_{k=1}^n -(p-k)=(-1)^n \prod_{k=1}^n (p-k)=(-1)^n \prod_{k=1}^n (2n+1-k)=(-1)^n \prod_{k=n+1}^{2n} k\pmod p.\]

And hence we have

\[n!\cdot n! \equiv n!\cdot (-1)^n \prod_{k=n+1}^{2n} k=(-1)^n\cdot (2n)!=(-1)^n\cdot (p-1)!\equiv (-1)^n\cdot (-1)=(-1)^{n+1}\pmod p\]

and the last congruence step follows by Wilson's theorem. Jubayer Nirjhor · 2 years, 10 months ago

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