# Modulus proof with factorials

Prove that (2pâˆ′1)!â‰،p(mod p2)(2pâˆ’1)! â‰، p \pmod {\space p^2}
Hint: pp divides (2p-1)!2p-1)!

Any help would be much appreciated.

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Another hint: (p-1)! = -1 \pmod{p}(p-1)! = -1 \pmod{p} (assuming pp is prime of curse)
– Simon
2 days ago

It’s not true. Check p = 4p = 4.
– Felix Marin
2 days ago

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The case p=2p=2 is easy to check by hand, hence we may assume p>2p>2.
\frac{(2p-1)!}{p}= 1\cdot 2\cdots (p-1)\cdot(p+1)\cdot(p+2)\cdots (2p-1)=\prod_{k=1}^{p-1}(p^2-k^2) \frac{(2p-1)!}{p}= 1\cdot 2\cdots (p-1)\cdot(p+1)\cdot(p+2)\cdots (2p-1)=\prod_{k=1}^{p-1}(p^2-k^2)
is equivalent \!\!\pmod{p}\!\!\pmod{p} to:
(-1)^{p-1}\left[(p-1)!\right]^2 (-1)^{p-1}\left[(p-1)!\right]^2
that by Wilson’s theorem is 1\pmod{p}1\pmod{p}.

Just to give another way. Noticed that (p+1)(p+2)….(2p-1)=mp^2+(p-1)!(p+1)(p+2)….(2p-1)=mp^2+(p-1)! It follows, using Wilson, (2p-1)!=(p-1)!p(p+1)(p+2)….(2p-1)=p(pn-1)(mp^2+(p-1)!)=(p^2n-p)(mp^2+pn-1)=Mp^2+p(2p-1)!=(p-1)!p(p+1)(p+2)….(2p-1)=p(pn-1)(mp^2+(p-1)!)=(p^2n-p)(mp^2+pn-1)=Mp^2+p

Thus (2pâˆ’1)! â‰، p \pmod {\space p^2}(2pâˆ’1)! â‰، p \pmod {\space p^2}

Why ?: \left(2 \times \color{#f00}{4} – 1\right)! = 7! = 5040 = 5036 + \color{#f00}{4} = {5036 \over 16}\,\color{#f00}{4}^{2} + \color{#f00}{4} = {1259 \over 4}\,\color{#f00}{4}^{2} + \color{#f00}{4} \left(2 \times \color{#f00}{4} – 1\right)! = 7! = 5040 = 5036 + \color{#f00}{4} = {5036 \over 16}\,\color{#f00}{4}^{2} + \color{#f00}{4} = {1259 \over 4}\,\color{#f00}{4}^{2} + \color{#f00}{4}
– Felix Marin
2 days ago