# Elementary proof of Wilson's Theorem

Proof of Wilson's Theorem

In the elementary proof here, we solve by pairing an element with its inverse. Why do we know necessarily that this will always happen?

• Hint. You know every element has an inverse. When can an element be its own inverse (remember your modulus is prime)? – Ethan Bolker Feb 16 '16 at 14:43
• when it is congruent to -1 or 1! thanks – feng Feb 16 '16 at 14:45
• You didn't read the article correctly. It explains that at length. – Yves Daoust Feb 16 '16 at 14:56

1. Every integer relatively prime to $p$ has an inverse mod $p$.
2. If $a$ has inverse $a^{-1}$ then $a^{-1}$ has inverse $a$.
3. What about elements which are their own inverse? This amounts to solving $x^2 \equiv 1 \pmod p$, whose only solutions are $x \equiv \pm 1 \pmod p$.