Lagrange's Theorem: let p be any prime and [tex]f(x) = a_nx^n +a_{n-1}x^{n-1} + ... + a_1x + a_0[/tex] with [tex]a_n[/tex] ≡/≡ 0 (mod p). Then f(x) ≡ 0 (mod p) has at most n solutions.(adsbygoogle = window.adsbygoogle || []).push({});

Use the above theorem to prove Wilson's theorem.

Hint: Let [tex]f(x) = (x-1)(x-2)...(x-(p-1)) - (x^{p-1} - 1)[/tex] for an odd prime p.

Proof:

Expanding,

[tex]f(x) = (x-1)(x-2)...(x-(p-1)) - (x^{p-1} - 1)[/tex]

= [tex]a_{p-2}x^{p-2} +...+ a_1x + a_0[/tex] where the [tex]a_i[/tex] are some coefficients

By the above theorem, f hasat mostp-2 roots mod p IF [tex]a_{p-2}[/tex] ≡/≡ 0 (mod p). (*)

But by Fermat's theorem, for a=1,2,...,p-1, [tex]a^{p-1} -1[/tex] ≡ 0 (mod p).

So for a=1,2,...,p-1, f(a) ≡ 0 (mod p).

So f hasat leastp-1 roots mod p. (**)

(*) and (**) contradict unless f(x) ≡ 0 (mod p). Therefore, we must have f(x) ≡ 0 (mod p).

=> f(0)=(-1)(-2)...(-(p-1)) - (-1) ≡ 0 (mod p)

=> [tex](-1)^{p-1} (p-1)! + 1[/tex] ≡ 0 (mod p)

For odd primes p, p-1 is even, so (p-1)! ≡ -1 (mod p)

(For p=2, check directly.)

=========================================

I don't understand the two lines in red.

I understand that there is a contradiction, but why does this imply that f(x) ≡ 0 (mod p)? Why in this case, there will be no contradiction? I'm totally lost here...

Also, f(x) ≡ 0 (mod p) doesn't necessarily mean it holds for EVERY integer x, so why can we substitute x=0 and say that f(0) ≡ 0 (mod p)? What is thejustificationfor this step?

I hope someone can explain this proof.

Thank you very much!

[also under discussion in math help forum]

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Prove Wilson's theorem by Lagrange's theorem

Loading...

Similar Threads - Prove Wilson's theorem | Date |
---|---|

I Proving that an operator is unbounded | Feb 8, 2018 |

I Proving a set is linearly independant | Apr 14, 2017 |

I Proving a property when elements of a group commute | Mar 29, 2017 |

Wilson's Theorem | Jun 13, 2012 |

**Physics Forums - The Fusion of Science and Community**