Facing a small obstacle while trying to proof Wolstenholme’s theorem 1 using multiplicative inverse in (mod p)

algebraic-number-theorynumber theoryprime numbersproof-verification

Wolstenholme's theorem states that for a prime $p>3$
and
$$ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+……+\frac{1}{(p-1)^2}=\frac{a}{b}.$$
$ a\equiv 0\pmod p $

I think that I found a reasonable way to prove this theorem but I face a small obstacle at the end of the proof.

My solution steps are as follows.

  1. make common factor for the fractions.
  2. for each term in the numerator take (mod p) and simplify the result using the multiplicative inverse in (mod p)
    for example the first term in numerator is
    $ 2^2 \times 3^2 \times ….. \times (p-1)^2 $
    in this quantity each term has its multiplicative inverse (mod p) multiplied by it so this quantity is only $ \equiv 1\pmod p $.
    and after doing the same thing to other terms you get at the end that
    the numerator is $ \equiv (1^2+1^2+2^2+3^2+….(p-2)^2)\pmod p $ .
    after using the formula for summing n squared numbers and simplifying I got
    $$a \equiv \frac{2p^3-9p^2+13p}{6}\pmod p $$

this is the question :
for p>3 (p is a prime number)

how to prove that this quantity :
$$ \frac{2p^3-9p^2+13p}{6} $$ is divisible by p .

or equivalently
$$ \frac{2p^2-9p+13}{6} $$ is Integer ( belongs to Z ) .

I appreciate any attempts for help.

Best Answer

HINTS

  1. In your numerator, $2p^2$ is always even and the other terms are always odd, and the difference of odd terms is always even, so the numerator is always even.

  2. Since $p$ is a prime, we cannot have $p \equiv 0 \pmod{3}$. Thus, $p = 3n \pm 1$. The middle term clearly is divisible by $3$ and the corner ones yield $$ \begin{split} 2p^2+13 &= 2(3n\pm 1)^2+13\\ &= 2\left(9n^2+6n+1\right)+13\\ &= 18n^2+12n+15, \end{split} $$ and all coefficients are divisible by 3.

Can you finish this?

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