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Peter
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#1
A 23
[GhEw pp.104]
(Wolstenholme's Theorem) Prove that if 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1} is expressed as a fraction, where p \ge 5 is a prime, then p^{2} divides the numerator.

PostPosted: Fri May 25, 2007 2:24 am
TomciO
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#2
We will treat rational numbers, which have their denominator relatively prime to p, as residues mod p.
From the Fermat's Little Theorem: \frac{1}{i} \equiv i^{p-2} \mod{p}
So we see that:
\sum_{i=1}^{p-1} \frac{1}{i} \equiv \sum_{i=1}^{p-1} i^{p-2} = \sum_{i=1}^{\frac{p-1}{2}} i^{p-2} + (p-i)^{p-2} \equiv \sum_{...
The last congruence comes from the fact that the rest term vanishes mod p^2. We are left with proving that:
\sum_{i=1}^{\frac{p-1}{2}} i^{p-3} \equiv 0 \mod{p}
or (one more time from FTL):
\sum_{i=1}^{\frac{p-1}{2}} \frac{1}{i^2} \equiv 0 \mod{p}
Now if i, j \leq \frac{p-1}{2} and \frac{1}{i^2} \equiv \frac{1}{j^2} \mod{p} then also (i-j)(i+j) \equiv 0 \mod{p} but since i, j \leq \frac{p-1}{2} we must have i=j. So the numbers: \frac{1}{i^2} for i=1,2,...,\frac{p-1}{2} are all different quadratic residues (and there are exactly \frac{p-1}{2} quadratic residues), as well as the numbers i^2 for i=1,2,...,\frac{p-1}{2} so:
\sum_{i=1}^{\frac{p-1}{2}} \frac{1}{i^2} \equiv \sum_{i=1}^{\frac{p-1}{2}} i^2 = \frac{p(p-1)(p+1)}{24} \equiv 0 \mod{p}
which ends the proof.

PostPosted: Fri May 25, 2007 2:24 am
Ilthigore
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#3
I love it. This is a beautiful proof.

PostPosted: Fri May 25, 2007 2:24 am
darij grinberg
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#4
Re: A 23
[GhEw pp.104]
Peter wrote:
(Wolstenholme's Theorem) Prove that if
1 + \frac {1}{2} + \frac {1}{3} + \cdots + \frac {1}{p - 1}
is expressed as a fraction, where p \ge 5 is a prime, then p^{2} divides the numerator.


This is a slightly edited version of an older MathLinks post of mine, and I am submitting it here since I will refer to it in a number of other proofs.

I will generalize the problem, but first I explain some preliminaries about primes.

1. p-adic evaluation

Let p be an arbitrary prime. We denote by \mathbb{Z}_{p} the field of residues of integers modulo p.

For any rational number x, we will now define a so-called p-adic evaluation of x. This evaluation will be denoted by v_{p}\left(x\right), and it is defined as follows: If x\neq 0, then v_{p}\left(x\right) is the greatest integer k such that \frac {x}{p^{k}} can be written as a fraction of two integers with the denominator not being divisible by p. Besides, we set v_{p}\left(0\right) = + \infty, where + \infty is just a symbol satisfying + \infty > a and + \infty + a = + \infty for any integer a (what we will mainly need is that v_{p}\left(0\right) > v_{p}\left(x\right) for any nonzero x).

We can easily see how to compute v_{p}\left(x\right) for rationals x\neq 0:
If x is a nonzero integer, then v_{p}\left(x\right) is the greatest integer k such that p^{k} divides x (particularly, v_{p}\left(x\right) = 0 if p doesn't divide x); if x is a fraction of two nonzero integers, say x = \frac {a}{b} with integers a and b, then v_{p}\left(x\right) = v_{p}\left(a\right) - v_{p}\left(b\right).

Note that there is a simple way to interpret the sign of the p-adic evaluation v_{p}\left(x\right) of a rational number x: If we write x as a reduced fraction (i. e. a fraction with numerator and denominator coprime; if x is an integer, then just write it in the form x = \frac {x}{1}), and it turns out that the numerator is divisible by p, then v_{p}\left(x\right) > 0; if neither the numerator nor the denominator is divisible by p, then v_{p}\left(x\right) = 0; if the denominator is divisible by p, then v_{p}\left(x\right) < 0.

It is rather easy to prove that for any two rational numbers x and y, we have v_{p}\left(xy\right) = v_{p}\left(x\right) + v_{p}\left(y\right) and v_{p}\left(x + y\right)\geq\min\left(v_{p}\left(x\right);\;v_{p}\left(y\right)\right).

2. Working with fractions modulo p

We will also need some familiarity with fractions in \mathbb{Z}_{p}. The main idea is to extend the notion of congruence modulo p from integers to rational numbers with nonnegative p-adic evaluation: If x and y are two rational numbers such that v_{p}\left(x\right)\geq 0 and v_{p}\left(y\right)\geq 0, then we say that x\equiv y\mod p if and only if v_{p}\left(x - y\right) > 0. Thus we have defined an equivalence relation (that it is an equivalence relation is easy to prove - do it for yourself) on the set of all rational numbers with nonnegative p-adic evaluation. Now one could expect that, in this way, we would obtain a kind of new group of remainders, say \mathbb{Q}_{p} - but in fact, we get nothing but our old familiar \mathbb{Z}_{p}, since for every rational number x, there is one and only one integer n satisfying 0\leq n < p such that x\equiv n\mod p (this is again easy to prove). Hence, you can work with rational numbers whose p-adic evaluation is nonnegative in the same way as you work with remainders modulo p. This also means that you can uniquely divide modulo p. [A nice application of this is problem 4 of the IMO 2005 - see Fedor Petrov's proof in post #2.]

3. Generalization of Wolstenholme's Theorem

Now, the problem A 23 asks us to show that if p is a prime number such that p\geq 5, then v_{p}\left(\sum_{k = 1}^{p - 1}\frac {1}{k}\right)\geq 2. This is a particular case (u = 1) of the following result:

Theorem 1. If p is a prime number and u is an odd integer such that p\geq u + 3, then

v_{p}\left(\sum_{k = 1}^{p - 1}\frac {1}{k^{u}}\right)\geq 2.

Proof of Theorem 1. We will first work in \mathbb{Q} and only later come over into \mathbb{Z}_{p}. Note that p > 2 (since p\geq u + 3). We start with the Gauss trick:

2\sum_{k = 1}^{p - 1}\frac {1}{k^{u}} = \sum_{k = 1}^{p - 1}\frac {1}{k^{u}} + \sum_{k = 1}^{p - 1}\frac {1}{\left(p - k\righ....

Now denote s_k=\sum_{i=2}^u\left(-1\right)^{u-i}\binom{u}{i}p^{i-2}k^{u-i} for every k\in\left\{1,2,...,p-1\right\}. Obviously, s_k is an integer for every k. We can expand \left(p - k\right)^{u} according to the binomial formula:

\left(p - k\right)^{u}=\sum_{i=0}^u\left(-1\right)^{u-i}\binom{u}{i}p^ik^{u-i}
=\left(-1\right)^{u-0}\binom{u}{0}p^0k^{u-0}+\left(-1\right)^{u-1}\binom{u}{1}p^1k^{u-1}+\sum_{i=2}^u\left(-1\right)^{u-i}\bi...
=\left(-1\right)^uk^u+\left(-1\right)^{u-1}upk^{u-1}+p^2\sum_{i=2}^u\left(-1\right)^{u-i}\binom{u}{i}p^{i-2}k^{u-i}
=\left(-1\right)^uk^u+\left(-1\right)^{u-1}upk^{u-1}+p^2s_k=\left(-1\right)k^u+1upk^{u-1}+p^2s_k (since u is odd, so that \left(-1\right)^u=-1 and \left(-1\right)^{u-1}=1)
=-k^u+upk^{u-1}+p^2s_k.

Thus, k^u+\left(p-k\right)^u = upk^{u-1}+p^2s_k for every k\in\left\{1,2,...,p-1\right\}, and therefore

2\sum_{k = 1}^{p - 1}\frac {1}{k^{u}} = \sum_{k = 1}^{p - 1}\frac {k^{u} + \left(p - k\right)^{u}}{k^{u}\left(p - k\right)^{u...
= p \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}}.

But since p > 2, we have v_{p}\left(2\right) = 0, so that

v_{p}\left(2\sum_{k = 1}^{p - 1}\frac {1}{k^{u}}\right) = v_{p}\left(2\right) + v_{p}\left(\sum_{k = 1}^{p - 1}\frac {1}{k^{u....

On the other hand,

v_{p}\left(2\sum_{k = 1}^{p - 1}\frac {1}{k^{u}}\right) = v_{p}\left( p \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left...
= 1 + v_{p}\left( \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}} \right).

Thus,

v_{p}\left(\sum_{k = 1}^{p - 1}\frac {1}{k^{u}}\right) = 1 + v_{p}\left( \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\lef....

Hence, instead of showing v_{p}\left(\sum_{k = 1}^{p - 1}\frac {1}{k^{u}}\right)\geq 2, it will be enough to prove v_{p}\left( \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}} \right)\geq 1. This is equivalent to v_{p}\left( \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}} \right) > 0, i. e. to \sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}} \equiv 0\mod p. Now we start working in \mathbb{Z}_{p}. In fact, we can consider all the fractions \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}} in \mathbb{Z}_{p} since they have nonnegative p-adic evaluations (because their denominators, k^{u}\left(p - k\right)^{u}, are not divisible by p, since 1\leq k\leq p - 1 and p is a prime). The numerators uk^{u-1}+ps_k of these fractions can be simplified to uk^{u - 1}, since p\equiv 0\mod p, and thus all terms containing p can be omitted. Also, the denominators can be simplified: Since p - k\equiv - k\mod p, we have k^{u}\left(p - k\right)^{u}\equiv k^{u}\left( - k\right)^{u} = \left( - 1\right)^{u}k^{2u}. Hence,

\sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}}\equiv\sum_{k = 1}^{p - 1}\frac {uk^{u - 1}}{\left( - ....

But we are working in \mathbb{Z}_{p}, and thus, by Fermat's small theorem, k^{p - 1}\equiv 1\mod p for every integer k such that 1\leq k\leq p - 1. Thus, k^{ - u - 1}\equiv k^{ - u - 1}k^{p - 1} = k^{p - u - 2}\mod p, and this sum becomes

\sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}}\equiv\frac {u}{\left( - 1\right)^{u}}\sum_{k = 1}^{p ....

Now, p - u - 2\geq 1 (since p\geq u + 3) and p - u - 2\leq p - 2. Also, p is an odd prime (since p is a prime and p > 2). Now, according to http://www.problem-solving.be/pen/viewtopic.php?t=676 part a) (and also http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 ), we have:

Theorem 2. If p is an odd prime and \rho is an integer such that 1\leq\rho\leq p - 2, then p\mid\sum_{k = 1}^{p - 1}k^{\rho}.

Applying this Theorem 2 to our prime p and \rho = p - u - 2 (we know that p is an odd prime and \rho = p - u - 2 satisfies 1\leq p - u - 2\leq p - 2), we get p\mid\sum_{k = 1}^{p - 1}k^{p - u - 2}, so that \sum_{k = 1}^{p - 1}k^{p - u - 2}\equiv 0\mod p, and thus

\sum_{k = 1}^{p - 1} \frac {uk^{u-1}+ps_k}{k^{u}\left(p - k\right)^{u}}\equiv\frac {u}{\left( - 1\right)^{u}}\sum_{k = 1}^{p ....

This completes the proof of Theorem 1.

[EDIT: Made the proof of Theorem 1 slightly more formal. See the other post for the old version.]

Darij

PostPosted: Fri May 25, 2007 2:24 am
Last edited by darij grinberg on Wed Nov 19, 2008 4:26 pm; edited 1 time in total
kdano
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#5
TomciO wrote:

\sum_{i = 1}^{p - 1} \frac {1}{i} \equiv \sum_{i = 1}^{p - 1} i^{p - 2}


Why is this true mod p^2?

PostPosted: Thu Jul 03, 2008 6:11 pm
TomciO
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#6
Yes, you are right. Thank you for pointing out the mistake. I have corrected the proof accordingly since it's not a big difference.

We will treat rational numbers, which have their denominator relatively prime to p, as residues mod p.
From the Euler's Theorem: \frac {1}{i} \equiv i^{p(p-1) - 1} \mod{p^2}
So we see that:
\sum_{i = 1}^{p - 1} \frac {1}{i} \equiv \sum_{i = 1}^{p - 1} i^{p(p-1) - 1} = \sum_{i = 1}^{\frac {p - 1}{2}} i^{p(p-1) - 1}...
The congruence come from the fact that the rest term vanishes mod p^2. We are left with proving that:
\sum_{i = 1}^{\frac {p - 1}{2}} i^{p(p-1) - 2} \equiv 0 \mod{p}
or from FTL:
\sum_{i = 1}^{\frac {p - 1}{2}} \frac {1}{i^2} \equiv 0 \mod{p}
Now if i, j \leq \frac {p - 1}{2} and \frac {1}{i^2} \equiv \frac {1}{j^2} \mod{p} then also (i - j)(i + j) \equiv 0 \mod{p} but since i, j \leq \frac {p - 1}{2} we must have i = j. So the numbers: \frac {1}{i^2} for i = 1,2,...,\frac {p - 1}{2} are all different quadratic residues (and there are exactly \frac {p - 1}{2} quadratic residues), as well as the numbers i^2 for i = 1,2,...,\frac {p - 1}{2} so:
\sum_{i = 1}^{\frac {p - 1}{2}} \frac {1}{i^2} \equiv \sum_{i = 1}^{\frac {p - 1}{2}} i^2 = \frac {p(p - 1)(p + 1)}{24} \equi...
which ends the proof.

PostPosted: Thu Jul 03, 2008 6:40 pm
manjil
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#7
Fields
Maybe we can use fields to solve this in a more efficient way!
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Manjil P. Saikia

PostPosted: Thu Aug 14, 2008 4:15 am
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