Science:Math Exam Resources/Courses/MATH437/December 2011/Question 04

MATH437 December 2011

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Question 04
Let $\displaystyle p\geq 5$ be a prime number. Let $\displaystyle a,b\in \mathbb {N}$ such that (i) $\displaystyle \gcd(a,b)=1$ (ii)$\displaystyle \sum _{i=1}^{p-1}{\frac {1}{i}}={\frac {a}{b}}$. Prove that $\displaystyle p^{2}\mid a$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First show that the problem is equivalent to the problem of showing that $\displaystyle p^{2}\mid \sum _{i=1}^{p-1}{\frac {(p-1)!}{i}}$.

Hint 2
Pair up the terms by looking at $\displaystyle {\frac {1}{i}}+{\frac {1}{p-i}}$.

Hint 3
In the words of Tom Hanks, Willlllsonnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn. ... Try to use Wilson's Theorem.

Hint 4
Lastly, you will want to show that $\displaystyle S:=1^{2}+...+\left({\frac {p-1}{2}}\right)^{2}\equiv 0\mod {p}$ Do this by showing that $\displaystyle S+S\equiv 0\mod {p}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, notice that $\displaystyle \sum _{i=1}^{p-1}{\frac {1}{i}}={\frac {a}{b}}={\frac {a{\tfrac {(p-1)!}{b}}}{(p-1)!}}$. Now, if we are to show that $\displaystyle p^{2}\mid a$ then it suffices to show that $\displaystyle p^{2}\mid a{\tfrac {(p-1)!}{b}}$ since the term given by $\displaystyle {\frac {(p-1)!}{b}}$ is corpime to p. Cross multiplying shows us that it suffices to show that $\displaystyle p^{2}\mid \sum _{i=1}^{p-1}{\frac {(p-1)!}{i}}$. Let's now pair terms with their additive inverses: $\displaystyle \sum _{i=1}^{p-1}{\frac {(p-1)!}{i}}=(p-1)!\sum _{i=1}^{(p-1)/2}\left({\frac {1}{i}}+{\frac {1}{p-i}}\right)=(p-1)!\sum _{i=1}^{(p-1)/2}\left({\frac {p-i+i}{i(p-i)}}\right)=p\sum _{i=1}^{(p-1)/2}\left({\frac {(p-1)!}{i(p-i)}}\right)$ and thus it now suffices to show that the sum $\displaystyle \sum _{i=1}^{(p-1)/2}{\frac {(p-1)!}{i(p-i)}}\equiv \sum _{i=1}^{(p-1)/2}(-i^{2})^{-1}(p-1)!\equiv \sum _{i=1}^{(p-1)/2}-(-i^{2})^{-1}\equiv \sum _{i=1}^{(p-1)/2}i^{-2}\mod {p}$ satisfies $\displaystyle S:=\sum _{i=1}^{(p-1)/2}i^{-2}\equiv 0\mod {p}$. To do this, we use the trick common in evaluating Gauss sums and looking at $\displaystyle 2S=S+S$. this gives $\displaystyle {\begin{aligned}S+S&=\sum _{i=1}^{(p-1)/2}i^{-2}+\sum _{i=1}^{(p-1)/2}i^{-2}\\&=\sum _{i=1}^{(p-1)/2}i^{-2}+\sum _{i=1}^{(p-1)/2}(-i)^{-2}\end{aligned}}$ Now note that $\displaystyle \{1^{-1},2^{-1},...,(p-1)^{-1}\}$, $\displaystyle \{1,2,...,p-1\}$ and $\displaystyle \{\pm 1,\pm 2,...,\pm (p-1)/2\}$ all form complete residue systems for $\displaystyle (\mathbb {Z} /p\mathbb {Z} )^{*}$. Thus, the above sum is $\displaystyle {\begin{aligned}S+S&=\sum _{i=1}^{(p-1)/2}i^{-2}+\sum _{i=1}^{-(p-1)/2}i^{-2}\\&=\sum _{i=1}^{p-1}i^{-2}\\&=\sum _{i=1}^{p-1}i^{2}\\&={\frac {p(p-1)(2p-1)}{6}}\end{aligned}}$ Here is where we use the fact that $\displaystyle p\geq 5$. Notice that as p is odd, we have that $\displaystyle 2\mid (p-1)$. Further, we know that either $\displaystyle p\equiv 1\mod {3}$ or $\displaystyle p\equiv 2\mod {3}$. If $\displaystyle p\equiv 1\mod {3}$ then $\displaystyle 3\mid (p-1)$. If $\displaystyle p\equiv 2\mod {3}$ then $\displaystyle 2p-1\equiv 2(2)-1\equiv 0\mod {3}$ and so $\displaystyle 3\mid (2p-1)$. In either case, we see that $\displaystyle {\frac {(p-1)(2p-1)}{6}}\in \mathbb {Z}$ and thus $\displaystyle {\frac {p(p-1)(2p-1)}{6}}\equiv p\left({\frac {(p-1)(2p-1)}{6}}\right)\equiv 0\mod {p}$. Hence, we see that $\displaystyle 2S\equiv 0\mod {p}$. Again as p is odd, we have that $\displaystyle S\equiv 0\mod {p}$. This completes the proof. For those of you wondering, this is also called Wolstenholme's Theorem.

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