Science:Math Exam Resources/Courses/MATH437/December 2011/Question 05

MATH437 December 2011

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Question 05
Let $\displaystyle m,n\in \mathbb {N}$ such that $\displaystyle m\phi (m)=n\phi (n)$. Prove that $\displaystyle m=n$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
If m and n differ, then there is a largest prime which they differ. Starting with that, argue using a descent argument.

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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. The claim is clearly true of one of m or n equals 1. Thus we suppose that both are strictly greater than 1. In this case, the Fundamental Theorem of Arithmetic tells us that we may write $\displaystyle m=\prod _{i=1}^{k}p_{i}^{\alpha _{i}}$ and $\displaystyle n=\prod _{j=1}^{\ell }q_{j}^{\beta _{j}}$ where each of the exponents is a positive integer at least 1. The left hand side of the equation in question thus reduces to $\displaystyle m\phi (m)=\prod _{i=1}^{k}p_{i}^{\alpha _{i}}\left(\prod _{i=1}^{k}p_{i}^{\alpha _{i}-1}(p_{i}-1)\right)=\prod _{i=1}^{k}p_{i}^{2\alpha _{i}-1}(p_{i}-1)$ and in total as we are given that $\displaystyle m\phi (m)=n\phi (n)$, we have $\displaystyle \prod _{i=1}^{k}p_{i}^{2\alpha _{i}-1}(p_{i}-1)=\prod _{j=1}^{\ell }q_{j}^{2\beta _{j}-1}(q_{j}-1)$ Now, suppose that m and n are distinct. Let $\displaystyle p_{0}$ be the largest prime where they differ and without loss of generality, suppose that $\displaystyle p_{0}^{\alpha _{0}}\parallel m$ but $\displaystyle p_{0}^{\alpha _{0}}\nmid n$. Then we know that we must have that $\displaystyle p_{0}\mid (q_{j_{0}}-1)$ for some $\displaystyle j_{0}$. Thus $\displaystyle q_{j_{0}}>p_{0}$. However, $\displaystyle p_{0}$ was chosen to be the largest such prime where m and n differ to any power. So $\displaystyle q_{j_{0}}^{\beta _{j_{0}}}=p_{i_{0}}^{\alpha _{i_{0}}}$ for some $\displaystyle i_{0}$ and thus $\displaystyle (q_{j_{0}}-1)=(p_{i_{0}}-1)$. Thus, we may cancel out these terms and repeat this argument to find another index $\displaystyle j_{0}$ and since there are only finitely many of these terms, we must eventually reach a contradiction. Hence m and n have all the same prime factors to the same powers and thus are equal. Food for thought: Is there an easier proof of this using the fact that $\displaystyle n\phi (n)=2\sum k$ where the above sum ranges over all integers $\displaystyle 1\leq k\leq n$ with $\displaystyle \gcd(k,n)=1$?

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