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If $m, n$ be positive integers, prove that $\phi(mn)=\phi((m,n))\phi([m,n])$, where $(m,n)=$ gcd of $m, n$ and $[m, n]=$ lcm of $m, n$.

I have no idea to solve this question. Please help me to solve the problem.

user1942348
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4 Answers4

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It is false. Take $m=n$.

You can prove that the statement holds if and only if we are in the trivial case $\text{gcd}(m,n)=1$.

To be precise, the claim should be modified to $$ \varphi(mn)=\mathrm{gcd}(m,n)\cdot \varphi(\mathrm{lcm}(m,n)). $$

Proof:

Since $mn=\mathrm{lcm}(m,n)\mathrm{gcd}(m,n)$ and a prime $p$ divides $mn$ if and only if it divides $\mathrm{lcm}(m,n)$ then $$ \frac{\varphi(mn)}{mn}=\prod_{p\mid mn}\left(1-\frac{1}{p}\right)=\prod_{p\mid \mathrm{lcm}(m,n)}\left(1-\frac{1}{p}\right)=\frac{\varphi(\mathrm{lcm}(m,n))}{\mathrm{lcm}(m,n))}. $$ It follows that $$ \frac{\varphi(mn)}{mn}=\frac{\varphi(\mathrm{lcm}(m,n))}{mn} \cdot \mathrm{gcd}(m,n) $$ hence the above claim..

Paolo Leonetti
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  • What is the meaning of φ(rad(gcd(m,n))) ? – user1942348 Apr 26 '15 at 15:02
  • gcd is the greatest common divisor; lcm is the least common multiple; $\varphi$ is the Euler's phi; $rad(x)$ is the product of distinct primes dividing $x$, i.e. its radical – Paolo Leonetti Apr 26 '15 at 15:09
  • Then how to prove the new form of the problem? – user1942348 Apr 26 '15 at 15:14
  • Can it be a right question: Prove that $\phi(m)\phi(n) = \phi(gcd(m, n))\phi(lcm(m, n))$ for all $m, n \in N$ – user1942348 Apr 26 '15 at 15:20
  • Again (6,9) is a counterexample. We are not trying to guess the correct question! If you dont know that it is the correct exercise, you should say it .... – Paolo Leonetti Apr 26 '15 at 15:54
  • The question given above is taken from a very std book. There is no typos when I have copied the question from that book. I don't know whether the problem given in the book is correct or wrong. But with (6,9) as a counterexample, it shows problem. If your claim that you have given above is correct, please prove that. – user1942348 Apr 26 '15 at 16:28
  • Give me the name of your book please, then i will add the proof – Paolo Leonetti Apr 26 '15 at 18:21
  • I cannot find it, but nevermind. Here is the proof – Paolo Leonetti Apr 27 '15 at 21:38
  • Thanks for your effort. Your last result shows $$ {\varphi(mn)}={\varphi(\mathrm{lcm}(m,n))} \cdot \mathrm{gcd}(m,n) $$ But the question is $$ {\varphi(mn)}={\varphi(\mathrm{lcm}(m,n))} \cdot \varphi(\mathrm{gcd}(m,n)) $$ – user1942348 Apr 29 '15 at 05:47
  • I already proved you that your claim is wrong... – Paolo Leonetti Apr 29 '15 at 05:57
  • If you dont like the counterexample proof, and you want to claim the both statements above are true, then it should be that $\varphi(x)=x$ where $x=\text{gcd}(m,n)$. Do you agree that this is possible if and only if $x=1$? – Paolo Leonetti Apr 29 '15 at 06:02
  • The question that I have given is also available in the book of Burton (Q No 14b, PP 136, PROBLEMS 7.2). Please check it. – user1942348 Apr 29 '15 at 15:44
  • I have the fifth edition.. Can you check where is it exactly? – Paolo Leonetti Apr 30 '15 at 16:01
  • Burton , 5th Edition (Q No 14b, PP 134, PROBLEMS section 7.2) ϕ(m)ϕ(n)=ϕ(gcd(m,n))ϕ(lcm(m,n)) – user1942348 Apr 30 '15 at 19:24
  • Indeed, he asks for $\varphi(m)\varphi(n)$, not $\varphi(mn)$ [recall that they are equal if and only if m and n are coprime, as expected..] – Paolo Leonetti May 01 '15 at 22:44
  • Yes, Prove that ϕ(m)ϕ(n)=ϕ(gcd(m,n))ϕ(lcm(m,n)) for all m,n∈N, I asked the same on Aprl 26, see few discussion lines above from here. . – user1942348 May 02 '15 at 01:27
  • Oh I see now that (6,9) is not a counterexample for the new problem. Wait, I put it in a new answer – Paolo Leonetti May 02 '15 at 07:48
  • It is true that the claim is false, take $m=n$, but your argument is not. – reuns Dec 06 '17 at 12:12
  • @reuns The question was different at the beginning. However, I edited it. – Paolo Leonetti Dec 11 '17 at 19:41
  • The first part is still not correct. The second part is correct : if $p^k | a$ then $\varphi(ap^k) = p^k \varphi(a)$. Thus $mn = \gcd(m,n) \text{lcm}(m,n)$ where $\gcd(m,n)\ |\ \text{lcm}(m,n) \implies \varphi(mn) = \gcd(m,n)\varphi(\text{lcm}(m,n))$ – reuns Dec 12 '17 at 03:56
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This is false.

To see this, let’s start with a lemma:

$$\phi(mn) = \frac{\phi(m) \phi(n)d}{\phi(d)}.$$ where $d = (m,n).$

Proof of lemma:

$$\frac{\phi(mn)}{mn} = \prod_{p \mid mn}(1-\frac{1}{p}) = \frac{\prod_{p \mid m}(1-\frac{1}{p}) \prod_{p \mid n}(1-\frac{1}{p})}{\prod_{p \mid (m,n)}(1- \frac{1}{p})} = \frac{\frac{\phi(m)}{m} \frac{\phi(n)}{n}}{\frac{\phi(d)}{d}}$$ and we are done.

Now, if we consider any integers $a,b$ then $m = (a,b), n = [a,b]$ and so $d = (m,n) = m$ since $m \mid n$ in this case. Thus, the lemma above yields

$$ \phi(ab) = \phi((a,b)[a,b]) = \frac{\phi((a,b)) \phi([a,b]) (a,b)}{\phi((a,b))} = \phi([a,b])(a,b)$$ and so equality with what you have holds only when $$(a,b) = \phi(a,b).$$ Now, the only integer $k$ that satisfies $\phi (k) = k$ is $k = 1$ (I leave this fact to you). This tells us that $(a,b) = 1$ by necessity for equality to hold. So, the integers must be coprime.

To see an example where equality doesn’t hold, consider $a = 2, b = 6$. Then $$\phi ((2,6)) \phi ([2,6]) = \phi(2) \phi (6) = 2 \neq 4 = \phi (12) = \phi ((2,6)[2,6]).$$

user328442
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First of all, we have $\operatorname{gcd}(a,b)\cdot\operatorname{lcm}(a,b)=ab$, then now you need to ascertain that Euler's Totient Function has the following property: $\phi(mn)=\phi(m)\phi(n)$. Hint: write a matrix with mn entries.

Roger
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    $\phi(mn)=\phi(m)\phi(n)$ holds only when m and n are co-prime. Then how to use this theorem here. Please write in details for me. Also how to write the matrix. – user1942348 Apr 26 '15 at 05:55
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Let $d=\text{gcd}(m,n)$, it is claimed that $$ \varphi(n)\varphi(m)=\varphi(d)\varphi(mn/d). $$ Dividing each side by $mn$, this is equivalent to $$ \prod_{p\mid n}\left(1-\frac{1}{p}\right)\prod_{p\mid m}\left(1-\frac{1}{p}\right)=\prod_{p\mid d}\left(1-\frac{1}{p}\right)\prod_{p\mid mn}\left(1-\frac{1}{p}\right). $$ This is true because each factor in the left hand side compares exactly once in the right hand: everything is collected in the last product, and the repeated terms between the first two factors appears in $\varphi(d)/d$.

Paolo Leonetti
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