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.
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.
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..
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 $$\phi ((2,6)) \phi ([2,6]) = \phi(2) \phi (6) = 2 \neq 4 = \phi (12) = \phi ((2,6)[2,6]).$$
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.
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 $\varphi(d)/d$.