Let $P(n)$ be the product of the positive integers which are $\leq n$ and relatively prime to $n$. Prove that   $$P(n) =  n^{\phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)}. $$
123871
Actually, I posted this long ago in MO but did not get a reply as it was unfit.   Now this is an exercise in some textbook (I think Apostol), and I would be happy to receive some answers.  Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime to $n$.  Prove that $$ \displaystyle P(n) = n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}.$$
6591
Please give me the proof for the following: Let $a_1,\,a_2,\,\dots\,a_n$ be $n$ positive real numbers whose product is equal to $1$. Prove that $$a_1+\cdots+a_n \geq n$$ and that the equality sign holds only if every $a_k$ is  equal to $1$.
109469
How to prove that the direct product of finitely many cyclic groups $C_{n_1}\times C_{n_2}\times\cdots\times C_{n_m}$ is cyclic if the $n_i$'s are pairwise relatively prime?
56053
If $\phi(n)$ is the Euler-totient function, how can I show that $\phi(n) \ge \sqrt{n}$?
36177
That is:  If $\phi(n) = n$ then $n = 1$  Could someone give me a clue?
103641
Given that $p$ is a prime and $p\mid a^n$, prove that $p^n\mid a^n$.   I know that the fundamental theorem of arithmetic states that any positive integer can be represented as a product of primes but how do I apply this to the proof?
1191
how to prove:   $$ \prod_{d|n} d= n^{\frac{\tau (n)}{2}}$$   $\prod_{d|n} d$ is product of all of distinct positive divisor of $n$,  $\tau (n)$ is number (count)of all of positive divisor of $n$
59840
How to prove $$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ and $$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
85137
Let $p_n$ denote the $n$'th prime number.  How would one go about proving that infinite products like: $$\prod_{k=1}^\infty1 - \frac{1}{(p_k)^2} = \frac{6}{\pi^2}$$  or  $$\prod_{k=1}^\infty\frac{{p_k}^2}{{p_k}^2 - 1} = \frac{\pi^2}{6}$$ are correct?  Is there any way to prove it except by exhaustion?
15235
I am asked to prove that:       For integers $n, x,y &gt; 0$, where $x,y$ are relatively prime,        every $n \ge (x-1) (y-1)$ can be expressed as $xa + yb$, with nonnegative integers $a,b \ge0$.   How should I approach this? I have very limited knowledge in number theory.
103535
Prove that if $p$ and $q$ are distinct primes such that $pq\mid n^2$ then $pq\mid n$.  My Attempt:  Assume that $pq\nmid n$. If $pq\nmid n$, then $p\nmid n$ or $q\nmid n$. Without loss of generality, assume that $p\nmid n$.  If $p\nmid n$, then $pq\nmid n^2$.
38542
The function $d(n)$ gives the number of positive divisors of $n$, including n itself. So for example, $d(25) = 3$, because $25$ has three divisors: $1$, $5$, and $25$.  So how do I prove that the product of all of the positive divisors of $n$ (including $n$ itself) is $n^{\frac{d(n)}{2}}$.  For example, the divisors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. $d(12)$ is $6$, and $1 · 2 · 3 · 4 · 6 · 12 = 1728 = 12^3 = 12^{\frac{6}{2}} = 12^{\frac{d(n)}{2}}$
6905
Let $(a_n)$ be a strictly increasing sequence of positive integers such   that: $a_2 = 2$ and $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Show that $a_n = n$, for every positive integer $n$.   This is a result apparently due to Paul Erdős, and  has a proof by induction.  I tried like this, $a_{10}=a_2a_5$. After this what we can do?    [Editor's comment] Possibly due to the apparent simplicity of the conditions it may be difficult to appreciate the subtleties of this question. If we try and construct a counterexample like $a_3=4$, $a_4=5$, $a_5=6$, then the requirements dictate $a_6=8$, $a_{10}=12$, $a_{15}=24$. At this point we realize that we have been speeding. For monotonicity forces $a_9\le 11$, and therefore $a_{18}\le22&lt;a_{15}$, violating the requirements. It is not obvious why something similar ruins all the modifications to the sequence $a_n=n$. [/comment, JL]
70092
I am trying to show that: \begin{equation} \frac{n}{\phi(n)} = \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)} \end{equation} where $\phi(n)$ is  and $\mu$ is the .  The identity clearly holds for $n=1$, so if we write $n=p_1^{a_1} \cdots p_k^{a_k}$ for the prime decomposition of $n$, the left-hand side becomes: \begin{eqnarray*} \frac{n}{\phi(n)} &amp; = &amp; \frac{p_1^{a_1} \cdots p_k^{a_k}}{\phi(p_1^{a_1}) \cdots \phi(p_k^{a_k})} \\ &amp; = &amp; \frac{p_1^{a_1} \cdots p_k^{a_k}}{(p_1^{a_1-1})(p_1-1) \cdots (p_k^{a_k})(p_k-1)} \\ &amp; = &amp; \frac{p_1 \cdots p_k}{(p_1-1)\cdots (p_k-1)} \end{eqnarray*}   Whereas the right-hand side becomes:  \begin{eqnarray*} \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)} &amp; = &amp; \sum_{\substack{d \mid n \\ p^2 \nmid d}} \frac{1}{\phi(d)} \\ &amp; = &amp; 1 + \frac{1}{\phi(p_1)} + \dots + \frac{1}{\phi(p_k)} + \frac{1}{\phi(p_1 p_2)} + \dots + \frac{1}{\phi(p_{k-1}p_k)} \\ &amp; + &amp; \dots + \frac{1}{\phi(p_1 \cdots p_k)} \\ &amp; = &amp; 1 + \frac{1}{p_1 - 1} + \dots + \frac{1}{p_k - 1} + \frac{1}{(p_1 - 1)(p_2 - 1)} + \dots + \frac{1}{(p_{k-1}-1)(p_k-1)} \\ &amp; + &amp; \dots + \frac{1}{(p_1 - 1) \cdots (p_k - 1)} \end{eqnarray*}  I am unsure of how to proceed from this point however. Any help would be much appreciated.
110294
Is it true that the product of $n&gt;1$ consecutive integers is never a $k$-th power of another integer for any $k \geq 2$?  I can see this is true in certain cases. For instance if the product ends on a prime, But how would one prove this in general?  Thanks for any help or suggestions.
85098
I need help!  I need to prove that for any $2 \le n,k$ positive integers $(n - 1)^2 \mid n^k -1$ if and only if $(n - 1) \mid k$  Thanks!
63022
Given that $a \equiv b \pmod n$ and $c\equiv d \pmod n$, I need to prove that $ac \equiv bd \pmod n$  So far, I've only managed to deduce that $a+b \equiv c+d \pmod n$. I don't know if this is usable, but it's there, at least.  Any hint would be greatly appreciated!