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QUESTION

\begin{description}

\item[(i)]
Give, without proof, a formula for Euler's function, $\phi(n)$, in
terms of the prime power factorisation of $n$.

\item[(ii)]
Let $m$ and $n$ be positive integers such that HCF$(m,n)=d$. Show
that

$$\phi(d)\phi(mn)=\phi(m)\phi(n)d.$$

\item[(iii)]
Hence show that

$$\phi(mn)\leq\phi(m)\phi(n)$$

with equality if and only if HCF$(m,n)=1$.

\item[(iv)]
Define what is meant by the multiplicative order of a congruence
class, $[x]\in U_m$, where $U_m$ denotes the group of units mod
($m$).

\item[(v)]
Suppose that there exists an integer, $x$, such that HCF$(x,m)=1$
and the order of $[x]$ in $U_m$ is equal to $m-1$. Using Euler's
Theorem, or otherwise, show that $m$ is prime.

\end{description}



ANSWER

\begin{description}

\item[(i)]
If $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ where the $p_i$'s are
distinct primes and the $a_i\geq1$ are integers then

$$\phi(n)=p_1^{a_1-1}p_2^{a_2-1}\ldots
p_k^{a_k-1}\prod_{j=1}^k(p_j-1)$$

\item[(ii)]
To prove this one we change notation a little. Let
$p_1,\ldots,p_k$ denote the set of distinct primes which appear to
a strictly positive exponent in the prime power factorisation of
at least one of $m$ or $n$. Hence we may write

$$n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k},m=p_1^{b_1}p_2^{b_2}\ldots
p_k^{b_k}$$

with each $a_i\geq0$ and $b_i\geq0$ but $a_i+b_i&gt;0$ for each $i$.
In this case

$$d=\prod_{i=1}^kp_i^{min(a_i,b_i)}$$

Hence the left hand side is equal to

$$\left(\prod_{i=1}^kp_i^{min(a_i,b_i)-1}\left(p_i-1\right)\right)
\left(\prod_{j=1}^kp_j^{a_j+b_j-1}(p_j-1)\right)$$

where in the first factor only $i$'s with $min(a_i,b_i)\geq1$
appear.

On the right hand side we have

$$\left(\prod_{i=1}^kp_i^{min(a_i,b_i)}\right)\left(\prod_{u=1}^k
p_u^{a_u-1}(p_u-1)\right)\left(\prod_{j=1}^kp_j^{b_j-1}(p_j-1)\right)$$

where in second factor only $u$'s with $a_u\geq1$ appear and in
the third factor only $j$'s with $b_j\geq1$ appear.

Now look at the occurrence of $p_i$ and $(p_i-1)$'s on both sides.
On the left we find

\begin{tabular}{ll}
$p_i^{min(a_i,b_i)-1}(p_i-1)p_i^{a_i+b_i-1}(p_i-1)$&amp;if
min$(a_i,b_i)\geq1$,\\ $p_i^{a_i+b_i-1}(p_i-1)$&amp;if
min$(a_i,b_i)=0$
\end{tabular}

On the right we find

\begin{tabular}{ll}
$p_i^{min(a_i,b_i)}p_i^{a_i-1}(p_i-1)p_i^{b_i-1}(p_i-1)$&amp;if
$a_i\geq1$ and $b_i\geq1$,\\
$p_i^{min(a_i,b_i)}p_i^{b_i-1}(p_i-1)$&amp;if $a_1=0$ and
$b_1\geq1$,\\ $p_i^{min(a_i,b_i)}p_i^{a_i-1}(p_i-1)$&amp;if $a_i\geq1$
and $b_i=0$.
\end{tabular}

These expressions are equal.

\item[(iii)]
If $d&gt;1$ then $\phi(d)&lt;d$ by the formula of (i) while $\phi(1)=1$.

\item[(iv)]
The order of a congruence class $[x]\in U_m$ is the least positive
integer, $k$, such that $x^k\equiv1$ mod ($m$).

\item[(v)]
By Euler's Theorem we know that $x^{\phi(m)}\equiv1$ mod ($m$) and
therefore that the order of $x$ divides $\phi(m)$. Hence $m$ is
such that $(m-1)|\phi(m)$. Since $\phi(m)&lt;m$ for $m&gt;1$ we must
have $\phi(m)=m-1$ which implies that $m$ is prime, by the formula
of (i).

\end{description}





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