Order of $ U(n) $

Question

Let $U(n)$ be group under multiplication modulo $n$. For $n=248$, find number of elements in $U (n)$.

As I tried to do this problem. The number of required elements are $\phi(n) $. So to calculate $\phi(248) $ I first write $248$ as product of powers of primes. So we have $248= 2^3\cdot 31$.

Since $\phi (n) = n (1- \frac{1}{p})(1-\frac{1}{q})$ , where $n= p^iq^j$,

So $\phi (248) =248 (1-\frac{1}{2})(1-\frac{1}{31}) =120$.

But book says answer is $180$. What's going wrong?

Answer 1

The euler totient function states that

$\phi(n) = n \ * \prod_{p|n} (1 - \frac{1}{p}) $

where the product is over the distinct prime numbers dividing n.

Now, by factorisation of 248 into product of primes, we have $248 = 2^{3} * 31$.

$ \therefore \ \{2,31\} $ are the distinct primes dividing 248.

So, $ \phi(248) = 248 \ * (1-\frac{1}{2}) \ * (1-\frac{1}{31})$

$ = 248 * (\frac{1}{2})*(\frac{30}{31}) = 120 $

"Understanding what a theorem means is a prerequisite to understanding its proof." - Contemporary Abstract Algebra - Gallian, Pg82, Theorem 4.3

Answer 2

The answer indeed is $120$ if calculated by the established formulae. Since $248 =(2^3)\times31$. Then order of group $U(248)$ is $$ \varphi \Big(\left(2^3\right)\times31\Big) = \varphi \left(2^3\right) \times \varphi (31) = 2^{3-1} \times (31-1) = 4 \times 30 = 120 $$