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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?

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    $\begingroup$ The book is wrong. wolframalpha.com/input/?i=phi%28248%29 $\endgroup$
    – lisyarus
    Jan 22 '15 at 6:44
  • $\begingroup$ Is the general formula for $ \phi (n) $ given in my question correct? $\endgroup$
    – Foggy
    Jan 22 '15 at 6:59
  • $\begingroup$ Yes, it is. You can find it here with some other formulas and their proofs. $\endgroup$
    – lisyarus
    Jan 22 '15 at 7:03
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    $\begingroup$ The even numbers are not invertible $\pmod{248}$ meaning that $\phi(248) \leq 124$. The book is clearly wrong. $\endgroup$
    – N. S.
    Oct 11 '20 at 4:01
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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

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

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