2
$\begingroup$

I’ve been working through some number theory proofs and have encountered a problem that’s giving me trouble. The proof involves Euler’s phi function and basically re-writing it in an alternative form. So I know that Euler’s phi function can be written as:

$$ \phi(n) = n \prod_{p | n, \text{p is prime}} ( 1 - \frac{1}{p})$$

But let's says the prime factorization of n can be written as $p^{a_1}_1 \cdot p^{a_2}_2 \ldots \cdot p_{m-1} \cdot p^{a_m}_{m}$ where each is a distinct prime. How would you shouw that:

$$ \phi(n) = p^{a_1 - 1}_1 \cdot p^{a_2 - 1}_2 \ldots \cdot p^{a_m - 1}_{m} \cdot \prod^{m}_{i = 1} (p_i -1) $$

So far just tried tinkering a bit with the original definition, but have failed thus far.

$\endgroup$
  • 4
    $\begingroup$ $p^a(1-\frac1p) = p^{a-1}(p-1)$. $\endgroup$ – lhf Feb 20 '13 at 0:41
3
$\begingroup$

To prove it, start by showing that $\phi$ is multiplicative, i.e., that if $\gcd(m, n) = 1$ then $\phi(m n) = \phi(m) \phi(n)$. The proof of this goes through the chinese remainder theorem, the pairs of residues modulo $m$ and $n$ that are relatively prime to each are in a biyection with the residues modulo $m n$ that are relatively prime to the later.

Then to prove $\phi(p^k) = p^{k - 1} (p - 1)$ for $p$ prime, note that the numbers that are not relatively prime to $p^k$ are exactly the $p^{k - 1}$ multiples of $p$ from 1 to $p^k$, so $\phi(p^k) = p^k - p^{k - 1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.