For semiprimes, computing the Euler totient function is equivalent to factoring. Indeed, if n = pq for distinct primes p and q, then φ(n) = (p-1)(q-1) = pq - (p+q) + 1 = (n+1) - (p+q). Therefore, if you can compute φ(n), then you can compute p+q. However, it's then easy to solve for p and q because you know their sum and product (it's just a quadratic equation).

If you believe factoring is hard for semi-primes, then so is computing the Euler totient function.

Update! Factoring and computing the Euler totient function are known to be equivalent for arbitrary numbers, not just semiprimes. One reference is "Riemann's hypothesis and tests for primality" by Gary L. Miller. There, the equivalence is deterministic, but assumes a version of the Riemann hypothesis. See also section 10.4 of "A computational introduction to number theory and algebra" by Victor Shoup for a proof of probabilistic equivalence.

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For semiprimes, computing the Euler totient function is equivalent to factoring. Indeed, if n = pq for distinct primes p and q, then φ(n) = (p-1)(q-1) = pq - (p+q) + 1 = (n+1) - (p+q). Therefore, if you can compute φ(n), then you can compute p+q. However, it's then easy to solve for p and q because you know their sum and product (it's just a quadratic equation).

If you believe factoring is hard for semi-primes, then so is computing the Euler totient function.