Is it true that $\varphi(n)$ is generally larger than $\lambda(n)$ for the same $n$? If so, can anyone give me a proof?

  • 1
    $\begingroup$ Is $n=p\cdot q$ with $p,q$ primes? $\endgroup$
    – Ievgeni
    Nov 14, 2021 at 12:01

1 Answer 1


If $n = p \cdot q$ and $p,q \in \mathbb P$ then $\varphi(n) = (p - 1)(q - 1)$ and $\lambda(n) = \operatorname{lcm}(p - 1, q - 1)$. The former multiplies $p - 1$ and $q - 1$, whereas the latter finds the least common multiple of the two. Naturally, $\forall a,b:\operatorname{lcm}(a,b) \le a\cdot b$, so we can say that $\forall n:\lambda(n) \le \varphi(n)$. In the case that the LCM for $a,b$ is $a$ and $b$ themselves, then they will be equal. Otherwise, the LCM will always be smaller. As $p$ and $q$ are always odd, we have $2\mid\gcd(p - 1, q - 1)$, thus $\lambda(n) < \varphi(n)$.

So yes, it is true.


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