Prove that φ(n) + d(n) ≤ n + 1

Question

Prove that φ(n) + d(n) ≤ n + 1.

d(n) is the number of positive divisors of n.

φ(n) is the Euler's Totient Function.

Attempt：

For a prime number n, φ(n) = n - 1 (all numbers less than n are relatively prime to n except for 1) and d(n) = 2 (1 and n), so φ(n) + d(n) = n - 1 + 2 = n + 1, which satisfies the inequality.
For a composite number n, φ(n) is less than $n - \sqrt{n}$ and d(n) is less than $2\sqrt{n}$.

Hint. Every divisor $d>1$ of $n$ is not relatively prime to $n$. — Kangyeon Moon, Oct 12 '23 at 13:18
FYI, I found the duplicate using an Approach0 search. Also, as indicated in Wikipedia's Divisor function's Definitions section, an alternate notation for $d(n)$ is $\tau(n)$. Finally, welcome to Math SE. — John Omielan, Oct 12 '23 at 13:29

score 2 · Accepted Answer · answered Oct 12 '23 at 13:21

2

We have

$$n = \sum_{d\mid n} \varphi(d)\,,$$

so

$$\varphi(n) + d(n) = n - \sum_{\substack{d\mid n \\ d < n}} \varphi(d) + \sum_{d \mid n} 1 = (n+1) - \sum_{\substack{d \mid n \\ 1 < d < n}}\bigl(\varphi(d) - 1\bigr)\,.$$

This is at most $n+1$.

answered Oct 12 '23 at 13:21

Dietrich Burde

140,055

for future reader: https://math.stackexchange.com/questions/715247/summation-involving-totient-function-sum-d-mid-n-varphid-n – comp.course.master Oct 13 '23 at 00:42

Prove that φ(n) + d(n) ≤ n + 1

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