
# 5.2: A Necessary Digression - Gauss’s Theorem on Sums of Euler’s Function


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Later in this chapter we will need a fact first proved by Gauss about Euler’s $$\phi$$ function:

Theorem $$\PageIndex{1}$$

For all $$n\in\NN$$, $$\displaystyle\sum_{\substack{d\in\NN\\s.t.\ d\mid n}} \phi(d) = n\ .$$

We’ll give two proofs, which illustrate different features of this situation:

Proof 1

Fix $$n\in\NN$$.

Let’s define some subsets of $$\{1,\dots,n\}$$, dependent upon a choice of a positive divisor $$d\mid n$$, as follows $S_d=\left\{k\in\NN\mid 1\le k\le n\ \text{and}\ \gcd(k,n)=d\right\}\ .$ These sets are disjoint since for each $$k\in\NN$$ such that $$1\le k\le n$$, $$d=\gcd(k,n)$$ has a specific value and $$k$$ is only in that $$S_d$$.

For $$k\in S_d$$, $$\gcd(k,n)=d$$ or, by Theorem 1.5.1, $$\gcd(k/d,n/d)=1$$. Thus $$\#(S_d)$$ is the number of elements $$\ell\in\NN$$ in the range $$1\le\ell\le n/d$$ which are relatively prime to $$n/d$$, i.e., $$\#(S_d)=\phi(n/d)$$.

Every $$k\in\ZZ$$ in the range $$1\le k\le n$$ is in exactly one of these $$S_d$$, for $$d$$ a positive divisor of $$n$$. Therefore $\{1,\dots,n\} = \bigcup_{\substack{d\in\NN\\s.t.\ d\mid n}} S_d$ so $n=\#\left(\{1,\dots,n\}\right) = \#\left(\bigcup_{\substack{d\in\NN\\s.t.\ d\mid n}} S_d\right) = \sum_{\substack{d\in\NN\\s.t.\ d\mid n}} \#\left(S_d\right) = \sum_{\substack{d\in\NN\\s.t.\ d\mid n}} \phi(n/d)\ .$ But as $$d$$ runs over the positive divisors of $$n$$, so does $$n/d$$; in other words $\left\{d\mid d\in\NN,\ 1\le d\le n\text{\ and\ }d\mid n\right\}=\left\{n/d\mid d\in\NN,\ 1\le d\le n\text{\ and\ }d\mid n\right\}$

We can therefore rewrite the last big equation as $n=\sum_{\substack{d\in\NN\\s.t.\ d\mid n}} \phi(n/d)=\sum_{\substack{d\in\NN\\s.t.\ d\mid n}} \phi(d)$ which is what we were trying to prove.

Proof 2

This approach will concentrate on the function $F(n)=\sum_{\substack{d\in\NN\\s.t.\ d\mid n}} \phi(d)\ ,$ which has some very nice properties.

For example, suppose $$p$$ is a prime and $$k\in\NN$$. Then the divisors of $$p^k$$ are $$1,p,p^2,\dots,p^k$$, so $F(p^k)=\phi(1)+\phi(p)+\phi(p^2)+\dots+\phi(p^k) =1+(p-1)+(p^2-p)+\dots+(p^k-p^{k-1}) = p^k\ .$

Furthermore, if $$p$$ and $$q$$ are distinct primes, then the divisors of $$pq$$ are $$1$$, $$p$$, $$q$$, and $$pq$$. Also, $$\gcd(p,q)=1$$, so by Theorem 2.5.2 \begin{aligned} F(pq)&=\phi(1)+\phi(p)+\phi(q)+\phi(pq)\\ &=\phi(1)+\phi(p)+\phi(q)+\phi(p)\phi(q)\\ &=(1+\phi(p))(1+\phi(q))\\ &=F(p)F(q)\ ,\end{aligned} From this fact, one can prove (it’s an exercise below) that $$F(ab)=F(a)F(b)$$ whenever $$a,b\in\NN$$ are relatively prime. This means that $$F$$ has the same sort of multiplicative property for relatively prime factors as does $$\phi$$.

We are ready to finish the proof. So let $$n\in\NN$$ be any number such that $$n>1$$ (for $$n=1$$, the theorem is trivial). Suppose the prime decomposition of $$n$$ is given by $n=p_1^{k_1}\cdot\cdots\cdot p_r^{k_r}\ ,$ where the primes $$\{p_1,\dots,p_r\}$$ are distinct. Therefore $$\gcd(p_i^{k_i},p_j^{k_j})=1$$ if $$i\neq j$$ and so we can use the multiplicative property of $$F$$ and our calculation of $$F$$ for prime powers to conclude \begin{aligned} F(n)&=F(p_1^{k_1}\cdot\cdots\cdot p_r^{k_r})\\ &=F(p_1^{k_1})\cdot\cdots\cdot F(p_r^{k_r})\\ &=p_1^{k_1}\cdot\cdots\cdot p_r^{k_r}\\ &=n\ ,\end{aligned} which is what we wanted to prove.

Exercise $$\PageIndex{1}$$

1. Finish the second proof of Gauss’s Theorem 5.2.1 by proving that $$F(ab)=F(a)F(b)$$ for all $$a,b\in\NN$$ which are relatively prime.