The first thousand values of φ(n). The points on the top line represent φ(p) when p is a prime number, which is p − 1.^[1]

In number theory, Euler's totient function counts the positive integers up to a given integer n that are relatively prime to n. It is written using the Greek letter phi as φ(n) or ϕ(n), and may also be called Euler's phi function. It can be defined more formally as the number of integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n, k) is equal to 1.^[2][3] The integers k of this form are sometimes referred to as totatives of n.

For example, the totatives of n = 9 are the six numbers 1, 2, 4, 5, 7 and 8. They are all relatively prime to 9, but the other three numbers in this range, 3, 6, and 9 are not, because gcd(9, 3) = gcd(9, 6) = 3 and gcd(9, 9) = 9. Therefore, φ(9) = 6. As another example, φ(1) = 1 since for n = 1 the only integer in the range from 1 to n is 1 itself, and gcd(1, 1) = 1.

Euler's totient function is a multiplicative function, meaning that if two numbers m and n are relatively prime, then φ(mn) = φ(m)φ(n).^[4][5]
This function gives the order of the multiplicative group of integers modulo n (the group of units of the ring ℤ/nℤ).^[6] It also plays a key role in the definition of the RSA encryption system.

History, terminology, and notation[edit]

Leonhard Euler introduced the function in 1763.^[7][8][9] However, he did not at that time choose any specific symbol to denote it. In a 1784 publication, Euler studied the function further, choosing the Greek letter π to denote it: he wrote πD for "the multitude of numbers less than D, and which have no common divisor with it".^[10] This definition varies from the current definition for the totient function at D = 1 but is otherwise the same. The now-standard notation^[8][11]φ(A) comes from Gauss's 1801 treatise Disquisitiones Arithmeticae,^[12] although Gauss didn't use parentheses around the argument and wrote φA. Thus, it is often called Euler's phi function or simply the phi function.

In 1879, J. J. Sylvester coined the term totient for this function,^[13][14] so it is also referred to as Euler's totient function, the Euler totient, or Euler's totient. Jordan's totient is a generalization of Euler's.

The cototient of n is defined as n − φ(n). It counts the number of positive integers less than or equal to n that have at least one prime factor in common with n.

Computing Euler's totient function[edit]

There are several formulas for computing φ(n).

Euler's product formula[edit]

It states

φ(n)=n∏p∣n(1−1p),{displaystyle varphi (n)=nprod _{pmid n}left(1-{frac {1}{p}}right),}

where the product is over the distinct prime numbers dividing n. (The notation is described in the article Arithmetical function.)

The proof of Euler's product formula depends on two important facts.

The function is multiplicative[edit]

This means that if gcd(m, n) = 1, then φ(mn) = φ(m) φ(n). (Outline of proof: let A, B, C be the sets of nonnegative integers, which are, respectively, coprime to and less than m, n, and mn; then there is a bijection between A × B and C, by the Chinese remainder theorem.)

Value for a prime power argument[edit]

If p is prime and k ≥ 1, then

φ(pk)=pk−1(p−1)=pk(1−1p).{displaystyle varphi (p^{k})=p^{k-1}(p-1)=p^{k}left(1-{frac {1}{p}}right).}

Proof: since p is a prime number the only possible values of gcd(p^k, m) are 1, p, p², ..., p^k, and the only way for gcd(p^k, m) to not equal 1 is for m to be a multiple of p. The multiples of p that are less than or equal to p^k are p, 2p, 3p, ..., p^{k − 1}p = p^k, and there are p^{k − 1} of them. Therefore, the other p^k − p^{k − 1} numbers are all relatively prime to p^k.

Proof of Euler's product formula[edit]

The fundamental theorem of arithmetic states that if n > 1 there is a unique expression for n,

n=p1k1⋯prkr,{displaystyle n={p_{1}}^{k_{1}}cdots {p_{r}}^{k_{r}},}

where p₁ < p₂ < ... < p_r are prime numbers and each k_i ≥ 1. (The case n = 1 corresponds to the empty product.)

Repeatedly using the multiplicative property of φ and the formula for φ(p^k) gives

φ(n)=φ(p1k1)φ(p2k2)⋯φ(prkr)=p1k1(1−1p1)p2k2(1−1p2)⋯prkr(1−1pr)=p1k1p2k2⋯prkr(1−1p1)(1−1p2)⋯(1−1pr)=n(1−1p1)(1−1p2)⋯(1−1pr).{displaystyle {begin{aligned}varphi (n)&=varphi left(p_{1}^{k_{1}}right)varphi left(p_{2}^{k_{2}}right)cdots varphi left(p_{r}^{k_{r}}right)\&=p_{1}^{k_{1}}left(1-{frac {1}{p_{1}}}right)p_{2}^{k_{2}}left(1-{frac {1}{p_{2}}}right)cdots p_{r}^{k_{r}}left(1-{frac {1}{p_{r}}}right)\&=p_{1}^{k_{1}}p_{2}^{k_{2}}cdots p_{r}^{k_{r}}left(1-{frac {1}{p_{1}}}right)left(1-{frac {1}{p_{2}}}right)cdots left(1-{frac {1}{p_{r}}}right)\&=nleft(1-{frac {1}{p_{1}}}right)left(1-{frac {1}{p_{2}}}right)cdots left(1-{frac {1}{p_{r}}}right).end{aligned}}}

This is Euler's product formula.

Example[edit]

φ(36)=φ(2232)=36(1−12)(1−13)=36⋅12⋅23=12.{displaystyle varphi (36)=varphi left(2^{2}3^{2}right)=36left(1-{tfrac {1}{2}}right)left(1-{tfrac {1}{3}}right)=36cdot {tfrac {1}{2}}cdot {tfrac {2}{3}}=12.}

In words, this says that the distinct prime factors of 36 are 2 and 3; half of the thirty-six integers from 1 to 36 are divisible by 2, leaving eighteen; a third of those are divisible by 3, leaving twelve numbers that are coprime to 36. And indeed there are twelve positive integers that are coprime with 36 and lower than 36: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, and 35.

Fourier transform[edit]

The totient is the discrete Fourier transform of the gcd, evaluated at 1.^[15] Let

F{x}[m]=∑k=1nxk⋅e−2πimkn{displaystyle {mathcal {F}}{mathbf {x} }[m]=sum limits _{k=1}^{n}x_{k}cdot e^{{-2pi i}{frac {mk}{n}}}}

where x_k = gcd(k,n) for k ∈ {1, …, n}. Then

φ(n)=F{x}[1]=∑k=1ngcd(k,n)e−2πikn.{displaystyle varphi (n)={mathcal {F}}{mathbf {x} }[1]=sum limits _{k=1}^{n}gcd(k,n)e^{-2pi i{frac {k}{n}}}.}

The real part of this formula is

φ(n)=∑k=1ngcd(k,n)cos⁡2πkn.{displaystyle varphi (n)=sum limits _{k=1}^{n}gcd(k,n)cos {2pi {frac {k}{n}}}.}

Note that unlike the other two formulae (the Euler product and the divisor sum) this one does not require knowing the factors of n. However, it does involve the calculation of the greatest common divisor of n and every positive integer less than n, which suffices to provide the factorization anyway.

Divisor sum[edit]

The property established by Gauss,^[16] that

∑d∣nφ(d)=n,{displaystyle sum _{dmid n}varphi (d)=n,}

where the sum is over all positive divisors d of n, can be proven in several ways. (see Arithmetical function for notational conventions.)

One way is to note that φ(d) is also equal to the number of possible generators of the cyclic group C_d; specifically, if C_d = ⟨g⟩, then g^k is a generator for every k coprime to d. Since every element of C_n generates a cyclic subgroup, and all subgroups of C_d ≤ C_n are generated by some element of C_n, the formula follows.^[17] In the article Root of unity Euler's formula is derived by using this argument in the special case of the multiplicative group of the nth roots of unity.

This formula can also be derived in a more concrete manner.^[18] Let n = 20 and consider the fractions between 0 and 1 with denominator 20:

120,220,320,420,520,620,720,820,920,1020,1120,1220,1320,1420,1520,1620,1720,1820,1920,2020{displaystyle {tfrac {1}{20}},,{tfrac {2}{20}},,{tfrac {3}{20}},,{tfrac {4}{20}},,{tfrac {5}{20}},,{tfrac {6}{20}},,{tfrac {7}{20}},,{tfrac {8}{20}},,{tfrac {9}{20}},,{tfrac {10}{20}},,{tfrac {11}{20}},,{tfrac {12}{20}},,{tfrac {13}{20}},,{tfrac {14}{20}},,{tfrac {15}{20}},,{tfrac {16}{20}},,{tfrac {17}{20}},,{tfrac {18}{20}},,{tfrac {19}{20}},,{tfrac {20}{20}}}

Put them into lowest terms:

120,110,320,15,14,310,720,25,920,12,1120,35,1320,710,34,45,1720,910,1920,11{displaystyle {tfrac {1}{20}},,{tfrac {1}{10}},,{tfrac {3}{20}},,{tfrac {1}{5}},,{tfrac {1}{4}},,{tfrac {3}{10}},,{tfrac {7}{20}},,{tfrac {2}{5}},,{tfrac {9}{20}},,{tfrac {1}{2}},,{tfrac {11}{20}},,{tfrac {3}{5}},,{tfrac {13}{20}},,{tfrac {7}{10}},,{tfrac {3}{4}},,{tfrac {4}{5}},,{tfrac {17}{20}},,{tfrac {9}{10}},,{tfrac {19}{20}},,{tfrac {1}{1}}}

First note that all the divisors of 20 are denominators. And second, note that there are 20 fractions. Which fractions have 20 as denominator? The ones whose numerators are relatively prime to 20 (1/20, 3/20, 7/20, 9/20, 11/20, 13/20, 17/20, 19/20). By definition this is φ(20) fractions. Similarly, there are φ(10) = 4 fractions with denominator 10 (1/10, 3/10, 7/10, 9/10), φ(5) = 4 fractions with denominator 5 (1/5, 2/5, 3/5, 4/5), and so on.

In detail, we are considering the fractions of the form k/n where k is an integer from 1 to n inclusive. Upon reducing these to lowest terms, each fraction will have as its denominator some divisor of n. We can group the fractions together by denominator, and we must show that for a given divisor d of n, the number of such fractions with denominator d is φ(d).

Note that to reduce k/n to lowest terms, we divide the numerator and denominator by gcd(k, n). The reduced fractions with denominator d are therefore precisely the ones originally of the form k/n in which gcd(k, n) = n/d. The question therefore becomes: how many k are there less than or equal to n which verify gcd(k, n) = n/d? Any such k must clearly be a multiple of n/d, but it must also be coprime to d (if it had any common divisor s with d, then sn/d would be a larger common divisor of n and k). Conversely, any multiple k of n/d which is coprime to d will satisfy gcd(k, n) = n/d. We can generate φ(d) such numbers by taking the numbers less than d coprime to d and multiplying each one by n/d (these products will of course each be smaller than n, as required). This in fact generates all such numbers, as if k is a multiple of n/d coprime to d (and less than n), then k/​ⁿ⁄_d will still be coprime to d, and must also be smaller than d, else k would be larger than n. Thus there are precisely φ(d) values of k less than or equal to n such that gcd(k, n) = n/d, which was to be demonstrated.

Möbius inversion gives

φ(n)=∑d∣nμ(d)⋅nd=n∑d∣nμ(d)d,{displaystyle varphi (n)=sum _{dmid n}mu left(dright)cdot {frac {n}{d}}=nsum _{dmid n}{frac {mu (d)}{d}},}

where μ is the Möbius function.

This formula may also be derived from the product formula by multiplying out

∏p∣n(1−1p){displaystyle prod _{pmid n}left(1-{frac {1}{p}}right)}

to get

∑d∣nμ(d)d.{displaystyle sum _{dmid n}{frac {mu (d)}{d}}.}

Some values of the function[edit]

The first 143 values (sequence A000010 in the OEIS) are shown in the table and graph below:^[19]

Graph of the first 100 values

φ(n) for 1 ≤ n ≤ 143

+	0	1	2	3	4	5	6	7	8	9	10	11
0	N/A	1	1	2	2	4	2	6	4	6	4	10
12	4	12	6	8	8	16	6	18	8	12	10	22
24	8	20	12	18	12	28	8	30	16	20	16	24
36	12	36	18	24	16	40	12	42	20	24	22	46
48	16	42	20	32	24	52	18	40	24	36	28	58
60	16	60	30	36	32	48	20	66	32	44	24	70
72	24	72	36	40	36	60	24	78	32	54	40	82
84	24	64	42	56	40	88	24	72	44	60	46	72
96	32	96	42	60	40	100	32	102	48	48	52	106
108	36	108	40	72	48	112	36	88	56	72	58	96
120	32	110	60	80	60	100	36	126	64	84	48	130
132	40	108	66	72	64	136	44	138	48	92	70	120

The top line in the graph, y = n − 1, is a true upper bound. It is attained whenever n is prime. There is no lower bound that is a straight line of positive slope; no matter how gentle the slope of a line is, there will eventually be points of the plot below the line. More precisely, the lower limit of the graph is proportional to n/log log n rather than being linear.^[20]

Euler's theorem[edit]

This states that if a and n are relatively prime then

aφ(n)≡1modn.{displaystyle a^{varphi (n)}equiv 1mod n.}

The special case where n is prime is known as Fermat's little theorem.

This follows from Lagrange's theorem and the fact that φ(n) is the order of the multiplicative group of integers modulo n.

The RSA cryptosystem is based on this theorem: it implies that the inverse of the function a ↦ a^e mod n, where e is the (public) encryption exponent, is the function b ↦ b^d mod n, where d, the (private) decryption exponent, is the multiplicative inverse of e modulo φ(n). The difficulty of computing φ(n) without knowing the factorization of n is thus the difficulty of computing d: this is known as the RSA problem which can be solved by factoring n. The owner of the private key knows the factorization, since an RSA private key is constructed by choosing n as the product of two (randomly chosen) large primes p and q. Only n is publicly disclosed, and given the difficulty to factor large numbers we have the guarantee that no-one else knows the factorization.

Other formulae[edit]

• a∣b⟹φ(a)∣φ(b){displaystyle amid bimplies varphi (a)mid varphi (b)}


• n∣φ(an−1)for a,n>1{displaystyle nmid varphi (a^{n}-1)quad {text{for }}a,n>1}


• φ(mn)=φ(m)φ(n)⋅dφ(d)where d=gcd⁡(m,n){displaystyle varphi (mn)=varphi (m)varphi (n)cdot {frac {d}{varphi (d)}}quad {text{where }}d=operatorname {gcd} (m,n)}

Note the special cases

• φ(2m)={2φ(m) if m is evenφ(m) if m is odd{displaystyle varphi (2m)={begin{cases}2varphi (m)&{text{ if }}m{text{ is even}}\varphi (m)&{text{ if }}m{text{ is odd}}end{cases}}}


• φ(nm)=nm−1φ(n){displaystyle varphi left(n^{m}right)=n^{m-1}varphi (n)}

• φ(lcm⁡(m,n))⋅φ(gcd⁡(m,n))=φ(m)⋅φ(n){displaystyle varphi (operatorname {lcm} (m,n))cdot varphi (operatorname {gcd} (m,n))=varphi (m)cdot varphi (n)}

Compare this to the formula

• lcm⁡(m,n)⋅gcd⁡(m,n)=m⋅n{displaystyle operatorname {lcm} (m,n)cdot operatorname {gcd} (m,n)=mcdot n}

(See least common multiple.)

• 
  φ(n) is even for n ≥ 3. Moreover, if n has r distinct odd prime factors, 2^r | φ(n)
  


• For any a > 1 and n > 6 such that 4 ∤ n there exists an l ≥ 2n such that l | φ(aⁿ − 1).


• φ(n)n=φ(rad⁡(n))rad⁡(n){displaystyle {frac {varphi (n)}{n}}={frac {varphi (operatorname {rad} (n))}{operatorname {rad} (n)}}}

where rad(n) is the radical of n.

• 
  ∑d∣nμ2(d)φ(d)=nφ(n){displaystyle sum _{dmid n}{frac {mu ^{2}(d)}{varphi (d)}}={frac {n}{varphi (n)}}} ^[21]
  


• ∑1≤k≤n(k,n)=1k=12nφ(n)for n>1{displaystyle sum _{1leq kleq n atop (k,n)=1}!!k={tfrac {1}{2}}nvarphi (n)quad {text{for }}n>1}


• 
  ∑k=1nφ(k)=12(1+∑k=1nμ(k)⌊nk⌋2)=3π2n2+O(n(log⁡n)23(log⁡log⁡n)43){displaystyle sum _{k=1}^{n}varphi (k)={tfrac {1}{2}}left(1+sum _{k=1}^{n}mu (k)leftlfloor {frac {n}{k}}rightrfloor ^{2}right)={frac {3}{pi ^{2}}}n^{2}+Oleft(n(log n)^{frac {2}{3}}(log log n)^{frac {4}{3}}right)} (^[22] cited in^[23])

• 
  ∑k=1nφ(k)k=∑k=1nμ(k)k⌊nk⌋=6π2n+O((log⁡n)23(log⁡log⁡n)43){displaystyle sum _{k=1}^{n}{frac {varphi (k)}{k}}=sum _{k=1}^{n}{frac {mu (k)}{k}}leftlfloor {frac {n}{k}}rightrfloor ={frac {6}{pi ^{2}}}n+Oleft((log n)^{frac {2}{3}}(log log n)^{frac {4}{3}}right)} ^[22]
  


• 
  ∑k=1nkφ(k)=315ζ(3)2π4n−log⁡n2+O((log⁡n)23){displaystyle sum _{k=1}^{n}{frac {k}{varphi (k)}}={frac {315,zeta (3)}{2pi ^{4}}}n-{frac {log n}{2}}+Oleft((log n)^{frac {2}{3}}right)} ^[24]
  


• 
  ∑k=1n1φ(k)=315ζ(3)2π4(log⁡n+γ−∑p primelog⁡pp2−p+1)+O((log⁡n)23n){displaystyle sum _{k=1}^{n}{frac {1}{varphi (k)}}={frac {315,zeta (3)}{2pi ^{4}}}left(log n+gamma -sum _{p{text{ prime}}}{frac {log p}{p^{2}-p+1}}right)+Oleft({frac {(log n)^{frac {2}{3}}}{n}}right)} ^[24]
  

(where γ is the Euler–Mascheroni constant).

• ∑gcd⁡(k,m)=11≤k≤n1=nφ(m)m+O(2ω(m)){displaystyle sum _{stackrel {1leq kleq n}{operatorname {gcd} (k,m)=1}}!!!!1=n{frac {varphi (m)}{m}}+Oleft(2^{omega (m)}right)}

where m > 1 is a positive integer and ω(m) is the number of distinct prime factors of m.^[25]

Menon's identity[edit]

In 1965 P. Kesava Menon proved

∑gcd(k,n)=11≤k≤ngcd(k−1,n)=φ(n)d(n),{displaystyle sum _{stackrel {1leq kleq n}{gcd(k,n)=1}}!!!!gcd(k-1,n)=varphi (n)d(n),}

where d(n) = σ₀(n) is the number of divisors of n.

Formulae involving the golden ratio[edit]

Schneider^[26] found a pair of identities connecting the totient function, the golden ratio and the Möbius function μ(n). In this section φ(n) is the totient function, and ϕ = 1 + √5/2 = 1.618… is the golden ratio.

They are:

ϕ=−∑k=1∞φ(k)klog⁡(1−1ϕk){displaystyle phi =-sum _{k=1}^{infty }{frac {varphi (k)}{k}}log left(1-{frac {1}{phi ^{k}}}right)}

and

1ϕ=−∑k=1∞μ(k)klog⁡(1−1ϕk).{displaystyle {frac {1}{phi }}=-sum _{k=1}^{infty }{frac {mu (k)}{k}}log left(1-{frac {1}{phi ^{k}}}right).}

Subtracting them gives

∑k=1∞μ(k)−φ(k)klog⁡(1−1ϕk)=1.{displaystyle sum _{k=1}^{infty }{frac {mu (k)-varphi (k)}{k}}log left(1-{frac {1}{phi ^{k}}}right)=1.}

Applying the exponential function to both sides of the preceding identity yields an infinite product formula for e:

e=∏k=1∞(1−1ϕk)μ(k)−φ(k)k.{displaystyle e=prod _{k=1}^{infty }left(1-{frac {1}{phi ^{k}}}right)^{frac {mu (k)-varphi (k)}{k}}.}

The proof is based on the two formulae

∑k=1∞φ(k)k(−log⁡(1−xk))=x1−xand∑k=1∞μ(k)k(−log⁡(1−xk))=x,for 0<x<1.{displaystyle {begin{aligned}sum _{k=1}^{infty }{frac {varphi (k)}{k}}left(-log left(1-x^{k}right)right)&={frac {x}{1-x}}\{text{and}};sum _{k=1}^{infty }{frac {mu (k)}{k}}left(-log left(1-x^{k}right)right)&=x,qquad quad {text{for }}0<x<1.end{aligned}}}

Generating functions[edit]

The Dirichlet series for φ(n) may be written in terms of the Riemann zeta function as:^[27]

∑n=1∞φ(n)ns=ζ(s−1)ζ(s).{displaystyle sum _{n=1}^{infty }{frac {varphi (n)}{n^{s}}}={frac {zeta (s-1)}{zeta (s)}}.}

The Lambert series generating function is^[28]

∑n=1∞φ(n)qn1−qn=q(1−q)2{displaystyle sum _{n=1}^{infty }{frac {varphi (n)q^{n}}{1-q^{n}}}={frac {q}{(1-q)^{2}}}}

which converges for |q| < 1.

Both of these are proved by elementary series manipulations and the formulae for φ(n).

Growth rate[edit]

In the words of Hardy & Wright, the order of φ(n) is “always ‘nearly n’.”^[29]

First^[30]

limsupφ(n)n=1,{displaystyle lim sup {frac {varphi (n)}{n}}=1,}

but as n goes to infinity,^[31] for all δ > 0

φ(n)n1−δ→∞.{displaystyle {frac {varphi (n)}{n^{1-delta }}}rightarrow infty .}

These two formulae can be proved by using little more than the formulae for φ(n) and the divisor sum function σ(n).

In fact, during the proof of the second formula, the inequality

6π2<φ(n)σ(n)n2<1,{displaystyle {frac {6}{pi ^{2}}}<{frac {varphi (n)sigma (n)}{n^{2}}}<1,}

true for n > 1, is proved.

We also have^[20]

liminfφ(n)nlog⁡log⁡n=e−γ.{displaystyle lim inf {frac {varphi (n)}{n}}log log n=e^{-gamma }.}

Here γ is Euler's constant, γ = 0.577215665..., so e^γ = 1.7810724... and e^−γ = 0.56145948....

Proving this does not quite require the prime number theorem.^[32][33] Since log log (n) goes to infinity, this formula shows that

liminfφ(n)n=0.{displaystyle lim inf {frac {varphi (n)}{n}}=0.}

In fact, more is true.^[34][35][36]

φ(n)>neγlog⁡log⁡n+3log⁡log⁡nfor n>2{displaystyle varphi (n)>{frac {n}{e^{gamma };log log n+{frac {3}{log log n}}}}quad {text{for }}n>2}

and

φ(n)<neγlog⁡log⁡nfor infinitely many n.{displaystyle varphi (n)<{frac {n}{e^{gamma }log log n}}quad {text{for infinitely many }}n.}

The second inequality was shown by Jean-Louis Nicolas. Ribenboim says "The method of proof is interesting, in that the inequality is shown first under the assumption that the Riemann hypothesis is true, secondly under the contrary assumption."^[36]

For the average order, we have^[22][37]

φ(1)+φ(2)+⋯+φ(n)=3n2π2+O(n(log⁡n)23(log⁡log⁡n)43)as n→∞,{displaystyle varphi (1)+varphi (2)+cdots +varphi (n)={frac {3n^{2}}{pi ^{2}}}+Oleft(n(log n)^{frac {2}{3}}(log log n)^{frac {4}{3}}right)quad {text{as }}nrightarrow infty ,}

due to Arnold Walfisz, its proof exploiting estimates on exponential sums due to I. M. Vinogradov and N. M. Korobov (this is currently the best known estimate of this type). The "Big O" stands for a quantity that is bounded by a constant times the function of n inside the parentheses (which is small compared to n²).

This result can be used to prove^[38] that the probability of two randomly chosen numbers being relatively prime is 6/π².

Ratio of consecutive values[edit]

In 1950 Somayajulu proved^[39][40]

liminfφ(n+1)φ(n)=0andlimsupφ(n+1)φ(n)=∞.{displaystyle {begin{aligned}lim inf {frac {varphi (n+1)}{varphi (n)}}&=0quad {text{and}}\[5px]lim sup {frac {varphi (n+1)}{varphi (n)}}&=infty .end{aligned}}}

In 1954 Schinzel and Sierpiński strengthened this, proving^[39][40] that the set

{φ(n+1)φ(n),n=1,2,⋯}{displaystyle left{{frac {varphi (n+1)}{varphi (n)}},;;n=1,2,cdots right}}

is dense in the positive real numbers. They also proved^[39] that the set

{φ(n)n,n=1,2,⋯}{displaystyle left{{frac {varphi (n)}{n}},;;n=1,2,cdots right}}

is dense in the interval (0,1).

Totient numbers[edit]

A totient number is a value of Euler's totient function: that is, an m for which there is at least one n for which φ(n) = m. The valency or multiplicity of a totient number m is the number of solutions to this equation.^[41] A nontotient is a natural number which is not a totient number. Every odd integer exceeding 1 is trivially a nontotient. There are also infinitely many even nontotients,^[42] and indeed every positive integer has a multiple which is an even nontotient.^[43]

The number of totient numbers up to a given limit x is

xlog⁡xe(C+o(1))(log⁡log⁡log⁡x)2{displaystyle {frac {x}{log x}}e^{{big (}C+o(1){big )}(log log log x)^{2}}}

for a constant C = 0.8178146....^[44]

If counted accordingly to multiplicity, the number of totient numbers up to a given limit x is

|{n:ϕ(n)≤x}|=ζ(2)ζ(3)ζ(6)⋅x+R(x){displaystyle {Big vert }{n:phi (n)leq x}{Big vert }={frac {zeta (2)zeta (3)}{zeta (6)}}cdot x+R(x)}

where the error term R is of order at most x/(log x)^k for any positive k.^[45]

It is known that the multiplicity of m exceeds m^δ infinitely often for any δ < 0.55655.^[46][47]

Ford's theorem[edit]

Ford (1999) proved that for every integer k ≥ 2 there is a totient number m of multiplicity k: that is, for which the equation φ(n) = m has exactly k solutions; this result had previously been conjectured by Wacław Sierpiński,^[48] and it had been obtained as a consequence of Schinzel's hypothesis H.^[44] Indeed, each multiplicity that occurs, does so infinitely often.^[44][47]

However, no number m is known with multiplicity k = 1. Carmichael's totient function conjecture is the statement that there is no such m.^[49]

Perfect totient numbers[edit]

Applications[edit]

Cyclotomy[edit]

In the last section of the Disquisitiones^[50][51] Gauss proves^[52] that a regular n-gon can be constructed with straightedge and compass if φ(n) is a power of 2. If n is a power of an odd prime number the formula for the totient says its totient can be a power of two only if n is a first power and n − 1 is a power of 2. The primes that are one more than a power of 2 are called Fermat primes, and only five are known: 3, 5, 17, 257, and 65537. Fermat and Gauss knew of these. Nobody has been able to prove whether there are any more.

Thus, a regular n-gon has a straightedge-and-compass construction if n is a product of distinct Fermat primes and any power of 2. The first few such n are^[53]

2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40,... (sequence A003401 in the OEIS).

The RSA cryptosystem[edit]

Setting up an RSA system involves choosing large prime numbers p and q, computing n = pq and k = φ(n), and finding two numbers e and d such that ed ≡ 1 (mod k). The numbers n and e (the "encryption key") are released to the public, and d (the "decryption key") is kept private.

A message, represented by an integer m, where 0 < m < n, is encrypted by computing S = m^e (mod n).

It is decrypted by computing t = S^d (mod n). Euler's Theorem can be used to show that if 0 < t < n, then t = m.

The security of an RSA system would be compromised if the number n could be factored or if φ(n) could be computed without factoring n.

Unsolved problems[edit]

Lehmer's conjecture[edit]

If p is prime, then φ(p) = p − 1. In 1932 D. H. Lehmer asked if there are any composite numbers n such that φ(n) | n − 1. None are known.^[54]

In 1933 he proved that if any such n exists, it must be odd, square-free, and divisible by at least seven primes (i.e. ω(n) ≥ 7). In 1980 Cohen and Hagis proved that n > 10²⁰ and that ω(n) ≥ 14.^[55] Further, Hagis showed that if 3 divides n then n > 10^1937042 and ω(n) ≥ 298848.^[56][57]

Carmichael's conjecture[edit]

This states that there is no number n with the property that for all other numbers m, m ≠ n, φ(m) ≠ φ(n). See Ford's theorem above.

As stated in the main article, if there is a single counterexample to this conjecture, there must be infinitely many counterexamples, and the smallest one has at least ten billion digits in base 10.^[41]

See also[edit]

• Carmichael function


• Duffin–Schaeffer conjecture


• Generalizations of Fermat's little theorem


• Highly composite number


• Multiplicative group of integers modulo n
  


• Ramanujan sum

Notes[edit]

References[edit]

External links[edit]

• 
  Hazewinkel, Michiel, ed. (2001) [1994], "Totient function", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
  


• Euler's Phi Function and the Chinese Remainder Theorem — proof that φ(n) is multiplicative


• Euler's totient function calculator in JavaScript — up to 20 digits


• Dineva, Rosica, The Euler Totient, the Möbius, and the Divisor Functions
  


• Plytage, Loomis, Polhill Summing Up The Euler Phi Function