An analytical proof for Lehmer's totient conjecture using Mertens' theorems
aa r X i v : . [ m a t h . G M ] A ug An analytical proof for Lehmer’s totient conjectureusing Mertens’ theorems
Ahmad Sabihi
Teaching professor and researcher at some universities of Iran
Abstract
We make an analytical proof for Lehmer’s totient conjecture. Lehmer con-jectured that there is no solution for the congruence equation n − ≡ mod φ ( n )) with composite integers, n , where φ ( n ) denotes Euler’s totientfunction. He also showed that if the equation has any composite solutions, n must be odd, square-free, and divisible by at least 7 primes. Several peo-ple have obtained conditions on values , n , and number of square-free primesconstructing n if the equation can have composite solutions. Using Mertens’theorems, we show that it is impossible that the equation can have any com-posite solution and implies that the conjecture should be true for all thepositively composite numbers. Keywords:
Lehmer’s totient conjecture; Mertens’ theorems; Euler’s totientfunction
MSC 2010 :11P32;11N05
Email address: [email protected] (Ahmad Sabihi)
Preprint submitted to journal January 25, 2018 . Introduction
Lehmer’s totient conjecture was stated by D.H. Lehmer in 1932 [1]. Lehmerconjectured that there are no composite solutions, n , for the equation n − ≡ mod φ ( n )) . We know that this conjecture is true for every prime num-bers. He also proved that if any such , n , exists, it must be odd, square-free,and divisible by at least seven primes [1]. Pinch calls such an n a Lehmernumber and defines the
Lehmer index of n to be the ratio n − φ ( n ) [2]. As weshould know every exponent λ ( n ) of the multiplication group ( Z / N ) ∗ mustdivide n − n must be square-freewith at least three prime factors, and p − | n − p dividing n .Conversely, any such n must be a Carmichael number. Since the exponent λ ( n ) of the multiplicative group divides its order φ ( n ), a Lehmer numbermust be a Carmichael number. Lieuwens [3] showed that a Lehmer numberdivisible by 3 must have index at least 4 and hence must have at least 212prime factors and exceeds 5 . . Kishore [4] proved that a Lehmer numberof index at least 3 must have at least 33 prime factors and exceeds 2 . .Cohen and Hagis [5] showed that a Lehmer number divisible by 5 and ofindex 2 must have at least 13 prime factors and if we have any compositesolution , n , to the problem, then n > and number of prime factors mustbe greater than or equal 14. We firstly show that using Mertens’ theorems,we are able to asymptotically prove that the equation n − mφ ( n ) = 1 withodd composite number , n , and having k square-free prime factors cannothave any solutions. We also investigate about the equation n − mφ ( n ) = − n , must be odd, and square-free as Lehmer showedbefore, but by another method. To prove our theorems, we make use ofMertens’ theorems on the density of primes and re-prove some of them.
2. Theorems
Theorem 1: Mertens’ 2nd theorem [6]
Let p be a prime and x > every real number, then X p ≤ x p = log log x + a + O ( 1log x ) (2.1) where a possible value of ” a ” can be a = 0 . ... Theorem 2: Mertens’ 3rd theorem [6]
Let p be a prime and x > every real number, then Y p ≤ x (1 − p ) ∼ e − γ log x (2.2) where the notation f ( x ) ∼ g ( x ) means that limitation f ( x ) g ( x ) = 1 when x tendsto infinity. γ denotes Euler’s constant.2.3. Corollary 1:
Let p be a prime, x > every real number, and c > an absolute constant,then Y p ≤ x (1 − p ) > c log x (2.3) where ” c ” can be 0.3 for x ≥ and 0.09 for x ≥ in this paper. .4. Theorem 3:
Let p i to p k be all of the prime factors including only odd square-free primefactors of the odd number n and sufficiently so large integers or all of primefactors values tend to infinity versus the number of them, then the equation n − m Y p i ≤ p ≤ p k ( p −
1) = ± does not any solution. m denotes a positive integer.2.5. Theorem 4:
Let p to p k be all of the prime factors including only odd square-freeprime factors of the odd number n ,all of them be existed, and p k sufficientlyso large integer or tends to infinity, then the equation n − m Y p ≤ p k ( p −
1) = ± does not any solution. m denotes a positive integer.2.6. Theorem 5:
Let p i to p k be all of the prime factors including only odd square-free primefactors of the odd number n and p k sufficiently so large integer or tends toinfinity, then n − m Y p i ≤ p ≤ p k ( p −
1) = ± does not any solution. m denotes a positive integer. .7. Theorem 6:
Let p i to p k be all of the prime factors including only odd square-free primefactors of the odd number , n , and none of them be so large and unbounded(all of them be bounded), then the equation n − m Y p i ≤ p ≤ p k ( p −
1) = 1 (2.7) does not any solution, but the equation n − m Y p i ≤ p ≤ p k ( p −
1) = − may have solutions. m denotes a positive integer.
3. Proofs
Proof of Theorem 1
As is well-known, Mertens himself has proven this theorem but we giveanother method for making its proof. We really reprove (reformulate) theproof. The proof can be made by applying three times the Abel summationformula to the the series X p ≤ x p = X p ≤ x log pp . p (3.1)Firstly, we apply it to the series P p ≤ x log p and reach to θ ( x ) = x + o ( x log x ).Secondly, P p ≤ x log pp . Let P p ≤ x log p = θ ( x ) and φ ( x ) = x and substitudethem into the Abel summation formula as follows: X p ≤ x log pp = θ ( x ) φ ( x ) + Z x θ ( u ) u du (3.2)5hen, we have X p ≤ x log pp = 1 + log( x ) + o (log log x ) (3.3)Thirdly, we apply it to the entire series. Let A ( x ) = P p ≤ x log pp and Φ( x ) = x ) into the Abel summation formula X p ≤ x p = A ( x )Φ( x ) + Z x A ( u ) . Φ ′ ( u ) .du = { x + o (log log x ) } . ( 1log x )+ Z x { u + o (log log u ) } u (log u ) du = 1 + log x + o ( log log x log x ) + Z x duu (log u ) + Z x duu log u + Z x o ( log log uu (log u ) ) = 1 + 1log 2 − log log 2 + log log x + o ( log log x log x ) + d + o ( log log x log x + 1log x ) = 1 + 1log 2 − log log 2 + d + log log x + o ( 1log x ) (3.4)Where d denotes all unknown constant values created in (3.4). Since ac-cording to the properties of small ”o” and big ”O” notations, we have o ( x ) = O ( x ),then X p ≤ x p = log log x + a + O ( 1log x ) (3.5)where a = 1+ − log log 2+ d . Although, precisely calculating a is difficult,but our attempts to calculate the value a using directly processing data bysubstituting into (3.5) gave us an approximate value about 0.261497... Proof of Theorem 2
The proof can be found in the Mertens’ paper [6].6 .3.
Proof of Corollary 1
The proof can easily be made by appealing to the Riemann Zeta Functionand Euler’s product [7] as follows: ζ − ( s ) = Y p (1 − p s ) = ∞ X n =1 µ ( n ) n s (3.6)Putting s = 1 in (3.6), we have Y p (1 − p ) = ∞ X n =1 µ ( n ) n (3.7)and trivially checking gives us Y p ≤ x (1 − p ) > Y p (1 − p ) = ∞ X n =1 µ ( n ) n (3.8)Abel Summation Formula gives us again that assuming P n ≤ x µ ( n ) = o ( x )[7], we have ∞ X n =1 µ ( n ) n = e − γ log x + o (log( x )) (3.9)Combining (3.9) with theorem 2 and (3.8) we find Y p ≤ x (1 − p ) > Y p (1 − p ) = ∞ X n =1 µ ( n ) n = e − γ log x + o (log( x )) > c log( x ) (3.10)If we let c < e − γ , then inequality and the theorem is completed. We choose c = 0 . x = 3, we can choose c = 0 .
09 since the term e − γ (1 − ) =0 . ... . Therefore, we choose a new lower bound for c i.e. c = 0 . e − γ (1 − x ) ) > e − γ (1 − ) > .
09 for all the odd primes ≥
3. Also, Dusart [9] in 2010, stated the Theorem 6.12 giving a new boundfor all x ≥ x ≥ c = 0 . .4. Proof of Theorem 3
If we divide both of sides of the equation (2.4) by Q p i ≤ p ≤ p k ( p − Q p i ≤ p ≤ p k (1 − p ) − m = ± Q p i ≤ p ≤ p k ( p −
1) (3.11)Since n = p i ...p k is odd and p i to p k are also odd square-free prime factors of n ,then trivially all of them must be ≥ Q pi ≤ p ≤ pk ( p − < if thenumerator of right side be (+1) and − Q pi ≤ p ≤ pk ( p − > − if the numerator ofright side be (-1). On the other hand, the left side of (3.11) should be greaterthan zero for the plus sign and less than zero for the minus sign. Therefore,for plus sign we have 1 Q p i ≤ p ≤ p k (1 − p ) − < m < Q p i ≤ p ≤ p k (1 − p ) (3.12)and for minus sign 1 Q p i ≤ p ≤ p k (1 − p ) < m < Q p i ≤ p ≤ p k (1 − p ) + 18 (3.13)Since our assumption says us that all p i to p k tend to infinity versus thenumber of primes within the interval ( p i , p k ), the relations (3.12) and (3.13)change to 1 + ε − < m < ε (3.14)1 + ε < m < ε + 18 (3.15)when ε tends to zero. This is due to if we let M denotes the number of primesfrom p i to p k (note: there may not exist all of consecutive primes within theinterval ( p i , p k )) then we find(1 − p i ) M ≤ Y p i ≤ p ≤ p k (1 − p ) ≤ (1 − p k ) M (3.16)8ince according to our assumption, p i to p k are so large versus M , then allthe fractions Mp i to Mp k tend to zero andlim p i −→∞ (1 − p i ) M = lim p i −→∞ { (1 − p i ) p i } Mpi = lim p i −→∞ ( 1 e ) Mpi = 1 (3.17)and in the similar way lim p k −→∞ (1 − p k ) M = 1 (3.18)Then the inequality (3.16) gives uslim p i to p k −→∞ Y p i ≤ p ≤ p k (1 − p ) = 1 (3.19)This means that the integer number m can only be 1 when (3.14) holds andcannot be any integer number when (3.15) holds. If m = 1, it is impossibleto hold by appealing to Lehmer’s paper [1] since m = 1 if and only if n isprime. This completes the proof. Proof of Theorem 4
If we divide both of sides of (2.5) by n = p ...p k and substitute Q p (1 − p ) = e − γ log x + o (log( x )) from (3.8) and (3.9) into it, then let x = p k − m { e − γ log p k + o (log( x )) } = ± p ...p k (3.20)Since p k −→ ∞ then also x −→ ∞ and (3.20) changes to1 − me γ log p k = ± p ...p k (3.21)The right side tends to zero since p k −→ ∞ . This means that the left sideshould also tend to zero and m is of order e γ log x . Since m is a positiveinteger, it could be of the form m = [ e γ log p k ] = e γ log p k − α (3.22)9or plus sign since the left side of (3.21) should be closed to 0 + or m = [ e γ log p k ] + j = e γ log p k + j − α (3.23)for minus sign since the left side of (3.21) should be closed to 0 − where j ≥ α denotes the fractional part of e γ log x . Therefore, the relation (3.21) canbe changed into αe γ log p k = 1 p ...p k or j − αe γ log p k = 1 p ...p k (3.24)for when p k , x −→ ∞ . Since the denominator of the right side fraction of(3.24) is of the order more than p k and the denominator of the left rightfraction is of order log p k , α and j − α are also bounded, then these two sidescannot be equal for when p k is tending to infinity and the equation (3.24),(3.21), and finally (2.5) cannot have any solutions. Proof of Theorem 5
The proof of this theorem also likes to Theorem 4. Consider all primes p i to p k exist or missing some of them, then regarding Theorem 2 A ( p ) . Y p i ≤ p ≤ p k ,p k −→∞ (1 − p ) = e − γ log p k (3.25)where A ( p ) denotes a function of prime numbers before p k or some before p k and some between p i and p k for completing and converting Q p i ≤ p ≤ p k (1 − p )to Q p ≤ p k (1 − p ),which may be a constant value or variative one.Similarly to(3.21), we have 1 − mA ( p ) e γ log p k = ± p ...p k (3.26)10here A ( p ) = Y p ≤ p ( i − (1 − p ) or A ( p ) = Y p ≤ p ( i − (1 − p ) . Y p i ≤ p m ≤ p k (1 − p m ) (3.27)and p m denotes primes missing within the interval ( p i , p k ). Trivially, A ( p ) <
1. Similarly to the proof of Theorem 4, we find m = [ A ( p ) e γ log p k ] = A ( p ) e γ log p k − β (3.28)for plus sign since the left side of (3.26) should be closed to 0 + or m = [ A ( p ) e γ log p k ] + j = A ( p ) e γ log p k + j − β (3.29)since the left side should be closed to 0 − for when p k −→ ∞ . Therefore,(3.26) changes to βA ( p ) e γ log p k = 1 p ...p k (3.30)or j − βA ( p ) e γ log p k = 1 p ...p k (3.31)Where 0 ≤ β <
1. To being better closed to zero in relation (3.31), we shouldchoose j = 1. The arguments are similar to the arguments of Theorem 4 andthe proof is completed. Proof of Theorem 6
Regarding Corollary 1,we have c log p k < Y p ≤ p k (1 − p ) = Y p ≤ p ≤ p ( i − (1 − p ) . Y p i ≤ p m ≤ p k (1 − p m ) . Y p i ≤ p ≤ p k (1 − p )(3.32)11here p m denotes primes missing within the interval ( p i , p k ). Let A ( p ) = Q p ≤ p ≤ p ( i − (1 − p ) . Q p i ≤ p m ≤ p k (1 − p m ) and knowing A ( p ) < cA ( p ) log p k < Y p i ≤ p ≤ p k (1 − p ) (3.33)and multiplying the left side by a coefficient l k >
1, we find an equation cl k A ( p ) log p k = Y p i ≤ p ≤ p k (1 − p ) (3.34)Similarly to (3.26), we have1 − mcl k A ( p ) log p k = ± p i ...p k (3.35)As the Theorems 4 and 5 arguments, we have m = [ A ( p ) log p k cl k ] = A ( p ) log p k cl k − ψ (3.36)for plus sign since the left side of (3.35) should be closed to 0 + or m = [ A ( p ) log p k cl k ] + j = A ( p ) log p k cl k + j − ψ (3.37)for minus sign to be closed to 0 − (According to Lehmer’s, Cohen’s, Kishore’s,and Lieuwens’ arguments, the number of prime numbers to have a compositesolution should be more than 7,14,33, or 212. Thus,the value p i ...p k shouldbe certainly closed to zero due to being large the value p i ...p k ). 0 ≤ ψ < ψcl k A ( p ) log p k = 1 p i ...p k (3.38)and ( j − ψ ) cl k A ( p ) log p k = 1 p i ...p k (3.39)12s we know, log p k isn’t an integer number and since cl k > .
09 regardingCorollary 1 and A ( p ) tends to zero by increasing the number of primes andbeing larger p ( i − and p k , then the value A ( p ) cl k gets smaller and smaller andlog p k larger and larger, thus the fractional part of A ( p ) log p k cl k gets closer tothe number 1. This means that ψ gets closer to the number 1 to that of zero.Therefore, since ψcl k A ( p ) gets larger than 1,then (3.38) cannot have any solution,but may (3.39) have solution since ( j − ψ ) gets closer to zero with j = 1 and (1 − ψ ) cl k A ( p ) gets closer to zero as well. As Lehmer [1], Kishore [4], Cohen [5], andspecifically Lieuwens [3] showed that if the case n − m Q p i ≤ p ≤ p k ( p −
1) = 1has composite solution,then the number of prime factors should be at least7, 14, 33 or 212, therefore, we see that the order of magnitude of n must bevery large and our hypothesis can be more precise. Example:
Lehmer showed that n = 3 . . .
257 is a composite solution for the equa-tion n − m Q p i ≤ p ≤ p k ( p −
1) = −
1. We compute the values 1 − ψ , A ( p ), cl k ,and log p k assuming c = 0 .
09 and substitute them into (3.38) and (3.39) asfollows:Here, we have p i = 3, p i +1 = 5 , p i +5 = 17, p k = 257. For computing A ( p ),one should compute all of other missing primes as: A ( p ) = (1 −
17 )(1 −
111 )(1 −
113 )(1 −
119 )(1 −
123 ) ... (1 − − − − − − − − . Q p i ≤ p ≤ p k (1 − p ) = 0 . p k = log 257 =5 . l k = 12 . ψ = 0 . ψcl k A ( p ) log p k = 0 . j = 1 we find(1 − ψ ) cl k A ( p ) log p k = 1 . × − (3.42)If we compute the right side of each of two relations (3.38) and (3.39)1 p i ...p k = 13 × × ×
257 = 1 . × − (3.43)and compare to the corresponding left sides (the relations (3.41) and (3.42)),then we find that the equation (2.8) has a solution since the left and rightsides are very close to each other, but the equation (2.7) (same Lehmer’sconjecture) does not any solution since the left and right sides are far fromeach other. Another example can be made by other composite number n =3 . . . . p k = 65537 ≥ c = 0 . Lehmer’s totient conjecture
We discuss about Lehmer’s totient conjecture here. Firstly, we know if n is a prime number , p , then φ ( n ) = p − n − ≡ mod φ ( n )). Conversely, if we have n − ≡ mod φ ( n )),then let n = p t ...p t k k be prime factors decomposition of n . This means that φ ( n ) = p t ...p t k k Q p ≤ p ≤ p k (1 − p ). If t , ..., t k ≥
2, then we find that p , ..., p k | both φ ( n ) and n .On the other hand, regarding our assumption n − ≡ mod φ ( n )) and we14hould have p , ..., p k | ( n − p , ..., p k | gcd ( n, n −
1) = 1,which is impossible to occur. Hence, n can have neither square prime factorsnor can be an even number(this is a Lehmer’s theorem, which we prove ithere by other method). This means that t ...t k ≤
1, thus some of t , ..., t k must be 0 or 1 or all of them be 1. Also, all of prime factors must be oddnumbers. Certainly, if all of t , ..., t k be zero but one, then n = p and theproblem is solved. If the number of square-free prime factors are greaterthan or equal 2, then using Theorems 3 to 6 of this paper, we find out theequation n − m Y p i ≤ p ≤ p k ( p −
1) = 1 (3.44)does not any solution and Lehmer’s totient conjecture is proven.
Acknowledgment
This paper was submitted to a journal and quickly took some commentsfor its amendment. The author would like to thank anonymous referee forhis/her nice comments. Also, thanks for Mr. Alexander Zujev, researchscholar, University of California at Davis for his interesting comments on thepaper. All the needed comments have been taken into account to the paper.
References [1] D.H. Lehmer, On Euler’s totient function, Bull. Amer. Math.Soc.38(1932),745-751.[2] R. G. E. Pinch, A note on Lehmer’s totient problem,personal web site,(2006) p.1. 153] E. Lieuwens, Do there exist composite numbers for which kφ ( M ) = M − n for which φ ( n ) | ( n − n if φ ( n ) | ( n − Prime Numbers , Wiley-Interscience,a divisionof John Wiley and Sons.Inc.New York (1985).[8] J.B. Rosser, L.Schoenfeld,Approximate formulas for some functions ofprime numbers, Illinois J.Math.6