On Cohn′s conjecture concerning the Diophantine equation (an-1)(bn-1)=x2

DONG Zhongmin，LI Xiaoxue

(1.School of Information Engineering,Xi′an University,Xi′an 710065,China;2.School of Mathematics,Northwest University,Xi′an 710127,China)



On Cohn′s conjecture concerning the Diophantine equation (an-1)(bn-1)=x2

DONG Zhongmin1，LI Xiaoxue2

(1.School of Information Engineering,Xi′an University,Xi′an 710065,China;2.School of Mathematics,Northwest University,Xi′an 710127,China)

Abstract:Let a, b be two distinct positive integers. Aimed at the Cohn′s conjecture that the equation (an-1)(bn-1)=x2 has no positive integer solutions (x,n) with n>4. Using the method of elementary number theory and the properties of exponential Diophantine equation,it is proved that if the parity of a and b are opposite, then the equation has no positive integer solutions (x,n) satisfying n>4 and 2|n.

Key words:exponential Diophantine equation; Cohn′s conjecture; elementary number theory method

1Introduction and main result

Let N be the set of all positive integers. Leta,bbe two distinct positive integers. In recent years, there are many papers investigated the solutions (x,n) of the equation[1-7]

(an-1)(bn-1)=x2,x,n∈N.

(1)

Otherrelevantcontentcanbefoundinthereferences[8-15].In2002,J.H.E.Cohn[3]conjecturedthat(1)hasnosolutions(x,n)withn>4.Thisproblemisnotresolvedasyet.

Inthispaper,usingthemethodofelementarynumbertheoryandthepropertiesofexponentialDiophantineequation,thefollowingresultcanbeproved:

Theorem 1If the parity ofaandbare opposite, then (1) has no solutions (x,n) satisfyingn>4 and 2|n.

2Preliminaries

To complete the proof of Theorem 1, we need several lemmas as follows.

Lemma 1Letdbe a positive integer which is not a square.Then the equation

u2-dv2=1,u,v∈N

(2)

(3)

Then(u,v)=(uk,vk) (k=1,2,…)areallsolutionsof(2).

ProofSee Theorems 10.9.1 and 10.9.2 of reference[8].

Lemma 2Let (u,v)=(ur,vr) and (us,vs) be two solutions of (2), whererandsare positive integers. If 2|urand 2⫮us,then 2⫮rand 2|s.

ProofWhen 2|r, by (3), we have

(4)

(5)

and hence,uris odd. It implies that if 2|ur, then

2|r.

(6)

By (3) and (6), we have

(7)

(8)

Therefore, we see from (7) and (8) that if 2|ur, then

2|u1.

(9)

When 2⫮s, by (3) and (9), we haveu1|usand 2|us. It implies that if 2⫮us, then

2|s.

(10)

Thus, by (6) and (10), the lemma is proved.

Lemma 3Ifa

ProofSee Result 2 of reference[3].

Lemma 4The equation

X2+1=Ym,X,Y,m∈N,m>1

(11)

hasnosolutions(X,Y,m).

ProofSee reference[9].

Lemma 5pis an odd prime，X,Y∈N,X>1,theequation

Xp+1=2Y2

hasonlythesolution(p,X,Y)=(3,23,78).

ProofSee Proposition 8.1 of reference[10].

3Proof of the theorem

Letaandbbe two positive integers which parity are opposite. By the symmetry ofaandbin (1), we may assume that

2|a,2⫮b.

(12)

We now assume that (x,n) is a solution of (1) satisfyingn>4 and 2|n. By Lemma 3, we have 4⫮n. It implies thatn/2 is an odd integer withn/2≥3, andn/2 has an odd prime divisor. So

n=2pl,pis an odd prime,l∈N, 2⫮l.

(13)

By(1),wehave

an-1=dy2,bn-1=dz2,x=dyz,d,y,z∈N.

(14)

(u,v)=(an/2,y),(bn/2,z).

(15)

ApplyingLemma1to(15),wehave

(an/2,y)=(ur,vr),(bn/2,z)=(us,vs),r,s∈N.

(16)

Further,usingLemma2,wefindfrom(12)and(16)thatrandssatisfy(6)and(10)respectively.

By(3), (5)and(10),wehave

(17)

Substitute(17)into(16),by(13),weget

(18)

Sinceb>1by(1),applyingLemma5to(18),weobtain

p=3, l=1, n=6, b=23, us/2=78.

(19)

d=6 083,u1=78, v1=1.

(20)

By(6), (7), (16), (19)and(20),weget

(21)

Since78=2×3×13,weseefrom(21)that78|aanda=78c,wherecisapositiveinteger.Therefore,by(21),weget

(22)

But,since2⫮rby(6), (22)isimpossible.Thus,iftheparityofaandbareopposite,then(1)hasnosolutions(x,n)satisfyingn>4and2⫮n.Thetheoremisproved.

References:

[1]SZALAY L.On the Diophantine equation (2n-1)(3n-1)=x2[J].Publicationes Mathematicae Debrecen,2000,57(1/2):1-9.

[2]HAJDU L,SZALAY L.On the Diophantine equations (2n-1)(6n-1)=x2and (an-1)(akn-1)=x2[J].Periodica Mathematica Hungarica,2000,40(2):141-145.

[3]COHN J H E.The Diophantine equation (an-1)(bn-1)=x2[J].Periodica Mathematica Hungarica,2002,44(2):169-175.

[4]LUCA F,WALSH P G.The product of like-indexed terms in binary recurrences[J].Journal of Number Theory,2002,96(1):152-173.

[5]LE M H.A note on the exponential Diophantine equation (an-1)(bn-1)=x2[J].Publicationes Mathematica Debrecen,2009,74(3/4):401-403.

[6]LI L,SZALAY L.On the exponential Diophantine equation (an-1)(bn-1)=x2[J].Publicationes Mathematica Debrecen,2010,77(3/4):465-470.

[7]GUO X Y.A note on the Diophantine equation (an-1)(bn-1)=x2[J].Periodica Mathematica Hungarica,2013,66(1):87-93.

[8]HUA L K.An introduction to number theory[M].Beijing:Science Press,1979.

[9]LEBESGUE V A.Sur I′impossibilité en nombers entiers de I′équationxm=y2+1[J].Nouvelles Annales de Mathematiques,1850,9(1):178-181.

[10]BENNETT M A,SKINNER C M.Ternary Diophantine equation via Galois representations and modular forms[J].Canadian Journal of Mathematics,2004,56(1):23-54.

[11]GE J.On the exponential Diophantine equation (an-1)(bn-1)=x2[J].Journal of Xi′an Shiyou University:Natural Science Edition,2012,27(3):206-107.

[12]HE G R.A note on the Diophantine equation (am-1)(bn-1)=x2[J].Pure and Applied Mathematics,2011,27(5)：581-585.

[13]LI Ling,LI Xiaoxue.The Diophantine equationpx+qy=z2(in Chinese)[J].Basic Sciences Journal of Textile Universities,2015,28(1):42-44.

[14]TANT Min.A note on the exponiential Diophantine equation (am-1)(bn-1)=x2[J].Journal of Mathematical Research & Exposition,2011,31(6):1064-1066.

[15]WU Lei,LI Zhaojun.On the Diophantine equation (an-1)(bn-1)=x2(in Chinese)[J].Pure Mathematics,2011(1):172-176.

编辑、校对：师琅

DOI：10.13338/j.issn.1006-8341.2016.02.003

Received date：2015-10-08

Foundation item:This work is supported by the Foundation of Shaanxi Educational Committee(14JK2124)

Corresponding author:DONG Zhongmin(1943—),male,native of Yancheng,Jiangsu province,associate professor of Xi′an University,research area is number theory.E-mail:lxx20072012@163.com

CLC number:O 156.7

Document code:A

关于Diophantine方程(an-1)(bn-1)=x2的Cohn猜想

董忠民1，李小雪2

(1.西安文理学院 信息工程学院,陕西 西安 710065；2.西北大学 数学系，陕西 西安 710127)

摘要:设a和b是两个不相等的正整数.针对Cohn 猜想，即方程(an-1)(bn-1)=x2没有正整数解(x,n),其中n>4.利用初等数论方法和指数Diophantine方程的性质,得到了如果a和b具有相反的奇偶性,那么方程没有满足n>4和2|n的正整数解(x,n).

关键词:指数Diophantine方程;Cohn猜想;初等数论方法

Article ID:1006-8341(2016)02-0148-04

Citation format：DONG Zhongmin，LI Xiaoxue.On Cohn′s conjecture concerning the Diophantine equation (an-1)(bn-1)=x2[J].纺织高校基础科学学报,2016,29(2)：148-151.

DONG Zhongmin，LI Xiaoxue.On Cohn′s conjecture concerning the Diophantine equation (an-1)(bn-1)=x2[J].Basic Sciences Journal of Textile Universities,2016,29(2):148-151.