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IOSR Journal of Mathematics (IOSR-JM)

e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 7, Issue 4 (Jul. - Aug. 2013), PP 01-03 www.iosrjournals.org

Finding Numbers Satisfying The Condition An+Bn=Cn.

T. Unnikrishnan

Academic Consultant, Department of Statistics, Kerala Veterinary and Animal Sciences University

Abstract: Here an attempt is made to find positive numbers a,b and c such that an+bn=cn. Two cases were considered. an+bn is not divisible by (a+b)2 and an+bn is divisible by (a+b)2. From this a solution for the case

n=3 is obtained as (9+ 5)3 + (9- 5)3 = 123. A condition for finding such numbers for any n is also reached. Key words: Fermat’s Last Theorem, Number Theory, Mathematics, Diophantine equation, Pythagorean triples

I.Introduction

Fermat’s last theorem states that “xn+yn=zn has no nonzero integer solution for x,y,z when n>2”. Pierre de Fermat wrote “I have discovered a truly wonderful proof which this margin is too small to contain”. He almost certainly wrote the marginal note around 1630, when he first studied Diaphantus “Arithmetica”. Unfortunately no proof has been found and his theorem remained unsolved as one of the most baffling unresolved problems of mathematics for over three and a half centuries. This theorem has a long history of inducing mathematicians to develop more and more theories to solve it. Volumes and volumes of higher mathematics have been created in an effort to prove or disprove the assertion. It has the classic ingredients of a problem to catch the imagination of a wide public – “a simple statement widely understood but a proof which defeats greatest intellects”. In 1955 Yutaka Thaniyama asked some questions about elliptic curves. Further work by Weil & Shimura produced a conjucture now known as Shimura-Thaniyama-Weil conjucture. Further work by other mathematicians showed that a counter example to Fermat’s last theorem would provide a counter example to Shimura-Thaniyama-Weil conjucture. During 21-23, June 1993 Andrew wiles, a British mathematician working at Princeton in U.S.A. gave a series of three lectures at Isaac Newton Institute for Mathematical Sciences in Cambridge. On Wednesday 23/06/93 at around 10:30 A.M. Wiles announced his proof of Fermat’s last theorem as a corollary to his main results. This however is not the end of the story. On 4 December 1993 Andrew wiles made a statement in view of his speculation & withdrew his claims to have a proof. On 6 October Wiles sent the new proof to three colleagues including Faltings. Taylor lectured at the British Mathematical Colloquium in Edinburgh in April 1995 and the recently accepted proof of Wiles contains over 100 pages. No proof of the complexity of this can easily be guaranteed to be correct.

II.Results

To find numbers satisfying xn+yn=zn , it need only to consider odd prime n only. Consider the case in which x,y,z are prime to each other and n as an odd prime. If an+bn=cn, where a,b,c are integers such that a+b is positive, then c can be considered as below

 

c (a+b)

= d( a+b), where d and e are prime to each other.

 

a+b

 

e

Surely a+b should be a multiple of e

Let a+b = kem, where k and e are positive integers such that k is not divisible by e. an+bn = (a+b)(an-1-an-2b+an-3b2-……….-abn-2 + bn-1 ) ,if n>2 is odd.

Now consider the two cases an+bn is divisible by (a+b)2 or not.

 

Case 1 : If an+bn is not divisible by (a+b)2

 

Here an-1-an-2b+an-3b2-……….-abn-2 + bn-1 will not be divisible by a+b.

 

Hence (dn/en) (kem)n-1 will not be divisible by kem.

 

Here mn-m-n should be less than m.

 

i.e, n < 2m/(m-1) which is ≤ 4, if m ≥ 2.

 

Since the case n=3 had proved early for positive integers a and b , an+bn ≠ cn, if m ≥ 2.

 

when 1 ≤ m < 2 , a+b = k1e1+(r/s) = ke , where k = k1e(r/s)

 

The case 3, a+b =ke will be considered in general later.

 

Case 2 : If an+bn is divisible by (a+b)2

 

In this case an-1-an-2b+an-3b2-……….-abn-2 + bn-1 will be divisible by (a+b).

 

Since a ≡ -b(mod(a+b)),b ≡ -a(mod(a+b)) & an-1-an-2b+an-3b2-...….-abn-2 + bn-1

≡ 0(mod(a+b)),

nan-1 ≡ 0(mod(a+b)) and nbn-1 ≡ 0(mod(a+b)).

 

But as a and b are prime to each other a+b should be either 1 or n.

 

case 2:(a), a+b = 1 can be proved as corollary to the case 3 which will be discussed later.

 

 

 

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Finding Numbers Satisfying The Condition An+Bn=Cn.

Case 2:(b), a+b = n.

Here replace b by n-a in an-1-an-2b+an-3b2-……….-abn-2 +bn-1 = (dn/en) nn-1 .(nC1)an-1 – (nC2) an-2n+ (nC3 )an-3n2-……..- (nCn-1)ann-2+nn-1= (dn/en) nn-1.

Here if e>1 , LHS is an integer while RHS is not. It is impossible.

If e=1, an-1≡ 0(mod n), if n>2, which will contradict the assumption that a and b are co-prime pairs as a+b = n.

Thus in this case an+bn ≠ cn.

 

Case 3 :Let a+b = ke.

 

Replacing b by ke-a in an-1-an-2b+an-3b2-……….-abn-2 +bn-1

= (dn/en) (ke)n-1 ==(dn/e) kn-1, the equation becomes,

nan-1-( nC2)an-2 ke + …… -(nCn-1) a(ke)n-2 + (ke)n-1 = dnkn-1 /e

…. ….. (1)

Hence nan-1 - dnkn-1 /e ≡ 0(Mod (ke))

 

Here if kn-1 /e is an integer larger than 1, any divisor of it will divide nan-1 also and the divisor will be n.

Hence kn-1/e is either 1 or n.

 

When kn-1 /e =1, From (1) we see that dn ≡ 1(Mod n)

 

Since in this case k ≢ 0(Mod n) , kn-1 ≡ 1(Mod n), i.e, e ≡ 1(Mod n).

Thus in this case d ≡ 1(Mod n) and e ≡1(Mod n).

 

Also when kn-1 /e =1, a + b = ke = kn

 

∴ a + b = 1 is a special case of this.

 

But as a+b, c-a and c-b cannot be 1 simultaneously, an+bn ≠ cn , in this case

When kn-1 /e = n, from(1) it can be sen that, dn ≡1(Mod n). Now as kn-1 /e = n, k ≡ 0(Mod n) & so e ≡ 0(Mod n). Thus in this case d ≡1(Modn) and e ≡ 0(Mod n).

Also if kn-1 /e = n, c is a multiple of n and hence a & b will not be a multiple of n.

Now consider the case x, y, z N and xn + yn = zn

Let x < y < z and x, y, z are prime to each other.

We have xn + yn = zn =zn (x + y)n/(x+y)n=(d1 / e1)n(x + y)n, Where d1,e1are co-prime pairs.

zn- xn = yn = yn (z - x)n /(z - x)n=(d2 / e2)n(z - x)n, Where d2,e2 are co-prime pairs. zn - yn = xn = xn (z - y)n/ (z - y)n = (d3 / e3)n(z-y)n, Where d3,e3are co-prime pairs.

From these equations, x + y = k1e1m1, z – x = k2e2m2, z – y = k3e3m3. Now from cases 1,2 and 3, it can be seen that m1 = m2 = m3 =1 and

k1n-1 / e1 =1or n, k2n-1 / e2 =1 or n, k3n-1/ e3 =1 or n.

Here x + y = k1e1, z – x = k2e2, z – y = k3e3 and z = d1k1, y = d2k2, x = d3k3.

If x, y, z are not a multiple of n, k1n-1 / e1 = k2n-1 / e2 = k3n-1 / e3 =1.

If one and only one of x, y, z is a multiple of n, with out loss of generality assume that z is a multiple of n. Hence k1n-1 / e1 = n & k2n-1 / e2 = k3n-1 / e3 =1.

Consider the case k1n-1 / e1 = k2n-1 / e2 = k3n-1 / e3 = 1. Here if k1 = k2 + k3 , k1n = (k2 + k3)n => k1n – (k2n + k3n) = (nC1) k2n-1 k3 + ……. + ((nCn-1) k2k3n-1

=> (x + y) – (z – x + z - y) = k2k3 ((nC1) k2n-2 + …… + ((nCn-1)) k3n-2) => 2k1h1n = 2k2h2n = 2k3h3n = k2k3 ((nC1)k2n-2 + ……. + ((nCn-1))k3n-2)

Here LHS is even. So RHS should be even. But both k2 or k3 cannot be even. For this both d2k2 = y and d3k3 = x will be even, which is impossible.

If one of them is even and the other is odd then cancelling the even ki

by equating the corresponding LHS

equation to RHS, a contradiction is arrived.

 

If both are odd, then k1=k2 + k3 will be even, then as k3 = k1 – k2 ,

 

k1n-(k2n+k3n) = ((nC1)k1n-1k2 – ……….-((nCn-1) k1k2n-2.

 

=> 2 h1n = k2[((nC1) k1n-2 -……………-((nCn-1)k2n-2].

 

Here as k1 is even, a similar contradiction is arrived.

 

Thus in this case if x,y,z are not a multiple of n, xn+yn ≠ zn.

 

Here from case 3 it can be seen that ei ≡ (Mod n), di ≡1(Mod n), i =1, 2, 3.

 

 

 

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Finding Numbers Satisfying The Condition An+Bn=Cn.

Let d1=g1n+1, d2=g2n+1, d3=g3n+1, e1=(g1 + h1)n+1, e2=(g2 - h2)n+1, e3=(g3-h3)n+1, where g1, g2, g3, h1, h2, h3 are all positive integers.

Now as x < y < z,

z – y < z – x < x + y

i.e, k3n < k2n < k1n.

Hence

k3 < k2< k1

 

 

But (x + y) - z = y - (z - x)

= x - (z - y).

 

 

i.e, k1e1 – d1k1 = d2k2 - k2e2 = d3k3 -k3e3.

 

 

i.e, k1h1 = k2h2 = k3h3.

 

Thus as k3 < k2 < k1 , h1< h2< h3.

 

Now x = k3n + nk1h1, y = k2n + nk1h1, z = k1n - nk1h1

 

Consider the case k1n-1/e1 = n, k2n-1/e2 = k3n-1/e3 = 1.

 

From case 3 we have d1 = g1n+1, d2 = g2n+1, d3 = g3n+1,

 

 

e1 = (g1 +h1)n, 2 = (g2-h2)n+1, e3 = (g3-h3)n+1, where gi and hi are positive integers.

Now

y-(z-x) = x-(z-y)

 

 

=> d2k2 – k2e2 = d3k3 – k2e3

 

=>

k2h2 = k3h3.

 

 

In this case x = k3n + nk2h2, y = k2n + nk2h2, z = ((k1n)/n) – nk2h2

III.Conclusion

The positive numbers a,b and c satisfying an+bn=cn. will be of the form

1. x = k3n + nk1h1, y = k2n + nk1h1, z = k1n - nk1h1 , if none of x,y,z is a multiple of n. (This can also be written in the simple form

x= (k1n - k2n + k3n)/2, y= (k1n +k2n - k3n)/2, z = (k1n +k2n + k3n)/2 )

2.x = k3n + nk2h2, y = k2n + nk2h2, z = ((k1n)/n) – nk2h2, if one and only one of x,y,z is a multiple of n

Acknowledgement

I am grateful to H.E. Dr.A.P.J.Abdul Kalam, Former President of India, Dr. N. Balakrishnan, Associate Director, IISc, Bangalore, Dr. R. Balasubrahmanian, Director, IMSc, Chennai, Dr. T. Thrivikraman Namboothiri, Former Head of the Department of Mathematics, Cochin University, Dr.V.K.Krishnan Namboothiri, Former Head of the Department of Mathematics, St. Thomas College, Thrissur and Prof. T.S.Balasubrahmanian, Former Head of the Department of Mathematics, Sreekrishna College, Guruvayur for their valuable comments and suggestions on my proof to find numbers a,b and c such that an+bn=cn .

References

[1].Stewart, I. and Tall, D. Algebraic Number Theory . (Chapman & Hall, 1994)

[2].www-gap.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html. Accessed on 02-08-2004. [online].

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