IOSR Journal of Mathematics
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: Fermatâ€™s Last Theorem, Number Theory, Mathematics, Diophantine equation, Pythagorean triples
I.Introduction
Fermatâ€™s last theorem states that â€œxn+yn=zn has no nonzero integer solution for x,y,z when n>2â€. Pierre de Fermat wrote â€œI have discovered a truly wonderful proof which this margin is too small to containâ€. He almost certainly wrote the marginal note around 1630, when he first studied Diaphantus â€œArithmeticaâ€. 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 â€“ â€œa simple statement widely understood but a proof which defeats greatest intellectsâ€. In 1955 Yutaka Thaniyama asked some questions about elliptic curves. Further work by Weil & Shimura produced a conjucture now known as
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 =
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 

Hence (dn/en) 

Here 

i.e, n < 

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

when 1 â‰¤ 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 

Since a â‰¡ 
â‰¡ 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
Here if e>1 , LHS is an integer while RHS is not. It is impossible.
If e=1,
Thus in this case an+bn â‰ cn. 

Case 3 :Let a+b = ke. 

Replacing b by 
= (dn/en) 
nan1( 
â€¦. â€¦.. (1) 
Hence 

Here if 

Hence 

When 

Since in this case k â‰¢ 0(Mod n) , 

Thus in this case d â‰¡ 1(Mod n) and e â‰¡1(Mod n). 

Also when 

âˆ´ a + b = 1 is a special case of this. 

But as a+b, 
When
Also if
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
zn xn = yn = yn (z  x)n /(z  x)n=(d2 / e2)n(z  x)n, Where d2,e2 are
From these equations, x + y = k1e1m1, z â€“ x = k2e2m2, z â€“ y = k3e3m3. Now from cases 1,2 and 3, it can be seen that m1 = m2 = m3 =1 and
Here x + y = k1e1, z â€“ x = k2e2, z â€“ y = k3e3 and z = d1k1, y = d2k2, x = d3k3.
If x, y, z are not a multiple of n,
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
Consider the case
=> (x + y) â€“ (z â€“ x + z  y) = k2k3 ((nC1)
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 â€“ k2 , 



=> 2 h1n = k2[((nC1) 

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 â‰ zn. 

Here from case 3 it can be seen that ei â‰¡ (Mod n), di â‰¡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,
Now as x < y < z, 
z â€“ y < z â€“ 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 â€“ d1k1 = d2k2  k2e2 = d3k3 



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 


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



e1 = (g1 +h1)n, 2 = 

Now 



=> d2k2 â€“ k2e2 = d3k3 â€“ k2e3 


=> 
k2h2 = k3h3. 


In this case x = k3n + nk2h2, y = k2n + nk2h2, z = ((k1n)/n) â€“ 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) â€“ 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].
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