Page 156 - ASBIRES-2017_Preceedings

Nimasha & Piyadasa

= + , (, ) = 1 (7) following equation (12) from the equation
3
3
3
(11).
If the equation is satisfied by non-trivial

− 2. ℎ
co-prime integers(, , ), we can obtain the − ℎ − −1 3
= 0 (12)
parametric equatio in the following.
− 2. 3 ℎ = 0 ( 8)
3
3
− ℎ − 3 3−1 3 where , ℎ, , are all co-prime
numbers and , ℎ are factors of ,

Where , h are factors of resp- respectively, and , are factors of .
is a factor of . It is
ectively and 3 3−1 3 3 −1
clear that It is clear that divides − ℎ and
therefore = ℎ + −1 t where and are co-
3
3
− ℎ ≡ 0(3 ) The following result prime. Now it is clear that ℎ + −1 t is a

holds on − ℎ . factor of −1

+ ℎ . By the Remainder

If − ℎ ≡ 0( ) − ℎ ≡ 0( −1 ) Theorem, we have
= 0 (13)
−

2 −3

( − ℎ ) = ( − ℎ)[ −1 + ℎ −2 + ℎ −1 −
+ ⋯ . . +ℎ −1 ]
( − ℎ ) = ( − ℎ)[ −1 − ℎ −1 + Which contradicts the condition (, ) =

2
ℎ( −2 − ℎ −2 ) + ℎ ( −3 − ℎ −3 + ⋯ . . +ℎ −1 ] 1. Hence, the equation (12) does not hold.
Now, consider the equation
Therefore ( − ℎ) ≡0(mod −1 ) and the term
− ℎ − − 2 ℎ = 0 (14)

2
[ −1 − ℎ −1 + ℎ( −2 − ℎ −2 ) + ℎ ( −3 Which is also related to Fermat’s last
− ℎ −3 ) + ⋯ . . +ℎ −1 ] Theorem and d is a factor of + − It is
≡ 0() (9) clear that

Now, it is clear that ℎ + = (ℎ + ) + (15)
− 2. 3 ℎ = 0
2
2
( − ℎ)[ + ℎ + ℎ ] − 3 3−1 3 Where(, ) = 1 since divides
2
− 2. 3 ℎ
( − ℎ)[( − ℎ) + 3ℎ] − 3 3−1 3 ,where c is non zero integer. Now, we

= 0 obtain − (ℎ + ) ≡ ( )

Since − ℎ ≡ 0(3 −1 ) let = ℎ + where ≥ 2.
3 −1 where is not divisible 3. Therefore − (ℎ + ) ≡ ( −1 )
3

− 2. 3 ℎ = 3 3−1 3 3 and let = ℎ + + −1 where is a non
+ ℎ
≡ 0((3 −1 + ℎ)) zero integer and we get
In other words ℎ + 3 −1 where (t, 3) = ℎ + = (ℎ + + −1 )[ −1 − 2 ℎ)

+ ℎ . By the
1, is a factor of 3 3−1 3 3
Remainder Theorem, we have Which follows from (14). Therefore, we
get from the Remainder Theorem,
= 0 (10)
3 3−1 3 3−3 3
− 3

And this leads to the contradiction that − ( + −1 ) = 0 (16)

(3, ) ≠ 1 . Therefore the equation (7)
does not hold. Which leads to the contradiction
that = 0. Therefore the equation (14)
Now, suppose that integers
, , satisfy the equation does not hold. In our study, we are able to
conjecture that the Diophantine equation

= + (11) = + 1, > 2 has only one rational

solution = 1 and = 0 .
Where any odd prime and 2 + 1 is
also a prime[3]. Then one derives the
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