Page 155 - ASBIRES-2017_Preceedings

SOLUTION TO SPECIAL DIOPHANTINE EQUATIONS USING THE REMAINDER THEOREM

ast theorem for = 3, First,consider the the Fermat’s last Theorem for any odd prime
equation. where 2 + 1 prime [3] and 2 + 1 is not a
prime.
3
3
− 2. 3 ℎ = 0 (1)
− ℎ − 3 3−1 3
2.1 New Proof of Remainder Theorem
Which is related to the Farmat’s last
theorem for = 3, where , ℎ, and 3 are co- Remainder theorem is useful to evaluate
prime integers when ≥ 3 and also, the polynomials ()which can be defined as
in the following.

− 2. 3 ℎ = 0 (2)

− ℎ − 3 −1 3

Where , ℎ, and p are co-prime () = + −1 + −2
0
2
1
integers also studied for integer solutions. In + −3 + ⋯ + −1 1
3
addition we consider the general equation +

− 2 ℎ = 0 (3)

− ℎ − −1

Which is related to Farmat’s last Now, if ( − ) is factor
theorem for odd prime [ 3 ]. Where 2 + 1 is of (), then () = 0. In studying above
also a prime Diophantine equations = substitution may
And the equation be problematic since numbers involved in
− ℎ − − 2 ℎ = 0 (4) equations are co-prime. Therefore the

following proof of the remainder theorem
In case of (4) above Diophantine
equations, the Remainder theorem is to be leads to the calculation without substituting
= .
applied where 2 + 1 is not a prime. In this

research, using the Remainder Theorem it () = + −1 + −2
1
0
2
will be shown that the above equations + −3 + ⋯ + −1 1
3
cannot have solutions satisfying the given +

restrictions. The proof of remainder theorem
is briefly discussed in the following. () = (( − ) + )
0
Let () is a polynomial of and if − + (( − ) + ) −1 + 2
1
, where is a constant, and let R be the + (( − ) + ) −3 + ⋯
3
1
remainder when () is divided by − . + −1 (( − ) + ) +

Then we can write
Using Binomial Expansion, We can
() = ( − )() + (5) write
It is obvious that () = . In obtaining () = ( − )() + (6)
we have substituted = in the equation
(5). This is true for all constant . But in Therefore, the Remainder of () when
case of Diophantine equations, we have to it is divided by ( − ) is given by
take into account that the constant numbers = () = + −1 + ⋯ +

1
also divided into groups using the fact co- 0 + (2.3) −1
prime or not as we discuss in the under next
topic. In this proof, we have shown that the
Remainder R can be obtained without
2 METHODOLOGY AND RESULT substituting = in the equation. If we
could not substitute = for remainder
Under this topic,first we prove the
remainder theorem using the binomial theorem this method will helps to solve the
expansion. After proving the Remainder equation. It is the advantage of this proof.
Theorem,It was applyed into special
Diophantine Equations which are consider 2.2 Application of Remainder theorem to
for this reseaech and show that they lead to special Diophantine Equations
apparent contradictions. The equations, we Consider the Diophantine equation
consider in this section, directly related to
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