JST Vol. 21 (1) Jan. 2013 - Pertanika Journal - Universiti Putra ...

S. Ismail and K. A. Mohd Atan
The following theorem gives the form of solutions to the following equation, when
= and gcd( abc , , ) > 1 .
x y
Theorem 1.2:
Suppose the triplet ( abc , , ) in which a= b and gcd( abc , , ) > 1 is a non-trivial solution to
3
4
gcd( abc , , ) > 1 . Then abc , , is of the form a= b= 4n
and c= 8n
for some integer, n .
Proof:
First, suppose that the triplet ( abc , , ) is a solution in which a= b.
Then, the equation will
become:
Clearly
4 3
2a = c
3
2|c which implies that c is even.
122 Pertanika J. Sci. & Technol. 21 (1): 283 - 298 (2013)
(1.1)
Hence, there exists q j in the prime power decomposition of such that q j = 2.
Rearranging the prime power decomposition of c , let q 1 = 2 , we will have the following:
f m f
1
j
c= 2 Õ q
j=
2 j where j
q are odd primes and 1
f j ³ for 1£ j£ m
Similarly, let p 1 = 2 in the prime power decomposition of a . Then, the following will be
obtained:
ei
e l
1 a 2 p
i=
2 i
= Õ in which j
q are odd primes, e 1 ³ 0 , e ³ 0 for 2 £ i£ l
Substituting the above expression for a and c into (1.1), the following will be obtained:
4 3
1 1 4 3 1
2 2
j
i
e + l e f m f
p
i i = q
= j=
j
2 ( ) 2 ( )
i
Õ Õ (1.2)
By uniqueness of the prime power decomposition of a and c , we will have for each i ,
the integer j in such a way that pi = qj
and vice versa for 2 £ i£ l,
2 £ j£ m,
and
l= m.
4e1 1 3f1
Also, we will have 2 2
+
= , which implies that 4e1+ 1= 3f1
3 f +- ( 4) e = 1
or 1 1
Since gcd(3, - 4) = 1,
many integral solutions to this equation do exist.
Meanwhile, we can see that e 1 = 1,
and f 1 = 3 is a particular solution that satisfies this
equation.
Thus, all the solutions for this equation are given by:
where t is the integer.
1
e = 2- 3t
1 1
Since e 1 and 1 f are positive integers, t 1 £ 0 .
f = 3- 4t
(1.3)
and 1 1