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

Integral Solutions of x 4 + y 4 = z 3
Now for each t 1 < 0 , the integer s 1 > 0 exists in such a way that t1=- s1.
Thus, from
(1.3), we will obtain the following:
e = 2+ 3s
and f1= 3+ 4s1
1 1
Similarly from (1.2), by uniqueness of prime power decomposition of a and c , we
will have by considering the odd prime factors of a and c that l , pi = q and 4e = 3f
j
for some i and j , where i, j= 2,3,... m.
From here, we can see that 4|3f j and 3|4e i .
Since gcd(4,3) = 1,
the integers j
It follows that
si = rj.
r and i
Pertanika J. Sci. & Technol. 21 (1): 283 - 298 (2013)
j j
i j
s exist in such a way that f = 4r
and e = 3s
.
Then, the prime power decomposition of a , b and c will respectively be:
Õ Õ
2+ 3s l
1 3si 2 s l
1
3si
3
i= 2 i i=
2 i
a= b= 2 p = 2 (2 p )
and
Õ Õ
3+ 4s l
1 4si
3 s l
1
si
4
i= 2 i i=
2 i
c= 2 p = 2 (2 p )
l s
i=
2 i
s1
i
Let n= 2 Õ p in this case.
3
Thus, a= b= 4n
and c 8n
4
= could be obtained as asserted, where n is an integer.
Now if t 1 = 0 , we will obtain e 1 = 2 and f 1 = 3 from (1.3).
Then, the prime power decomposition of a , b and c will respectively be as follows:
Õ Õ
2 l 3s 2 l
i 3si
3
i= 2 i i=
2 i
a= b= 2 p = 2( p )
and
Õ Õ
3 l 4s
3 l
i si
4
i= 2 i i=
2 i
c= 2 p = 2( p )
=Õ in this case.
l si
Let n p
i=
2 i
3
Thus, a= b= 4n
and c 8n
Hence, the triplet
x= y .
4
= could be obtained as asserted, where n is an integer.
3 3 4
(4 n ,4 n ,8 n ) is a solution to the equation
4 4 3
x + y = z , when
i i
k k 3
In Theorem 1.1 (i.e. where the solutions are of the form ( abc , , ) = ( un, vn, n ) and in
3 3 4
Theorem 1.2 [where the solutions of this equation in which x= y is ( abc , , ) = (4 n, 4 n,8 n)],
it is clear that gcd( abc , , ) > 1 . The subsequent Theorem 1.3 gives the forms of integral
4 4 3
solution to the equation x + y = z , in which x¹ y and gcd( abc , , ) > 1 also. The
cases in which gcd( xyz= , , ) 1 have been discussed by a number of earlier authors. For
example, Cohen (2002) showed that non-trivial integral solutions did not exist in such cases.
Thus, this assertion was reproduced in the following Lemma 1.1, the proof of which could be
found in Cohen (2002).
4k+ 1
123