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

Since
Integral Solutions of x 4 + y 4 = z 3
4 4
u + v = n,
we will have the following:
3k
dn = n or
3k1 d n -
=
3k1 From Theorem 1.3, we have then a un -
3k1 = , b vn -
4k1 = and c n -
= since
Pertanika J. Sci. & Technol. 21 (1): 283 - 298 (2013)
3k1 d n .
-
=
3k-1 3k-1 4k-1 Table 1 below shows some example solutions ( abc , , ) = ( un , vn , n ) that were
obtained for Theorem 1.3 for various values of k .
TABLE 1: Some examples of the integer solutions for the diophantine equation
x¹ y and gcd( xyz> , , ) 1 for various values of k .
3k1 k
a un -
=
1 2
a = un
2 5
a = un
3 8
a = un
4 11
a = un
5 14
a = un
3k1 b vn -
=
2
b = vn
5
b = vn
8
b = vn
11
b = vn
14
b = vn
4k1 c n -
=
3
c= n
7
c= n
11
c= n
15
c= n
19
c= n
3k1 d n -
=
2
d = n
5
d = n
8
d = n
11
d = n
14
d = n
4 4 3
x + y = z when
3k1 Note: The solution set obtained from Theorem 1.3, where d n -
= with k ³ 1 as represented in
Table 1 is also the solution set obtainable from the forms of solutions in Theorem 1.1, in which
k º 2(mod 3) . In this study, the assertions of Theorem 1.1, Theorem 1.2 and Theorem 1.3 have been
illustrated by showing the actual solutions for the integers ( xyz , , ) , as in Table 2 that satisfy this
equation using the C language in integer domain.
TABLE 2: Some examples of the integer solutions generated by the C language for the diophantine
equation
4 4 3
¹ .
x + y = z when x= y and x y
x y z
4 4 8
32 32 128
108 108 648
256 256 2048
578 4913
500 500 5000
4000 4000 80000
6724 20172 551368
18818 28227 912673
391876 1959380 245314376
821762 2054405 263374721
340707 454276 38272753
66049 264196 16974593
125