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

Integral Solutions of x 4 + y 4 = z 3
We will show that the solution set of the equation in which gcd( xyx> , , ) 1 is always
non-empty in the following theorem. It shows that a solution set can always be constructed
from a given pair of integers.
Theorem 1.1
4 4
Suppose u and v are integers and n= u + v . Let k be an integer such that k º 2(mod 3).
Then,
Proof:
k
a un
= , k
4k+ 1
3
b = vn and c= n is in the solution set of the equation
By multiplying both sides of the equation
or
4 4
n= u + v with
4k 4k 4 4k 4
n × n= ( n × u ) + ( n × v )
+
n = ( nu) + ( nv)
4k 1 k 4 k 4
Since k º 2(mod 3), we have 4k + 1º 0(mod 3) .
Thus, there exists integer m such that 4k+ 1= 3m.
It follows that
or
n = ( nu) + ( nv)
3m k 4 k 4
( n ) = ( nu) + ( nv)
m 3 k 4 k 4
k
k
m
Hence, a = un , b = vn and c= n satisfy the equation
4k+ 1
assertion by putting m = .
3
Pertanika J. Sci. & Technol. 21 (1): 283 - 298 (2013)
4 4 3
x + y = z .
4k
n , the following is obtained:
4 4 3
x + y = z . We obtain the
A corollary obtained from the above theorem is as shown below. The forms of solutions
for the triplet ( abc , , ) are obtainable when n is a cube in which case there exist infinitely
4 4 3
many solutions to the following equation, x + y = z .
Corollary 1.1
Suppose u and v are integers and
Let k be a positive integer. Then
4 4 3
x + y = z .
Proof:
4 4
n= u + v . Suppose
k
k
a = un , b = vn and
From the proof of Theorem 1.1, it can be shown that:
+
n = ( nu) + ( nv)
4k 1 k 4 k 4
Let
3
n= r , then the following will be obtained:
+
( r ) = ( nu) + ( nv)
4k 1 3 k 4 k 4
Hence,
cube.
k
k
a = un , b = vn and
4k+ 1
3
n r
c= n is in the solution set of
3
= for some integer, r .
4k+ 1
3
c= n is in the solution set of
4 4 3
x + y = z , if n is a
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