Sum of Three Biquadatics a Multiple of a n th Power , n = ( 2 , 3 , 4 , 5 , 6 , 7 , 8 & 9 )

Abstract: Consider the below mentioned equation: (A). Historically Leonard Euler has given parametric solution for equation (A) when w=1 (Ref. no. 9) and degree ‘n’=2. Also S. Realis has given parametric solution for equation (A) when ‘w’ equals 1 and degree ‘n’ =3. More examples can be found in math literature (Ref. no.6). As is known that solving Diophantine equations for degree greater than four is difficult and the novelty of this paper is that we have done a systematic approach and has provided parametric solutions for degree’s ‘n’ = (2,3,4,5,6,7,8 & 9 ) for different values of “w”. The paper is divided into sections (A to H) for degrees (2 to 9) respectively.


Summary of Background
Whereas in math literature, we find many examples of sums of equal powers, meaning the degree is same on both sides of equation, in this paper we have demonstrated that it is possible to equate parametrically, unequal powers. Meaning the degree on the left hand side of the equation is different from the right hand side of the equations for n = 2, 3, 5, 6, 7, 8 & 9

Section (A) Equation 1
We have the below mentioned equation,

Substituting equation (B) above in eqn. (A) we get
After some algebra we get,

Equation 3
For w=3 degree n=2 We have known solution Substituting (p,q,r, s)=(1,1,1,1) in equation (A) we get Hence we arrive at after substitution: Proof: we prove the case of w=2, since this is the simplest case.

Section (B)
Equation (4) substituting above values below we get: Similarly, we can obtain the parametric solutions of the case w=1 and w= 3.

For w=2
There are infinitely many solutions. Using identity , we can get a parametric solution.
One of the solution =1 is (x, y) = (1, 0), so we obtain a following parameter solution.

For, w=18
There are infinitely many solutions.
In the same way as w=2, we can get a parameter solution.

For, w=98
There are infinitely many solutions.
In the same way as n=2, we can get a parameter solution.

Equation (8)
We have the known identity given below, We have known solution = 2 Since Hence we We get After substituting this in equation (2) we get the below mentioned parametrization: Hence we get the new Identity: For m=2 in above eqn. we get:

Section (F) Equation (9)
Numerical solution for above is:

Equation (11)
A parametric solution of Let X=a, Y=b, Z=a+b and, We obtain a parametric solution as follows.

Equation (12)
Let X=p, Y=q, Z=p+q Hence = Then we arrive at: We obtain a parametric solution as follows: Numerical solution is: For (a, b) =(1,1)

Conclusions
This paper has analyzed the equation. , for n = (2,3,4,5,6,7,8,9). In the near future attempt can be made by others to find solution for degree n>9 and for different values of integer 'w'.