Universal Journal of Applied Mathematics 4(1) 22-31: 2016

http://www.hrpub.org

DOI:10.13189/Ujam.2016.040103

Sum of Three Biquadatics a Multiple of a nth Power,n = (2,3,4,5,6,7,8 & 9)
Seiji Tomita1, Oliver Couto2
1

Tokyo
Software Company (Inc),Tokyo, Japan
2
University of Waterloo, Waterloo, Canada
1
fermat@m15.alpha-net.ne.jp, 2samson@celebrating-mathematics.com
Copyright©2016 by authors, all rights reserved. Authors agree that this article remains permanently open access under theterms of the Creative Commons Attribution License 4.0 International License

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 forequation (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 donea 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.

Keywords: Quartic Equation, Diophantine Equations, Pure Math, Number Theory, Sums Of Powers

1. Summary of Background
Whereas in math literature, we find many examples of sums of equal powers, meaning the degree is same on both sides ofequation, 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

2. Section (A)Equation 1
We have the below mentioned equation,
Let, w=1, degree n=2
Let:Where) are known solutions

(1)We have solution given by Leonard Euler which is given below for,a=We have,) using maple soft

Substitute k and p to (1), hence we obtain above parametric solution.

We next show the parametric solutions of x4+y4+z4= w*t2 for several “w”.
In the case of w=2 & 3, parametric solutions and numerical examples are shown.
Identity by S. Realis: w=3, n=2()
Search for smallest solutions of x4+y4+z4= w*t2 in the case of w< 20 are as follows.
wxyzt

1

2015

12481

2

2

1

1

3

3

1

1

1

1

960

45

36

1443

11

5

5

3

11

17

22

17

6

137

18

2

1

1

1

19

15

15

173

Above equation doesn't have a solution in the case of:w =5,6,7,10,13,14,15(mod 16).
Equation (2)(B)
We have known solution, (p, q, r, s)=(2,1,1,3)(C)
Substituting equation (B) above in eqn. (A) we getAfter some algebra we get,( )( )Hence equation (B) above has general solution for any ‘w’ & n = 2
We have known solution for w=2, (p, q, r, s)=(2,1,1,3)
After substituting the above we get:()We have equations,

After substituting for (g, h, k, p, q, r) we get

For (a, b) = (1,0) 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 getHence we arrive at after substitution:

For (a,b)=(1,1) we get:Numerical Examples(a, b)=(a, b)=(a, b)=(a, b)=Taking, n=3 we get
We show the parametric solutions offor w = 1, 2 & 3.
In the case of w =1, 2, and 3, parametric solutions and numerical examples are shown below.
Smallest solutions for:w
In the case of “w” for are as follows:
w

xyz

T

176

72

4392

2

19

18

149

3

228

216

121176

5380

360

20

1960

657

54

3147

7

32

20

1256

9106

91

80

297

Proof:
we prove the case of w=2, since this is the simplest case.

Hence (-1) = 0 or (u-1)*(+u+1) = 0{}

3. 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.
Equation (5)We get, w=1, degree n=3
Since z = (x+y), we get

Since z = (x + y) we get,Numerical Examples:(a,b)=
For,(a, b) for, n=3(a, b)(a, b)=

4. Section (C)degree n=44

x +y4+z4=w*t4.
For degree ‘n’Consideration of x4+y4+z4=wt4
above equation has no solutions.

(mod 16). In the case of (n mod 16)=4,5,6,7,8,9,10,11,12,13,14,15,Consideration of x4+y4+z4=w*t4(mod 25)
In the case of (n mod 25) = 4, 5, 9, 10, 14, 15, 19, 20, 24 above equation has no solutions.
For w=2
There are infinitely many solutions.
Using identity
One of the solution, we can get a parametric solution.
=1 is (x, y) = (1, 0), so we obtain a following parameter solution.
(-1+k2)4+ (k2+2k)4 + (1+2k)4 = 2(1+k+k2)4

Numerical solutions for above for w=2 are given below:
kxyz

t2

83573

158713535

241131648

35

1343

721

16519880

63

1773

999

80

1991

2. For, w=18
There are infinitely many solutions.
In the same way as w=2, we can get a parameter solution.
x2+xy+y2=3 is (x, y)=(1,1)(-1+2k+2k2)4 + (-2-2k+k2)4+(1+4k+k2)4 = 18(1+k+k2)4
3. For, w=98
There are infinitely many solutions.
In the same way as n=2, we can get a parameter solution.
x2+xy+y2=7 is (x, y)=(2,1)(-3-2k+2k2)4+ (-1+4k+3k2)4+(2+6k+k2)4 = 98(1+k+k2)4

5. Section (D)w=1 & degree n=5
Equation (7)Numerical solution is;(a, b) = (1, 0)

6. Section (E)w=2 & degree n=6
Equation (8)We have the known identity given below,(2)We have known solution=2Since
Hence we
We getAfter substituting this in equation (2) we get the below mentioned parametrization:

Hence we get the new Identity:(B)
For m=2 in above eqn. we get:w =1 & degree n=7

7. Section (F)Equation (9)Numerical solution for above is:(a,b) =(a,b)=

8. Section (G)Degree n = 8

For w=2, n=8
Equation (10)(1)

then the right hand side of above equation (1)becomes after substitution of

For (m,n) = (2,1) we get numerical solution as (x,y,z,w) = (16,55,39,7)

9. Section (H)Degree n=9
For w=2, n=9
Equation (11)A parametric solution of
Let X=a, Y=b, Z=a+b

and,
We obtain a parametric solution as follows.

Numerical solution for (m, n) = (1,1) we get
w=1, n=9Equation (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)

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

REFERENCES[1]

Seiji Tomita, Self web page, Computation number theory: http://www.maroon.dti.ne.jp/fermat.

[2]

Published math paper, Oliver Couto & Seiji Tomita, Generalized parametric solution to multiple sums of powers, Universal Journalof applied mathematics, Dec 2016, Volume 3(5), pages 102-111,http://www.hrpub.org.

[3]

Published math paper, Oliver Couto & Seiji Tomita, Solution of Polynomial Equation of any degree ’n’ with Special Emphasis
for n= (2, 3, 4, 5 & 6), Journal “SOP transactions onApplied mathematics” ISSN(Print):
2373-8472, July 2015.http://www.scipublish.com/journals/am.

[4]

Published Math paper, Oliver Couto, Taxicab Equations for power (2, 3, 4 & 5), Journal, International Math Forum, Hikari ltd.,
Vol.9, Jan 2014 no.12, pages 561-577. http;//www.m-hikari.com/imf.html.

[5]

Oliver Couto, Self, web page on Mathematics, http://www.celebrating-mathematics.com.

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SeijiTomita,Fourthpowerpolynomialequation,
Computation number theory- web-page,http://www.maroon.dti.ne.jp/fermat/diop122e.html(7)SeijiTomita,sixthpowerpolynomialequation,Computation
Number theory-Web-page,http://www.maroon.dti.ne.jp/fermat/diop160e.html.
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9)Euler Leonhard, Opera Omnia, 1984.
10)Tito Piezas-Online collection of algebraic identities http://sites.google.com/site/tpiezas.
11)Ajai Choudhry, Symmetrical Diophantine equations, Journal of mathematics, Rocky mountain journal vol. 34,no.4,winter(2004), pg.
1261-1298.
12)Jaroslaw Wroblewski, Tables
www.math.uni.wroc.pl/~jwr/eslp.

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solutionsfordegree

three,four,six,seven&nine,13)L. E. Dickson, history of the theory of numbers, Vol.II, (Diophantine analysis), AMS Chelsea publication, year 2000.

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website,