Universal Journal of Applied Mathematics 5(1): 6-10, 2017
http://www.hrpub.org
DOI: 10.13189/ujam.2017.050102

Parametric solutions to equation ()where ‘n’ stands for degree 2,3,4,5,6,7,8 & 9Authors
1

Number theory theorist, Tokyo, Japan
Email: fermat@m15.alpha-net.ne.jp

2

University of Waterloo, Ontario, Canada
Email: samson@celebrating-mathematics.com

ABSTRACT) has been studied forHistorically equation () herein called equation (1)degree 2,3,4 etc., and equation (has been published for n=4 ,p=1,q=4 (Ref.no. 1) by Ajai Choudhry. Also Tito Piezas &others has discussed about equation (1) (Ref. no. 3 & 2) . While Ref. no. (1, 2 & 3) deals
with equation no. (1) for degree n=4 this paper has provided parametric solutions fordegree n=2,3,4,5,6,7,8 & 9. Also there are instances in this paper where parametricsolutions have been arrived at using different methods.
Keywords: Diophantine equations, Equal sums of powers, pure mathematics.

Parametric solutions to equation :(First we consider:
Degree two)

()

( )

For n=2

In equation (1) above;

We get ()Hence we have after simplification:

We takep=& q=

Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.
(p,q)=(w,u)
Hence we get the Identity given below:
()

()

()

( )

( )

()

Let: (u,v,w)=(5,3,2)(p,q,a,b,,c,d)=(2,5,8,1,2,5)
We get:( )Solution to:( )where (p,q)=(1,3),

We have:( )( )( )( )( )The above equation has numerical solution (a,b,c,d)=(1,5,7,3)Using the substitution: (a,b,c,d)=(1+t,5+t,7+kt,3+t) we get after simplifying()

Substituting value of ‘t’ in (a,b,c,d) we get the below mentioned parametric equation:
( )()

()

( )()

( )(( )( )( )( )( )For k=3 we have numerical solution as:( ))----------------------------------------(Next): Degree (n=3)
There is a parametric solution for:
()

()
()

()

a, b, p, q: arbitrary

( )
( )

()

()

()&Substitute m and t to (2), and we obtain a solution.
Case. (p,q,v)=(2,3,1) (After removing common factors) we get parametric form:()())(((())

()

()
())

()

For u=2, p=2, q=3 we get numerical solution:( )(--------------------------)

( )()

( )()

( )()

Alternate solution:
We have ()We get:Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.

()

()

()

(Where:()()Let: (u,v,w)=(3,2,1)(p,q,a,b,,c,d)=(31,126,5,1,1,3)
( )

( )

( )

( )

-------------------------Adittional method:
We have below mentioned equation:()

For (p,q)=(1,86) above equation becomes
( )

()

( )

The above has numerical solution (a,b,c,d)=(9,2,11,1)
We substitute (a,b,c,d)=(9+t,2+kt,11+t,1+kt)
We also have(b-d) = [(2+kt)-(1+kt)] = 1 and(c-a) = [(11+t)-(9+t)] = 2

After simplification we get()))

()After substituting for t in ( a, b, c, d) we get the below parameterization:

For k=1, we get numerical solution:
( )()

()()

( )()

()

-----------------------Degree (n=4)
We show solutions of quartic equation with coefficient ‘h’ for degree n=4--------------------- (3)Gerardin gave a solution of(Ref. no. 4)
()
()
( )
()
()
For
Let a = px+w, b = qx-1, c = px-w, d = qx+1, then (3) becomes (4) given below.
()() .................. 4)

To obtain a solution of (4), we have to find the rational solution of (5).
( )Let us consider eqn.(5) as a quadratic equation in h.
We find two solutions of (5) as follows.
h =( )( )Substitute h =(w/q) to (5), then()()()().

Let:,(then) ()From eqn. (4) above we get:We obtain a solution below: (Form A))()()
a=(b=c=()()()

d=
h=()

By using this new solution as a known solution, we obtain more new solutions.
Next we take h =,Similarly, (5) becomes to(()()()() ()())

Let w =Then()()

We obtain another solution below: (Form B)()()(-----------------------------------Alternate solution:
To get equation of the type below:)We put n=1 in the equation [Form (B) above] and then substitute m=x/y, giving us theparametric form below.
()

()
()

()

()

For x=1 & y=2 we get the numerical solution:
( )()

( )

( )()

()

---------------------------------------------Alternate solution:
We have ()We get:Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.

()

()

()

Where:((Let: (u,v,w)=(5,3,2) & we get:))(p,q,a,b,c,d)=(26,170,8,1,2,5)

()

( )

( )

( )

( )

---------------------(See Table below for parametric solutions)

Table (A) : (degree fourth power)()

For p=f(m) & q=f(m,n), where f(n) & f(m,n) are a function in variable ‘n’ & ‘(m,n)’ respectively
The following parametric solutions were found:()( ),f(m,n)]

()()

(,)

()
(((,)

()

()

()

()

()

()

()

()

)
)

()

,()

()

()

()

()

Next for fifth power:
Degree (n=5)(We have eqn.

)

We get:Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.

()

()

()

()

Where:()()

Let: (u,v,w)=(5,3,2) & we get:(p,q,a,b,,c,d)=(1562,16368,8,1,2,5)
( )

( )

( )

( )

------------------------------Degree (n=6)
We have ()We get:Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.

()

()

()

()

Where:()(())()

Let: (u,v,w)=(5,3,2)(p,q,a,b,c,d)=(1302,21840,8,1,2,5)
( )

( )

( )

( )

-----------------------------------------Degree (n=7)
We have ()We get:Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.

()

()

()

()

Where:()()

Let: (u,v,w)=(5,3,2) & we get for degree n=7 the below mentioned numerical solution:(p,q,a,b,c,d)=(19531,524256,8,1,2,5)( )Degree n=8( )

( )

( )

We have ()We get:Let a=u+v, b=v-w, c=u-v, d=v+wBy resolving above polynomial into factors, we get {p,q}.

()(Where;)(()(()()()))

Let: (u,v,w)=(5,3,2) & we get numerical solution:(p,q,a,b,c,d)=(4069,174760,8,1,2,5)
( )

( )

( )

( )

-------------------------------------------------------------Degree n=9:
We have ()We get:Let a=u+v, b=v-w, c=u-v, d=v+w
By resolving above polynomial into factors, we get {p,q}.

(Where:)()()(()(Let: (u,v,w)=(5,3,2)(p,q,a,b,c,d)=(976562,266304,8,1,2,5)

()

()))

( )

( )

( )

( )

On a side note, the interested reader may look at the below identities (A, B, C & D) that were
derived for the equation (2) given below:--------(2)Here the coefficients for eqn.(2) are different than that used for equation (1) at the top of thepage with the coefficients on the left hand side of equation (2) being different from thecoefficients on the right hand side.
(m, n, u, v)=(1,3,1,4 )[A]( )(----------------------– )

()

( )(–

(– )

()

( )(–)

(–)

(– )

)

(m, n, u, v)=(1,6,1,2 )[B](1)(----------------------)(m,n,u,v)=( 1,24,1,8)[C]( )(-----------------------– )

( )

( )()

(m,n,u,v)=(1,27,1,36 )[D](1)(– )()
( )(–)(-----------------------------------------------------[See below for numerical solutions for equation (1) for different degree ‘n’])Equation (1) -----()

for n=2,3,4,5,6,7,8 & 9Degree (n)pqabcd2

2

3111

193

1362

511345354

16525434

91249

6

19

542

1575473261298353

60

4

32

59729

194

316(See below for list of references)References:[1].Ajai Choudhry, A Note on the Quartic Diophantine Equation,a^4+ hb^4= c^4+ hd^4. https://arxiv.org/pdf/1608.06220v1[ 2].Farzali & Mehdi, ( Is the quartic equation (a^4 + hb^4 = c^4 + hd^4), solvable forany integer h?) Journal Arxiv,1701 Jan. 2017[3]. Tito Piezas-Online collection algebraic identities, equation a^4+ hb^4= c^4+ hd^4.
equation for fourth power, http://sites.google.com/site/tpiezas[4] Gerardin on the eqn. a^4+hb^4=c^4+hd^4, pg.648, vol. (20, Dickson, historyof theory of number theory, Chesea publication, year 2000
( )( ), Universal Journal of applied mathematics
Vol. (4),(3), pages 55-65, 2016, http://www.hrpub.org[6] Published math paper, Oliver Couto & Seiji Tomita,), Sum of three biquadratics
a multiple of a n th power, Universal Journal of applied mathematics ,Vol. (4)(1),
pages 22-31,2016, http://www.hrpub.org[7] Published math paper, Oliver Couto & Seiji Tomita, Generalized parametric solution
to multiple sums of powers, Universal Journal of applied mathematics,
Jan 2016, Volume 3(5), http://www.hrpub.org[8] Published math paper, Oliver Couto & Seiji Tomita, Solutionof Polynomial Equation of any degree ’n’ with Special Emphasis
For n= (2, 3, 4, 5 & 6), Journal “SOP transactions on applied mathematics”
ISSN (Print): 2373-8472, July 2015. http://www.scipublish.com/journals/am[9] Published math paper, Oliver Couto, Taxicab Equations for power(2, 3, 4 & 5), Journal, International Math Forum, Hikari ltd., Vol.9,
Jan 2014.n0.12, pages 561-577. http;//www.m-hikari.com/imf.html[10] Math paper, Seiji Tomita & Oliver Couto, Sum (k+3) & (k+5) bi-quadratics,
Scheduled to be posted on-line, Universal Journal of applied mathematics,
Vol. (4) (2), 2016, http://www.hrpub.org[11] Seiji Tomita, webpage http://www.maroon.dti.ne.jp/fermat
Web site “Computation number theory”[12] Oliver Couto, webpage, http://www.celebrating- mathematics.com
Web site “Number theory Math”[13] Ramanujan lost notebook, Narosa publishing house, New Delhi, India[14] Euler Leonhard, Opera Omnia books, 1984 publication.
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