Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

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July 2015

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SOP TRANSACTIONS ON APPLIED MATHEMATICS

Solution of Polynomial Equation of any Degree’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)
Oliver Couto

1*, Seiji Tomita

2

1 University of Waterloo, Mathematics Student, Ontario, Canada
2 Computation number Theory, Theorist, Tokyo, Japanemail
fermat@m15.alpha-net.ne.jp*Corresponding author emailsamson@celebrating-mathematics.com-------------------------------------------------------------------------------------------------------------Absract=----------------- (A)
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Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

1.

Theory & Methods

Degree Two

Put a0=2, a1=5, a2=3 we get

LetThis has parameterization in variables (m, n)Degree Three

Put= ===1 we getLet
Substituting above & after parameterization we get,()()()(Degree Four
(See below section (II) after degree six)
Degree Five2)Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

Substituting above & parameterization we get()

()
()

()

Degree Six---------------------(1)
( )

( )

( )

( )

()
()

()

( )
( )

(( ))

) ((()())(( )3)
( )

( )

Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

( )(((())))For t = 2 we get ( x , y, u, v ) = ( 9, 23, 7, 25 )2.

Proof And Solving( )=Where {,,-------,,,,} are coefficents of the above polynomial.

We fix the condition that (sum of even coefficients = sum of odd coefficients).
For quartic polynomial, we get after putting n=4 in eqn (A)Example for coefficients of degree four we have(4+ a2 + a0 = a3 +)Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

Let x=pt+1, y=qt-1, u=pt-1, v=qt+1
We take (, t) as parameters & for simplicity we call (Thus we have (-) = (d, e, f)) = (e – d - f) and equation (A) above becomes for n = 4

After some algebra we get after solving for (p, q),
p = 4d - e , q = 8d - 5e + 2f,= ( 4d - 3e + f ),=( 3d - 2e + f )----------Similarly for Degree fiveWe get for (p, q)
p = 5d-e
Also we get,and= ( 10d - 6e + 3f ),q = 15d-7e+2f= (15d - 8e + 3f),

= (6d - 3e + f)And so on for degrees ( 6, 7, 8, ------ etc ).
The coefficients ‘’for degree (3, 4, 5, 6 ) are= (d – e + f)

for degree three= (4d-3e+2f)

for degree four

= (10d-6e+3f)

for degree five= (20d-10e+4f)

for degree six

Thus the general form for the coefficient( ) = ( r*d - s*e + w*f)

Putting the above in tabular form we have,Coefficients ( )Degree three
Degree four
Degree five
Degree six
5Coefficient of d1
410
20

Coefficient of e13610

Coefficient of f1234

Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

The above three columns are in series, Hence they are represented as()(( ())())( )Since the 1st term is p = 1 and the degree is n = 3,(n, p) are related as(p = (n-2). After putting value of (p, r, s, w) in) = (r*d - s*e + w*f) we get

Thus we get coefficent ()=()( ()())In terms of degree ’n’Thus (p,q) has representation in terms of degree ‘n’ as
P = ( n*d -e) and q = [ n(n-2)d - (2n-3)e + 2f ]And the coefficients (a3, a4, a5, ---) as
()

( ()

(()))()(& so on for other coefficients (,,(()(()()()()(), -----,------))))()For integer m, value of coefficient (am) we arrive at,()()(((6)() ()()(()))() )( )Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

For coeeficent

put m=6 in the above & we get()(()(())( )())

Which simplyfies as,()(()(())())

So now we solve by the new method, the polynomial equation of degree six by using the known
coefficents above,

Substituting the values of (m, n) in equation (B) for various coefficients(,,,) we get,For example for,
we put m=3, n=6
we put m=4, n=6
we put m=5, n=6
we put m=6, n=6
Thus we get after the above substitution= (20d-10e+4f),= (45d-20e+6f) and= (36d-15e+4f),= (10d-4e+f)
We know that
7p = (n*d - e)&q = [ n(n-2)d - (2n-3)e + 2f ]Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

Hence for n=6 we get p = (6d - e), (q = 24d - 9e + 2f )We have,x = pt+1, y = qt-1, u = pt-1, v = qt+1

Let us take d=5 e=4 & f=3, t=2 & andSince (= d,=e &Hence we have (= f, t = t ), t)=(5,4,3,2) and

p=26, q=90, a3=72, a4=163, a5=172, a6=37
Thus the sum of even & sum of odd coefficents are(+a4+a2+)=(+a3+a1)(37+163+3+5) = (132+72+4) = 208
x=pt+1, y=qt-1, u=pt-1, v=qt+1
And we get (x, y, u, v) = (53,179,51,181)
After substituting the coefficients in the sixth degree polynomial we get the below mentioned
equation,The later equation can easily be verified that(x, y, u, v) = (53,179,51,181) is the solution to the sixth degree polynomial.

Degree, n=78

Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

For n=7 Using,= d,=e&= f, t = t&

Substituting the values of (m, n) in equation (B) for various coeficents (a3, a4, a5, a6) we get,For example for,
we put m=3, n=7
we put m=4, n=7
we put m=5, n=7
we put m=6, n=7
we put m=7, n=7.
Hence we get := (35d-15e+5f),= (105d-40e+10f)= (126d-45e+10f),= (70d-24e+5f),=(15d-5e+f)& we get p = (7d - e),p = 31, q=137,(q = 35d - 11e + 2f)=130, a4=395 ,a5=480, a6=269 & a7=58

x=pt+1, y=qt-1, u=pt-1, v=qt+1
Hence for (ao, a1, a2, t) = (5,4,3,2) we get
p=31, q=137,=130, a4=395, a5=480, a6=269,=58

And (x, y, u, v)=(63,273,61,275) & we get the seventh degree polynomial given below

The later equation can easily be verified that (x, y, u, v) = (63, 273, 61, 275) is the solution to theseventh degree polynomial.

9Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

3. Conclusions
The later method can be used to solve polynomial equations of any degree ‘n’ with only fourparameters (d, e, f, t). Also polynomial equations like the above can be generated which weknow have guaranteed integer solutions.
References:
1) Oliver Couto, Published Math paper, Taxicab Equations for power (2,3,4 & 5),
International Math Forum, Hikari ltd.,Vol.9,2014.n0.12, pages 561-577.
2) Seiji Tomita, Computation number theory- webpage,
http://www.maroon.dti.ne.jp/fermat3) Oliver Couto, Web page on Mathematics, http://www.celebrating-mathematics.com
4) Seiji Tomita, Fourth power polynomial equation, Computation number theorywebpage, http://www.maroon.dti.ne.jp/fermat/diop122e.html
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theory,75, (1999)
6) Seiji Tomita, sixth power polynomial equation, Computation number theorywebpage, http://www.maroon.dti.ne.jp/fermat/diop160e.html7) Ramanujan lost notebook, Narosa publishing house8) Euler Leonhard, Opera Omnia, 1984
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http://sites.google.com/site/tpiezas
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mountain journal vol 34,no.4,winter(2004), pg. 1261-1298.
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Solution of Polynomial Equation of any Degree ’n’ with Special Emphasis for n= (2, 3, 4, 5 & 6)

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