_Solving the Diophanitine equation by algebraic method
The very simple innovtive way to derive numerical examples for Diophantine equation of any order with any exponent is explained here. At first assume an algebraic expression with two or three unknowns as a general form for the required Diophantine euation. Then by assuming its dependance one can reduce the numer of unknowns to a minimum of 2. Giving a suitable value to an unknown the value of other unknown can be fixed,which finally provide a group of numerical examples for the Diophantine equation considered.
As an example . let us try with the Diophantine equation where the sum of two fourth powers is equal to sum of two other fourth powers A4 + B4 = C4 + D4
A4 + B4 = C4 + D4 -------> (a-x)4 + (b+y)4 = (a+y)4 + (b+x)4 , where x,y,a and b are some unknowns .The condition b = ka reduces one unknown.
- 4a3x + 4k3a3y = 4 a3 y + 4 k3 a3 x
6a2x2 + 6k2a2 y2 6a2 y2 + 6k2a2 x2
- 4ax3 + 4kay3 4 a y3 + 4kax3
Let us assume tht the first two terms in the RHS is equal to the first two terms in LHS which relates x with y.
- 4a3x + 4k3a3y = 4 a3 y + 4 k3 a3 x
k3 (y-x) = (y+x) or y = [ (k3 +1)/(k3 -1)] x
The sum of the remaining four terms must be equal.
3ax2 + 3k2a y2 - 2x3 + 2ky3 = 3a y2 + 3k2a x2 + 2 y3 + 2k x3
3ax2 - 2x3 + 2kx3 - 3k2 ax2 = 3a y2 - 3k2a y2 + 2 y3 - 2k y3
= 3a [(k3 +1)/(k3 -1)]2 x2 + 2[(k3 +1)/(k3 -1)]3 x3 - 2k [(k3 +1)/(k3 -1)]3 x3
- 3k2 a[(k3 +1)/(k3 -1)]2 x2
x2 is common and rearranging the terms with a and without a, we get
3a(k2 -1){[(k3 +1)2 - (k3 -1)2]/(k3 -1)2}
= 2x{ [ (k+1)(k3 -1)3 - (k-1) (k3 +1)3]/ (k3-1)3 }
which gives a = [2x/3(k2 -1)]{ [ (k+1)(k3 -1)3 - (k-1) (k3 +1)3]/ (k3 -1) [ (k3 +1)2 - (k3 -1)2] }
when k = 2; a = x (2/9)[300/(7x32)]=(25/84) x ; and y = (9/7)x
If a = 25, x = 84 , b = 2x = 50 and y =108 the sum of equal fourth power relation becomes 594 + 1584 = 1334 + 1344
When k =3; a = x (2/24){[4(26)3- 2(28)3]/ [26(282 - 262 )} = 275/351 x
If x = 351 ; a = 275; b = 3a =825 and y = (14/13)x =378 givws 764 + 12034 = 6534 + 11`764
