Pythagorean triples with given endings

Here is yet another method to identify Pythagorean triples with given endings.

The Pythagorean triples (a,b,c) satisfy the relation a² + b² = c² .Let us try to find a series of Phythagorean triples ending with 3,4 and 5 (greatest)

The most general form for this set is

(10x+3)² + (10y+4)² = (10z+5)². It gives a condition for the suitable values of x,y,and z.- 5(x² + y²) +3x +4y = 5z(z+1)

The other requirement for the Pythagorean triples is a+b = c+2α ,where α may be either odd or even. When we apply this condition to our general form ,it predicts α takes only certain allowed values 1,6,11,16,21… and so on.

 α = 1 ; x+y = z

α = 6 ; x+y-1= z

α = 11 ; x+y-2 = z

α = 16 ; x+y -3 = z

When α = 1

5(x² + y²) +3x +4y = 5 (x+y)(x+y+1) ,it is reduced to 10xy+2x+y = 0. The oly possibility is x=0;y=0 and z=0.

When α = 6

5(x² + y²) +3x +4y = 5(x+y-1)(x+y) ,the required condition becomes x = 9y/[2(5y-4)].If x and y are positive integers,the pre determined endings will not altered.For example when y=1 then x = 9/2 which gives

Irreducible triple 7² + 24² = 25^2; When y=3 ;x=27/22  which gives 84² + 187² = 205²

 When α = 11 ,the required condition becomes x = (19y-10)/(10y-18) with y being even.When y=2 ,x=14.It gives (143,24,145)

Y=4 gives (3,4,5)

Y=6 gives (583,1344,1465)

Y=8 gives (803,2604,2725)