Digital ending method of getting conditional Pythagorean triples
A Pythagorean triple (a,b,c) must satisfy the relation a^2 + b^2 = c^2
where c >a and c > b but a+b > c.
The values of a,b and c must be such that the digital endings of both sides must be same..
Table given below shows the digital ending of c for all possible digital endings of a and b.
Table.: Digital ending of c in a^2 + b^2 = c^2
a / b 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1 1 x 5 0 x 4,6 x 0 5 x
2 2 5 x x 0 3,7 0 x x 5
3 3 0 x x 5 2,8 5 x x 0
4 4 x 0 5 x 1,9 x 5 0 x
5 5 4,6 3,7 2,8 1,9 0 1,9 2,8 3,7 4,6
6 6 x 0 5 x 1,9 x 5 0 x
7 7 0 x x 5 2,8 5 x x 0
8 8 5 x x 0 3,7 0 x x 5
9 9 x 5 0 x 4,6 x 0 5 x
It is noted that certain digital ending combinations for a and b are forbidden. For example both a,and b
cannot be ending with identical digits except 5
It is found that a^2 - a = a(a-1) is always equal to some multiples of 2,hence
a^2 = a+ an(a)
b^2 = b + 2n(b)
c^2 = c+2n(c)
It gives a condition
a +b = c + 2n
According to the requirement of digital endings of the equal power relation ,one can assume the following
typical relation as an example,
(10x+7)^2 + (10y+4)^2 = (10z+5)^2
Since a+b = c+2n, 10(x+y)+11 = 10z+5 + 2n
It demands that n = (3+5k) with k = 0,1,2,3......and x+y = z+k
With k = 0 , x+y = z
(10x+7)^2 + (10y+4)^2 = [10(x+y)+5]^2
it gives y = 2(x+1)/10x+1
When x = 0; y=2,z=2 ; 7^2+24^2=25^2
when x=1 ,y=4/11,z=15/11; 187^2+84^2 = 205^2
when x=2 , y = 6/21,z=48,21; 567^2 +144^2= 585^2
when x = 3, y=8/31,z=101/31 ; 1147^2+204^2 = 1165^2
When k = 1, x+y= z+1
(10x+7)^2 + (10y+4)^2 = [10(x+y)-5]^2
It gives y = 12x+2/10x-9
when x=1 ,y=14,z=14 ;17^2+144^2=145^2
when x=2,y=26/11,z=37/11; 297^2+304^2 = 425^2
A Pythagorean triple (a,b,c) must satisfy the relation a^2 + b^2 = c^2
where c >a and c > b but a+b > c.
The values of a,b and c must be such that the digital endings of both sides must be same..
Table given below shows the digital ending of c for all possible digital endings of a and b.
Table.: Digital ending of c in a^2 + b^2 = c^2
a / b 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1 1 x 5 0 x 4,6 x 0 5 x
2 2 5 x x 0 3,7 0 x x 5
3 3 0 x x 5 2,8 5 x x 0
4 4 x 0 5 x 1,9 x 5 0 x
5 5 4,6 3,7 2,8 1,9 0 1,9 2,8 3,7 4,6
6 6 x 0 5 x 1,9 x 5 0 x
7 7 0 x x 5 2,8 5 x x 0
8 8 5 x x 0 3,7 0 x x 5
9 9 x 5 0 x 4,6 x 0 5 x
It is noted that certain digital ending combinations for a and b are forbidden. For example both a,and b
cannot be ending with identical digits except 5
It is found that a^2 - a = a(a-1) is always equal to some multiples of 2,hence
a^2 = a+ an(a)
b^2 = b + 2n(b)
c^2 = c+2n(c)
It gives a condition
a +b = c + 2n
According to the requirement of digital endings of the equal power relation ,one can assume the following
typical relation as an example,
(10x+7)^2 + (10y+4)^2 = (10z+5)^2
Since a+b = c+2n, 10(x+y)+11 = 10z+5 + 2n
It demands that n = (3+5k) with k = 0,1,2,3......and x+y = z+k
With k = 0 , x+y = z
(10x+7)^2 + (10y+4)^2 = [10(x+y)+5]^2
it gives y = 2(x+1)/10x+1
When x = 0; y=2,z=2 ; 7^2+24^2=25^2
when x=1 ,y=4/11,z=15/11; 187^2+84^2 = 205^2
when x=2 , y = 6/21,z=48,21; 567^2 +144^2= 585^2
when x = 3, y=8/31,z=101/31 ; 1147^2+204^2 = 1165^2
When k = 1, x+y= z+1
(10x+7)^2 + (10y+4)^2 = [10(x+y)-5]^2
It gives y = 12x+2/10x-9
when x=1 ,y=14,z=14 ;17^2+144^2=145^2
when x=2,y=26/11,z=37/11; 297^2+304^2 = 425^2
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