Vasanth

• Aug 10th, 2005

 

Two numbers, say a & b, form Pythogeran triplets, if  
 
a^2 + b^2 = (b+1)^2 
 
=> a^2 + b^2 = b^2 + 2b + 1 
 
=> a^2 = 2b - 1 
 
=> b = (a^2 - 1) / 2 
 
for (a^2 - 1) to be divisible by 2, a^2 should be odd and hence "a" should be odd. 
 
For every a = 3, 5, 7, 9, 11, .... we will have a "b" and "b+1" which form Pythagorean triplets. 
 
Ex: 3, 4, 5 
5, 12, 13 
7, 24, 25 
9, 40, 41 
11, 60, 61 
13, 84, 85 
 
...

Radeesh

• Aug 27th, 2005

 

Two numbers, say m & n, form Pythogeran triplets, if  
n=11 and m=60 
First condition n^2 + m^2 = (m+1)^2  
11^2+60^2=(60+1)^2 
Second condition n^2=2m+1 
11^2=2(60)+1 
So if we prove first condition to second the this must be true. 
Consider Second condition n^2=2m+1 
add m^2 on both sides 
n^2+m^2=m^2+2m+1 
n^2+m^2=(m+1)^2 

C. Sivasankaran

• May 16th, 2006

 

There are some other ways also

1.  Multiples of the Odd number triplets, say for example 3, 4, 5

     Its multiples also triplets i.e. 6,8,10 ;   9, 12, 15  ;  12, 16, 20 ; 15, 20, 25

2.  It can be extended to  even numbers also

    4^2 = 16 , then divide by 4 we get 4, 4-1 and 4+1 are its sets. By this method we can get

    4, 3,  5

    6, 8,  10

    8, 15,  17

    10, 24, 26

    12, 35,  37  and so on...

3.  Simillarlly we can divide a square number by 6 and subtracting and adding by 1.5 we get the triplets.  By this we can get the following.

9	12	15
15	36	39
21	72	75
27	120	123
33	180	183
39	252	255
45	336	339

     and so on...

stockball

 Profile Answers by stockball 

• May 24th, 2011

 

This is an overly simple algebra question.  The first form of the equation is x^2 = 2y + 1.  The second form of the equation is x^2 + y^2 = (y+1) ^2, which is just y^2 + 2y + 1.  Subtract the y^2 from both sides of the equation and you again have x^2 = 2y + 1.