Pythagorean Triple

Sohel Sahoo : Hello Guys,

Pythagorean triples are very useful for fast solving questions on applications in competitive exams because they are whole numbers that make the Pythagorean Theorem true. If you are looking for the length of a side of a right triangle, and you know the lengths of two sides, check first to see if you have a right triangle whose sides are a Pythagorean triple.

Pythagorean Triple:-

A Pythagorean triple consists of  three positive integers a, b and c such that a² + b² = c² which is based on Pythagorean theorem.They are commonly written as (a, b, c). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k.

We can describe it as the sum of the squares of the two smaller positive integers is equal to the square of largest positive integer. The smallest & best known Pythagorean triple is (3,4,5).

Geometrically, a Pythagorean triple represents a right angled triangle whose right angle adjacent sides are represented by a&b and hypotenuse is represented by c.

Clearly,you can count such unit squares of equal sizes to prove pythagorean theorem with example & find that 3²+ 4² =9+16=25 = 5²

Primitive Pythagorean Triple:- 

Primitive Pythagorean triplet  is the one in which a, b and c are pairwise coprime (that is, they have no common divisor larger than 1) because if any two is divisible by a number then other must be divisible by that number according to Remainder Theorem.

e.g. (3,4,5) , (5,12,13) , (8,15,17) etc.

Condition  of  Generating a  Primitive Pythagorean  triple:-

Choose any two positive integers m and n with m > n (Say) ;then, for finding infinite numbers of pythagorean triples the fundamental formula is given by :

a = m²-n² , b = 2mn and c = m²+n²

Notice that the triplets generated by this formula are primitive if and only if m and n are coprime and not both odd or even (i.e. One of them is odd while the other one is even).

Proof:-  Let, a, b, c be the numbers that form a primitive Pythagorean triple. where a² + b² = c² and a, b, c are coprime.

  Now, a² + b² = c² 

        => c²- a² = b²         

       => (c+a) (c-a) = b²

      => (c+a)/b = b/(c-a)

    Since, a,b&c are positive integers & sum,difference&product of 2 positive integers is also a positive integer; we simply set the above equality to equal with  in lowest terms as m&n don't have any common factor to cancel out.

      (c+a)/b = b/(c-a) = m/n

  => (c+a)/b = m/n  and   b/(c-a) = m/n

  => (c+a)/b = m/n and (c-a)/b = n/m

  => c/b+a/b = m/n .....(1)

 and c/b - a/b = n/m .....(2)

Adding equation (1)&(2) we have,

 2c/b = m/n + n/m

=> c/b = 1/2 [  m/n + n/m ]

=> c/b = (m²+n²)/2mn ....(3)

Subtracting equation(2) from (1) we get,

 2a/b = m/n - n/m

=> a/b = 1/2 [  m/n - n/m ]

=> a/b = (m²- n²)/2mn .....(4)

Clearly, compairing both numerators and denominators of equation (3)&(4) we obtain;

       a = m²- n², b = 2mn , c = m²+ n²

clearly, a²+b² = (m²-n²)² + (2mn)² = (m²-n²)² + 4m²n² =(m²+n²)² = c²

As m/n is fully reduced,m and n are coprime, and they cannot both be even(otherwise,2 will be a common factor).

Note:-  Any odd number can be represented as 2n -1 , where n is a positive integer.

       Now, (2n-1)² = 4n² – 4n+1

So,every  square of a odd number leaves 1 as remainder on dividing by 4.

If m&n are both odd the numerator of (m²- n²)/2mn i.e. (m²- n²) is a multiple of 4 as per remainder theorem.and the denominator 2mn would not be a multiple of 4. Since 4 would be the minimum possible even factor in the numerator and 2 would be the maximum possible even factor in the denominator, this would imply a to be even despite defining it as odd.So,by contadiction m&n can never be odd simultaneously.

Take m = 2 , n =1 as m>n here taken

  So, a = m²- n² = 2²- 1² = 4-1 = 3

       b = 2mn = 2x2x1= 4

       c =  m²+n²=  2²+ 1²=4+1 = 5

Here, we get (3,4,5) as primitive pythagorean triple.

Like this you can obtain any number of triplets.

For your kind consideration I have tabularized all 16 primitive triplets with c<100 .

Value of ‘m’	Value of ‘n’	Triple (a,b,c)
2	1	(3,4,5)*
3	2	(5,12,13)*
4	1	(8,15,17)*
4	3	(7,24,25)*
5	2	(20,21,29)
6	1	(12,35,37)
5	4	(9,40,41)*
7	2	(28,45,53)
6	5	(11,60,61)
8	1	(16,63,65)
7	4	(33,56,65)
8	3	(48,55,73)
7	6	(13,84,85)
9	2	(36,77,85)
8	5	(39,80,89)
9	4	(65,72,97)

 Note:-  I think you had an clear idea of enumeration that by taking any positive integer  m as constant & take all n less than m from 1 to m ;We get certain values of m&n from which you can get all pythagorean triplets.From this the necessary condition should be obeyed that if m&n is of opposite category(i.e.when one is odd other is even not both) then it gives us primitive pythagorean triplets.Other such pairs gives us multiples of these primitive triplets.

(* denoted triplets are common and should be memorized for competitive exam)

Non-primitive triplet:-

Any such triplet that are not primitive pythagorean triple are said to be non-primitive triplets. These are simply a number times primitive triplets.

So, non-primitive triplet can be obtain  by inserting an additional parameter K  to the formula of primitive pythagorean triple. The following will generate all Pythagorean triples uniquely:

$$

a = K x (m²-n²), b = K x 2mn and c =K x (m²+n²)

e.g. (9, 12, 15) is a non-primitive triplet.

(9, 12, 15) = 3 x (3,4,5)

Like this we can multiply any number to primitive pythagorean triple to make non-premitive triplets.

Hope,You enjoyed reading these concepts that are written with my efforts to make you understand the concepts easily with necessary proof.It would definitely help you alot to solve problems asked in exams &daily use activity.You can share this article to all school going children which can help them in mensuration&in seminar;moreover to grasp concepts in a better & magical way.

Let me know if you have any query & I will put forward my level best to solve your question to the extreme extent.Till then enjoy reading this over & over and try to apply in such problems.