# How do you calculate Pythagorean triples with 2 consecutive integers?Also, how do you find a pattern to generate infinitely many Pythagorean triples? A thorough and step-by-step...

How do you calculate Pythagorean triples with 2 consecutive integers?

Also, how do you find a pattern to generate infinitely many Pythagorean triples?

A thorough and step-by-step explanation is greatly appreciated.

sociality | High School Teacher | (Level 1) Valedictorian

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Pythagorean triplets are three numbers a, b and c such that a^2 + b^2 = c^2. If we have to find Pythagorean triplets with 2 numbers of the triplet made up by consecutive integers we can take them as n and n+1. Now the third number X is such that X = n^2 + (n+1) ^2

expanding (n+1)^2, we get

X = n^2 + n^2 + 2n +1

or X = 2n^2 + 2n +1

Therefore for any value of n, the Pythagorean triplet such that two of the numbers are consecutive integers is n, n+1 and 2n^2 + 2n + 1.

We can verify this with a few examples:

n= 3, we have the triplet 3, 4, and 2*3^2 +2*3 +1 = 25.

n= 8, we have the triplet 8, 9, and 2*8^2 + 2*8 +1 = 145

neela | High School Teacher | (Level 3) Valedictorian

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To generate the pythagorean triplets with consecutive integers.

We know any three positive numbers of the type a, b and c  are a Pythagorean if a^2+b^2 = c^2.

but our interest is to generate Pythagorean triplets.

We know  (a^2 +x^2 ) ^2 - (a^2-x^2)^2 = 4a^2x^2 is an identity valid for all a and x.

Or (a^2+x^2)^2 - (a^2-x^2) = (2ax)^2. Or

(a^2-x^2)^2 + (2ax)^2 = (a^2+x^2)^2  is an identity so it is true for all x and a.

Therefore (a^2-x^2) , 2ax and (a^2+x^2) is a Pythagorean triplet for all a and x.

Example: a= 13, x = 4. Then a^2-x^2 = 13^2 -4^2 = 153

2ax = 104 and a^2+x^2 =  185.

Therefore 153^2+104^2 = 34225.  but 185^2 = 34225.

Now to get successive integers as triplets:

We know (a^2+x^2)-(a^2-x^2) = 4a^2x^2....(1)

We put x^2 = 1^2 in (1):

(a^2+1)^2-(a^2-1)^2 = 4a^2.

Divide by 4:

{(a^2+1)/}^2 - {a^2-1)/2}^2 = a^2.

Therefore  a^2 + {a^2-1)/2}^2 = {a^2 +1)/2}^2.

Now we can generate two  consecutive integers and another integer as a Pythagorean triplet from the triplet:

a , (a^2-1)/2 and (a^2+1)/2. But  we have to take a as an odd integer.

Example; a = 25. Then (a^2-1)/2 = (25^2-1)/2 = 624/2 = 312.  (a^2+1)/2 = (626/2 = 323.

25^2 +312^2 = 97969

313^2 = 97969.

Therefore 25, 312 and 313 are the Pythagorean triplet with 312 and 313 as consecutive integers.

Example 2: a = 7. (a^2-1)/2 = (7^2-1)/2 = 24 , (a^2+1)/2- (7^2+1)/2 = 25.

7^2+24^2 = 625 which is 25^2.

Therefore 7,24 and 25 are Pythagorean triplet with 24 and 25 as cosecutive integer.

T

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We know that the right triangle whose side lengths have the greatest common divisor 1 is called primitive right triangle.

The Pythagorean triple example is (3,4,5).

GCD(3,4,5)  = 1

We'll note the cathetus lengths: b,c.

The hypothenuse length: a

We'll apply the Pythagorean theorem:

a^2 = b^2 + c^2

To obtain another Pythagorean triple, a number has to be divisible by 3, another has to be divisible by 4 and the 3rd, divisible by 5.

Example: (20,21,29) ; (11,60,61)

Remark the consecutive terms: 20,21 and 60,61

If we have the pattern (a,b, sqrt(a^2+b^2)), where a = odd number, b = consecutive even number, we can plug in values for a and obtain the Pythagorean triple.