Pythagorean Triangles Research Paper

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The Right- Angle Triangle and the Pythagorean Triples
Written by Nana Ekua Opoku

INTRODUCTION
It can be said that the Pythagorean Triple was derived by the Greek Philosopher and Mathematician Pythagoras and it is closely associated to the right angled triangle. The Pythagorean triple represents three positive integers namely a,b and c where a^2+b^2=c^2and they can be written as (a,b,c)
It can be said that the Pythagorean Triples were derived from Pythagoras’ theorem, a very simple theorem which is widely used when right-angled triangles are concerned and it states that the sum of the squares of the sides of a right-angled triangle is equal to the square of the hypotenuse (longest side). Thus the formula a^2=b^2+c^2 where is the hypotenuse. …show more content…

The Pythagorean Triples will represent three positive integers that will satisfy the formula a^2+b^2=c^2 thus the square of the hypotenuse (longest side) c will be equal to the square of the sides a and b.

9 + 16 = 25

3 4 5
Therefore:
3^2+4^2=5^2
9+16=25
Thus (3,4,5) are Pythagorean triples.
FIRST DERIVATION OF PYTHAGOREAN TRIPLES:
Generally when trying to derive the Pythagorean triple, we begin by finding the algebraic sum of the squares of the two smaller numbers which in turn will be used to derive the square of the hypotenuse that satisfy the condition of the square of the two smaller numbers.
The square of one of the Pythagorean Triples can either be an even or an odd number. Thus supposing a^2 is the square of one of the Pythagorean Triples and b is the second triple. Therefore:
Using a case where a^2 is odd: a^2=2b+1 When it is odd,
Making b the subject: a^2-1=2b Divide both sides by 2:
((a^2-1))/2=2b/2
∴b=((a^2-1))/2
Making b+1 the subject:
Dividing both sides by 2: a^2/2=2b/2+1/2 Adding 1/2 to both sides a^2/2+1/2=b+1/2+1/2 b+1=((a^2+1))/2
The hypotenuse should be:
a^2+b^2=b^2+2b+1 …show more content…

I realized a unique pattern about the table above; for all the triples obtained, when arranged in order of magnitude, have a difference of 1 between the last two numbers. Thus a triple like 9,12,15 will not be obtained even though 9 is an odd square number. This pattern was unique but was the limitation of my formula.
I have been able to derive a general formula which can help generate a number of triples; however, the formula can only be applied if at least one of the triples is known, like a^2 and this known triple must specifically be an odd square number. This makes my formula somewhat difficult to one that does not have prior knowledge to what a square odd number is. Therefore, to eliminate this limitation, I chose to generate another formula that can help derive triples without prior knowledge of one of the triples, meaning a formula that enables the generation of triples from any real number.
From my first formula, the triple b is restricted to even numbers, so b=2x, where x is a real

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