Pythagoras Theorem

Pythagoras of Samos

Pythagoras (569 B.C. - 479 B.C.) is a great Greek mathematician who left us few documents to work with. However, the geometric theorems he and his followers developed had certainly made a big impact on modern geometry. For a detailed description of his biography, visit this site.

His most well-known theorem in geometry

Pythagoras Theorem states that, for a right-angled triangle represented by three sides, a, b and c, where a & b form the right angle, and c is the hypotenuse, the equation:

a² + b² = c²

relates the three sides, and the inverse is also true.
For example, a triangle with sides 3, 4 and 5 is right-angled, since
3² + 4² = 5²

How do we find sets of integers that satisfy the equation?

Many formulas abound that allow us to find the triplets. One that I recalled from school days was to square an odd number, a, and calculate b=(a²-1)/2, and c=(a²+1)/2.
If a=5, b=(5²-1)/2=12, c=(5²+1)/2=13, which satisfies
5² + 12² = 13²

However, the greatest formula devised by Brahmagupta in the year 628, according to Heinz Becker Neumuenster, provides ALL the triplets involving a particular number greater than 2, whether it is odd or even. Interestingly, the number of triplets depends on the factors of the square of the number. A prime number will yield only one triplet, so do even numbers not divisible by 4 (4n+2). The following paragraphs describe how the formula works.

We are looking for a triplet of the form:

A² + B² = C²

Where A, B and C are integers (whole numbers).

Furthermore, we look for ALL possible combinations of B and C for a given value of A.

Step 1:
We look for all values of M that are factors of A², and where A-M= a positive even number.
If A is even, the quotient A²/M must also be even to have integral answers.
If A is odd, A²/M must be odd. Since all factors of A² are odd, therefore the quotient is always odd.

For example, for A=15, possible values of M are: 9 (225/9=25), 5 (225/5=45), 3 (225/3=75, and 1 (225/1=225).

For each value of m, calculate the value of B=(A²/M-M)/2
Thus B=(225/9-9)/2=8, or (225/5-5)/2=20, or (225/3-3)/2=36, or (225/1-1)/2=112

Similarly, calculate C=B+M, or
C=8+9=17, or 20+5=25, or 36+3=39, or 112+1=113

A by-product of the calculations is the radius of the inscribed circle (definition) that is completely inside the triangle, and that is tangential to all three sides,

R=(a-m)/2

By excluding M=13, 11 and 7 (not a factor of 225), ALL possible combinations involving A=15 (odd) are:

A	M	B=(A²/M-M)/2	C=B+M	R=(A+M)/2
15	9	8	17	3
15	5	20	25	5
15	3	36	39	6
15	1	112	113	7

For A=20 (even), M=18, 14, 12 and 6 (not a factor of 400) are excluded, M=16 is excluded because 20²/16=25 is not even.

A	M	B=(A²/M-M)/2	C=B+M	R=(A+M)/2
20	10	15	25	5
20	8	21	29	6
20	4	48	52	8
20	2	99	101	9

Thus we now know the formula to help us find all the Pythagoras triplets from a given integer.