## Pythagorean Triples

Introduction

A Pythagorean triple is a set of integers (a, b, c) that specifies the lengths of a right angle triangle, i.e.

a2 + b2 = c2

in which a is the shortest side, b the middle side and c is the hypotenuse or the longest side of the right angle triangle as shown in Figure 1.

The first set of triples is (3, 4, 5) because

32 + 42 = 52

The second set of triples is (5, 12, 13) because 52 + 122 = 132, the third set (7, 24, 25) because 72 + 242 = 252, and so on. Thus there are infinitely such sets of triples that exist in this way. I can tabulate some of the first few triples so that I can get my preliminary results.

Preliminary Results

Table 1. First five sets of Pythagorean triples

In Table 1, I have extended some more triples beyond the ones I have got earlier, namely (9, 40, 41) and (11, 60, 61). I got these triples based on the patterns I have noticed in the values of (a, b, c). These triples can also be verified whether or not they are really Pythagorean triples. For example,

92 + 402 = 412 ⇒ 81+ 1600 = 1681

In the table, n is the nth Pythagorean triple. For example, when n = 1, (3, 4, 5) will be the first triple, n = 2, (5, 12, 13) the second triple and so on. By observing the pattern of the sets of Pythagorean triples (a, b, c) in Table 1, I can derive formulas to find each side of any Pythagorean triple.

Finding a formula for the shortest side a

I find from Table 1 that the shortest side a in the second column follows a linear sequence. This is because each time the length of the side increases by 2 for every 1 increase in n. I have shown in Appendix A how one can obtain the formula for the nth term of a more general linear sequence algebraically.

Considering the a values as a linear sequence as:

## Comparing this with the ones derived in Appendix A, I find that

a = 2

and

a + b = 3

Substituting for a ⇒ 2 + b = 3 ⇒ b = 3 – 2 ⇒ b = 1

So the formula to find the shortest side a of the nth Pythagorean triple is:

Finding a formula for the middle side b

I find from Table 1 that the middle side b in the third column follows a quadratic sequence. This is because only the second difference is constant as shown below. I have shown in Appendix B how one can obtain the formula for the nth term of a more general quadratic sequence algebraically.

Considering the b values as a quadratic sequence as:

## Comparing this with the ones derived in Appendix B, I find that

2a = 4 (considering the 2nd difference)

a = 2

3a + b = 8 (considering the 1st term of the 1st difference)

Substituting for a in the above equation ⇒ 6 + b = 8 ⇒ b = 2

Considering the 1st term of the sequence Sn, i.e. S1, and substituting for a and b from above,

a + b + c = 4 ⇒ 2 + 2 + c = 4 ⇒ c = 4 – 4 = 0

So the formula to find the middle side b of the nth Pythagorean triple is:

Finding a formula for the longest side c

I find that the length of the hypotenuse or the longest side c (fourth column in Table 1) is always one more than its corresponding middle side b of the triple. Therefore the expression to find the hypotenuse c of the nth Pythagorean triple can be easily deduced as: