Show that when uv1 for au2v2 b2uv and cu2v2 we get a primiti

Show that when u=v+1 (for a=u^2-v^2, b=2uv, and c=u^2+v^2), we get a primitive Pythagorean triple where b and c differ by 1. Hence deduce that there are infinitely many such triples, and show that in this case b is always four times a triangular number.

Solution

Let us first write the given three equations in the questions.

a = u² - v², b = 2uv, c = u² + v²

Also, u = v+ 1 which we will put in above three equations and get the value of a, b and c in the form of v only instead of both u and v. We get:

a = 2v + 1

b = 2v² + 2v

c = 2v² + 2v + 1

which is a primitive Pythagorean triple.

Now we can see that b and c differ by 1 as 2v² + 2v will get cancelled out and only 1 will be left.

Hence proved.

By Schinzel and Sierpinski\'s Hypothesis, we then expect to see infinitely many triples with two prime entries. Here are the first few:

a = Prime leg

b= Even leg

c = Hypotenuse

5

12

13

11

60

61

19

180

181

29

420

421

59

1740

1741

61

1860

1861

71

2520

2521

Now from the above table, we can easily see that b is always a multiple of 3 and always four times a triangular number.