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Pythagorean Triples



The most famous Pythagorean Triangle (a right triangle with all integer sides) is the 3 , 4 , 5 where 3^2 + 4^2 = 5^2. The 'legs', a=3 and b=4, are
consecutive integers, making a nearly-isosceles right triangle. What are the next two smallest integral right triangles with consecutive legs?
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We have a^2 + b^2 = c^2; we are seeking solutions where b=a+1. So, a^2 + (a+1)^2 = c^2 => 2a^2 +2a + 1 - c^2 = 0.
Using the standard formula for solving a quadratic:
a = (1/4)*( -2 +or- sqrt(2^2 - 4*2*(1-c^2))) => a= (1/2)*(-1 + sqrt(2*c^2 - 1)) (eqn 1) [we only need pos sqrt]
For a solution in integers sqrt(2*c^2-1) is an integer, k. Then k=sqrt(2*c^2-1) => k^2 - 2*c^2 = -1.
This is a well known Pell equation which can be solved using continued fractions for sqrt(2). (see below)
Or at least, if we discover the first solution we can generate all others. First few integer solutions for (c,k) are: (1,1);(5,7);(29,41);(169,239);(985,1393);(5741,8119);(33461,47321);(195025,475807);(1136689,1607521);(6625109,9369319).
For the current problem we need only the first couple solutions (the first being trivial and the second given in the question)
Substituting the next four results into (eqn 1) gives:(c,a) = (29,20);(169,119);(985,696) and (5741,4059).
So, the required Pythag. triangles are: (a,b,c) = (20,21,29); (119,120,169);
You can generate the (a,b,c) by using m=2, n=1 ==> (3,4,5) ; then m=prev c = 5, n= prev m = 2 ==> (21,20,29), etc.
These give a ratio of successive perimeters approaching 5.828 = 3 + 2 sqrt[2].
This helps explain the reference to 'continued fraction solution of Pell's equation x^2 - d y^2 = +/- 1.
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