BMAD Posted July 19, 2013 Report Share Posted July 19, 2013 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? Quote Link to comment Share on other sites More sharing options...

0 k-man Posted July 19, 2013 Report Share Posted July 19, 2013 20,21,29 and 119,120,169 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted July 19, 2013 Report Share Posted July 19, 2013 5 12 13 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted July 19, 2013 Author Report Share Posted July 19, 2013 5 12 13 Hmm do they consider the hypotenuse a leg in America? Quote Link to comment Share on other sites More sharing options...

0 superprismatic Posted July 19, 2013 Report Share Posted July 19, 2013 I've exhausted all triangles whose two legs differ by 1 wherethe length of the shortest leg is less than or equal to 10^{10}.I found only 9:3 4 520 21 29119 120 169696 697 9854059 4060 574123660 23661 33461137903 137904 195025803760 803761 11366894684659 4684660 6625109 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted July 20, 2013 Report Share Posted July 20, 2013 5 12 13 Hmm do they consider the hypotenuse a leg in America? No, they don't. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted July 20, 2013 Author Report Share Posted July 20, 2013 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); Note: 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. Quote Link to comment Share on other sites More sharing options...

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