Given the Pythagorean triple, a = x 2 − 25 y b = 10 x y c = x 2 + 25 y so a 2 + b 2 = c 2 and y ≠ 0. We wish to set its area G = 5. Thus, G = 1 2 a b = 1 2 10 x y ( x 2 − 25 ) y = 5Or simply, (1) x 3 − 25 x = y 2 Can you help find two non-congruent right-angled triangles that have rational side lengths and area equal to 5?

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Given the Pythagorean triple,                    a                            =                                            x              2                        −            25                    y                                    b                            =                              10            x                    y                                    c                            =                                            x              2                        +            25                    y                    so       a    2    +      b    2    =      c    2   and   y  ≠  0. We wish to set its area   G  =  5. Thus,  G  =      1    2    a  b  =      1    2              10      x        y              (              x        2            −      25      )        y    =  5Or simply,                    (1)                              x          3                −        25        x        =                  y          2                    Can you help find two non-congruent right-angled triangles that have rational side lengths and area equal to 5?

Sawyer Day · Accepted Answer

It turns out that for rational   a  ,  b  ,  c, if      a    2    +      b    2    =      c    2              1      2        a  b  =  nthen   n is a congruent number. The first few are   n  =  5  ,  6  ,  7  ,  13  ,  14  ,  15  ,  20  ,  21  ,  22  ,  …...After all the terminology, I guess all the OP wanted was to solve (1). He asks, &quot;... find two non-congruent right-angled triangles that have rational side lengths and area equal to 5&quot;.We&#039;ll define congruence as &quot;... two figures are congruent if they have the same shape and size.&quot; Since we want non-congruent, then two such right triangles are,  a  ,  b  ,  c  =      3    2    ,        20    3    ,        41    6    a  ,  b  ,  c  =      1519    492    ,        4920    1519    ,        3344161    747348  However, since the elliptic curve,                    (1)                              x          3                −        25        x        =                  y          2                    has an infinite number of rational points, then there is no need to stop at just two. Two small solutions to (1) are       x    1    =      5    2    /      2    2    =      u    2    /      v    2   and       x    2    =  45. We can use the first one as an easy way to generate an infinite subset of solutions. Let,   x  =      u    2    /      v    2  , so,            u      2              v      6        (      u    4    −  25      v    4    )  =      y    2  or just,                    (2)                              u          4                −        25                  v          4                =                  w          2                    Here is a theorem by Lagrange. Given an initial solution       u    4    +  b      v    4    =      w    2  , then further ones are,      X    4    +  b      Y    4    =      Z    2  where,  X  ,  Y  ,  Z  =      u    4    −  b      v    4    ,    2  u  v  w  ,    (      u    4    +  b      v    4        )    2    +  4  b      u    4        v    4  Thus, using Lagrange&#039;s theorem recursively, we have the infinite sequence,  u  ,  v  ,  w  =  5  ,    2  ,    15  u  ,  v  ,  w  =  41  ,    12  ,    1519  u  ,  v  ,  w  =  3344161  ,    1494696  ,    535583225279and so on.

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