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Simultaneous Pythagorean Poser 2 (Posted on 2015-09-05) Difficulty: 3 of 5
Find all possible triplets (A, B, C) of positive integers, with A ≤ B, that satisfy this system of equations:

A2 + B2 = C2, and

A+B = 2C - 99

No Solution Yet Submitted by K Sengupta    
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Solution, with thanks to the classical Greeks | Comment 3 of 4 |
Take the parametric solution for Pythagorean equation found by the Greeks, a=k*2pq, b=k*(p^2 - q^2), c= k*(p^2 + q^2) and substitute into the second equation.

p^2 - 2pq +3q^2 - 99/k = 0 

Solving this for p requires discriminant = 396/k - 8q^2 which will be a square.

So k=1,3,9,11,33,99 and checking each in turn gives these solutions:

   a       b       c
 27     120   123
 55     132   143
 72     135   153
 84     135   159
 99     132   165
   

  Posted by xdog on 2015-09-05 15:53:14
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