5
$\begingroup$

Is there exists two pythagorean triples $(a,b,c)$ and $(b,c,d)$ such that $a < d$ ?

Any idea to prove or disprove ?

  • 0
    Euclid's formula, perhaps? – 2017-01-29
  • 1
    If not restricted to primitive, [**spiral of Theodorus**](https://en.wikipedia.org/wiki/Spiral_of_Theodorus) is the way. – 2017-01-29
  • 0
    Pythagorean triple only defines of integers – 2017-01-30
  • 0
    Well, it's obvious that if there exists a non-primitive solution, then there exist a primitive solution. – 2017-01-30
  • 2
    http://math.stackexchange.com/questions/1844034/for-which-n-can-a-nb-c-and-b-c-d-be-pythagorean-triples – 2017-01-30
  • 0
    Fermat's resolution of this question is given at https://en.wikipedia.org/wiki/Fermat's_right_triangle_theorem – 2017-01-30

1 Answers 1

3

I understand you want $a^2+b^2=c^2$ and also $b^2+c^2=d^2$ which means you really want 3 evenly spaced squares $a^2$, $c^2$ and $d^2$ that are separated by $b^2$. You can have evenly spaced squares like 49, 169 & 289 which are separated in that case by 120 but if you have evenly spaced squares they cannot be separated by a square. This would lead to an infinite regress and is sort of related to Fermat's Last Theorem. For more info see Wikipedia article on Congruum and also Fermat's Right Triangle Theorem.