Is there exists two pythagorean triples $(a,b,c)$ and $(b,c,d)$ such that $a < d$ ?
Any idea to prove or disprove ?
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Euclid's formula, perhaps? – svelaz Jan 29 '17 at 21:43
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1If not restricted to primitive, spiral of Theodorus is the way. – Ng Chung Tak Jan 29 '17 at 22:55
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Pythagorean triple only defines of integers – Kevin Jan 30 '17 at 03:38
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Well, it's obvious that if there exists a non-primitive solution, then there exist a primitive solution. – Kevin Jan 30 '17 at 03:47
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2http://math.stackexchange.com/questions/1844034/for-which-n-can-a-nb-c-and-b-c-d-be-pythagorean-triples – Saketh Malyala Jan 30 '17 at 07:27
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Fermat's resolution of this question is given at https://en.wikipedia.org/wiki/Fermat's_right_triangle_theorem – Gerry Myerson Jan 30 '17 at 08:20
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Using Euclid's formula, we take the smallest example $5,12,13$ and solve $B$ for $n$: b – poetasis Jul 05 '19 at 17:52
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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.
Uncle Buck
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