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The following Pythagorean hypotenuses have many possibilities of triangles. $125$ has three triangles $$35, 120, 125$$ $$44, 117, 125$$ $$75, 100, 125$$ the $365$ has $4$ triangles , $85$ has $4$ triangles , $1125$ has $3$ triangles ,$845$ has $7$ traingles, $1885$ has $13$ traingles , $2859545$ has $141$ triangles , and $2859547$ has only one.

We notice that the number of Pythagorean triangle not depend on a clear relation.So, I want to know if there is a specific Pythagorean hypotenuse which gives a maximum number of triangles?

E.H.E
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  • See http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples and http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity – lab bhattacharjee Nov 28 '14 at 17:36
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    One can get arbitrarily many such triangles, even primitive triangles, by using a product of many distinct primes of the form $4k+1$. If you don't care about primitive, you can use $5^n$ for large $n$. – André Nicolas Nov 28 '14 at 17:42
  • The larger the hypotenuse, the larger i the number of triples that $may$ contain it so thee is no hypotenuse with the $largest$ number of triples. – poetasis Aug 03 '19 at 11:16

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You got the most efficient case from Robert in your recent question. Take the primes $$ 5,13,17,29,37,41, \ldots $$ that are the $$ p_i \equiv 1 \pmod 4. $$ Then take hypoteneuse $$ H = 5 \cdot 13 \cdot 17 \cdots p_k. $$ If you allow nonprimitive triangles, the total count of nonzero, ordered, positive triples, giving triangles, is $$ (3^k - 1)/ 2. $$ If you restrict to primitive triangles, it is just $$ 2^{k-1} $$

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Will Jagy
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