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If the inradius=$2013$ of a right angled triangle with integer sides. Find the no. of possible right angled triangles that can be formed using the above information. I have tried $r(a+b+c)=ab$ and $a^2+b^2=c^2$ , but couldn't reach further. Thanks in anticipation

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One thing we can get is

$CF=CD=b-r$ and $BE=BD=c-r$. So, $BC=a=b+c-2r\quad (1)$.

By Euclide's formula we can look for a solution such that $\gcd(a,b,c)=1$ and $b=m^2-n^2$, $c=2mn$ and $a=m^2+n^2$ with, $\gcd(m,n)=1$ and not both odd.

We can then back to $(1)$ and get

$$a+2\cdot2013=b+c\\ m^2+n^2+2\cdot2013=m^2-n^2+2mn\\ n^2+2013=mn\to n(m-n)=2013$$

so, $n|2013$ and once $2013=3\cdot11\cdot61$ then we have the possibilities:

$$n\in\{3,11,61,3\cdot11,3\cdot61,11\cdot61,3\cdot11\cdot61\}$$

So you just have to try all possibilities for $n$ and find a suitable $m$.

For example, for $n=3$ you will find $m=674$ and then you have one solution which is $(a,b,c)=(674^2+3^2,674^2-3^2,2\cdot3\cdot674)$.

Now just find the others.

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  • $\begingroup$ you have shown the no. of possible primitive pythagoras triples and not in general. $\endgroup$ – Chandramauli Chakraborty Jun 1 '17 at 12:40
  • $\begingroup$ @ChandramauliChakraborty: I know that, but I think you have enough information to go on and find the others. $\endgroup$ – Arnaldo Jun 1 '17 at 12:46
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Answer to the question is easy: 8 primitive Pythagorean triples are there for the inradius 2013. There are no other triples (non-primitive Pythagorean triples).

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  • $\begingroup$ Sorry, there are also non-primitive pythagorean triples $\endgroup$ – Chandramauli Chakraborty Jun 12 '17 at 18:20
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My extended Answer to the question: 2013 has 3 prime factors such that 2013 = 3*11*61, thus, number of Pythagorean triples with inradius = 2^3 = 8. Therefore, there are 8 Pythagorean triples with the given inradius 2013. Ironically, all of them are primitive and no non-primitive Pythagorean triples present for the inradius of 2013.

Reference: Neville Robbins, On the number of primitive Pythagorean triangles with a given inradius, Fibonacci Quarterly 2006, 44(4), pp. 368–369.

For total Pythagorean triples, read: Tron Omland, How many Pythagorean triples with a given inradius?, Journal of Number Theory 2017, 170(1), 1–2.

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