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Given the question (from Burton):

"For an arbitrary positive integer $n$, show that there exists a Pythagorean triangle the radius of whose inscribed circle is $n$."

My solution is $3n$,$4n$,$5n$ while textbook hints at a solution of $2n+1$,$2n^2+2n$,$2n^2+2n+1$

The latter seems to describe a Pythagorean triangle with side lengths that form a primitive Pythagorean triple while mine obviously does not except for the case $n=1$. Is that what is expected if the question does not specify? How does one approach generating such a primitive solution?

The text only says " let us define a Pythagorean triangle to be a right triangle whose sides are of integral length"

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Your example fully answers the question as put.

For a primitive triple, suppose it has the standard representation $s^2-t^2,2st,s^2+t^2$ where $s$ and $t$ obey the usual conditions. Then the perimeter $p$ is $2st+2s^2$, and the area $A$ is $st(s^2-t^2)$.

If $r$ is the radius of the incircle, then $rp/2=A$, so $rp=2A$. For an incircle of radius $n$, we want $n(2st+2s^2)=2st(s^2-t^2)$. Cancellation simplifies this to $n=t(s-t)$.

The simplest way to achieve this is to let $s=n+1$ and $t=n$. That ensures $s$ and $t$ are relatively prime and of opposite parity. For $n$ odd we can let $t=1$ and $s=n+1$.

These are the only possibilities for $n=1$ or $n$ an odd prime. Other alternatives are available for $n$ with a more complicated multiplicative structure.

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  • $\begingroup$ That is very clear, thank you $\endgroup$ – topoquestion Jan 25 '16 at 0:42
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Jan 25 '16 at 0:51
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For a triangle with sides $a,b,c$ and area $E,$ and inradius $r,$ let $s =(a+b+c)/2.$ We have $E=rs.$ Now if $a^2+b^2=c^2$ then $E=a b/2$ and $r=E/s=a b/(a+b+c).$ When also $a,b,c$ are positive integers, there are positive integers $k,d,e$ with $\{a,b\}=\{2 k d e, k(d^2-e^2)\}$ and $c=k(d^2+e^2).$ $$\text {This gives}\quad r=\frac {a b}{a+b+c}=\frac {2 k^2 d e (d^2-e^2)}{2 k(d e +d^2)}=k e (d-e).$$ Now given positive integer $n$ let $n=2^p q$ where $p$ is a non-negative integer and $q$ is an odd integer.$$\text {Let }\; k=1,\;e=2^p\text { and}\; d=e+q=2^p+q.$$ Then $r=n$ and $\gcd (d,e)=1, $ while $d,e$ are not both odd, so $\{a,b,c\}=\{2 d e, d^2-e^2,d^2+e^2\}$ is a primitive Pythagorean triplet..... Remark: To show that in any triangle ABC with area E, we have $rs=E \;$: Let $O$ be the center of the incircle $\Sigma$ and let $\Sigma$ be tangent to $BC$ at $A^*$. Since $A^*O$ is perpendicular to $BC$, the area of triangle $BOC$ is $(1/2)(BC)(A^*O)=(1/2)r a.$ Similarly the areas of COA and AOC are $(1/2)r b$ and $(1/2)r c.$ Adding these we have area $E=(1/2)(r a+r b+ rc)=rs.$

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