Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$?

Question

This question was raised in comments of

Is the $(3,4,5)$ triangle the only rectangular triangle with this property?

and I was suggested to ask it as a separate one.

First some notation, let's write $(a,b,c)$ where $a<=b<c$ for a pythagorean triple, and let's write $(x,y,z)$ where $x<=y<z$ for the distances from the vertices to the center of the inner circle of the corresponding right triangle.

($XYZ$ should be lowercase but geogebra did not allow these labels)

The answer to above question proved (partially left to reader) the property:

$x * y = z$ if and only if $c - b = 1$

In comments was asked if a similar property would exist for $c - b = 2$

and it was confirmed (and proof was left to reader) that:

$x * y = 2 * z$ if and only if $c - b = 2$

Somewhat natural question then was raised (by me) if one can generalize for other values of $n >= 1$, that is:

for which $n$ (perhaps all) holds:

$x * y = n * z$ if and only if $c - b = n$

?

Thanks to @heropup for the suggestion (and the answers for $n=1,2$)

update

A simple computer programmed enumeration seems to confirm equivalence. At least for all $(a,b,c)$ with maximum $c <= 10000$. Note that it is not known to me (but perhaps it is known to others) if all $c - b$ cover all $n >= 1$.

So asking for all $n$ is a bit ambiguous since some $n$ might never occur.

An alternative, perhaps better, question rephrase is:

Prove equality

$$x * y = (c - b) * z$$

for all pythagorean triples.

update see also my answer below for non pythagorean triples

Answer 1

Let $(a,b,c)$ be a primitive Pythagorean triple with $a<b<c$. Then $\{a,b\}=\{2mn,m^2-n^2\}$ and $c=m^2+n^2$ for some coprime integers $m>n>0$ of different parity. If $2mn<m^2-n^2$, then $c-b=2n^2$ is twice a square. Moreover, if $m>3n$, then $2mn<(2/3)m^2<(8/9)m^2<m^2-n^2$, so for every $n$ there exists $(a,b,c)$ such that $c-b=2n^2$.

If $m^2-n^2<2mn$, then $c-b=(m-n)^2$, the square of an odd number. Let $k\ge7$ be odd, let $m=2k-2$, $n=k-2$; then $m,n$ are coprime of different parity, $m>n>0$, and $m-n=k$; also, $m^2-n^2=3k^2-4k$, $2mn=4k^2-12k+8$, and $m^2-n^2<2mn$ is equivalent to $k^2-8k+8>0$, which is $(k-4)^2>8$, which is true for $k\ge7$. So, for every odd $k\ge7$ there exists $(a,b,c)$ such that $c-b=k^2$. For $k=1,3,5$, respectively, we can take $(a,b,c)$ to be $(3,4,5)$, $(33,56,65)$, $(65,72,97)$ to get $c-b=k^2$.

Conclusion: for a primitive Pythagorean triple, $c-b$ can take on every odd square, and every twice-a-square, and only those values.

Given any positive integer $t$, the Pythagorean triple $(3t,4t,5t)$ has $c-b=t$, so, if non-primitive triples are allowed, then $c-b$ can take on any positive integer value.

Now, let's bring in $x,y,z$. Using the formulas at The distance from the incenter to an acute vertex of a right triangle, we have
$2x^2=(a+b-c)^2$
$2y^2=a^2+(c-b)^2$
$2z^2=b^2+(c-a)^2$
If $a=2mn<m^2-n^2=b$, $c=m^2+n^2$, then $$ 4(xy)^2=(2mn-2n^2)^2(4m^2n^2+4n^4)=(4n^2)(m-n)^2(4n^2)(m^2+n^2)=16n^4(m-n)^2(m^2+n^2) $$ so $xy=2n^2(m-n)\sqrt{m^2+n^2}$, and $$ 2z^2=(m^2-n^2)^2+(m-n)^4=2m^4-4m^3n+4m^2n^2-4mn^3+2n^4=2(m-n)^2(m^2+n^2) $$ so $z=(m-n)\sqrt{m^2+n^2}$. Thus, $xy=2n^2z=(c-b)z$, as requested.

If $a=m^2-n^2<2mn=b$, $c=m^2+n^2$, then $$ 4(xy)^2=(2mn-2n^2)^2((m^2-n^2)^2+(m-n)^4)=8n^2(m-n)^4(m^2+n^2) $$ so $xy=\sqrt2n(m-n)^2\sqrt{m^2+n^2}$, and $$ 2z^2=4m^2n^2+4n^4=4n^2(m^2+n^2) $$ so $z=\sqrt2n\sqrt{m^2+n^2}$. Thus, $xy=(m-n)^2z=(c-b)z$, again as requested.

If the triple is not primitive, the common factor cancels out in the end, and the conclusion remains.

Answer 2

Hint: For all primitive Pythagorean triples (right triangles with integer sides), $\space C-B=(2n-1)^2,n\in\mathbb{N}$

This can be seen at a glance using a formula I developed in $2009.$

\begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*}

This formula generates all triples where $\space GCD(A,B,C)\space$ is an odd square. This includes all primitives where $GCD(A,B,C)=1.$

For Pythagorean triples, it appears that $\quad z^2=(2n-1)^2C$

Answer 3

The confusion before comes from not ensuring which of $A$ and $B$ is used in calculations. For example, you thought $\space z^2=10\space$ for $\space (3,4,5)\space$ when the calculations below show that $\space y^2=10,\space$ and $\space z^2=5.\quad$ We begin with my formula for generating Pythagorean triples.

\begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*}

Let $\space A\space$ be the horizontal component and let $\space B\space$ be the vertical component.

\begin{align*} x^2&=r^2+r^2\\ y^2&=(B-r)^2+r^2\\ z^2&=(A-r)^2+r^2\\ \end{align*}

From these equations comes the following table.

By studying these figures we can see numerous relationships such as $\space x^2=2r^2,\space$ and some are unexpected such as how $\space r=(2n-1)k,\quad$ how $\space \dfrac{y^2}{C}=2k^2\,\quad$ and how $\space\dfrac{z^2}{C}=(2n-1)^2.\quad$ Algebraic manipulation of these functions should be able to provide a proof but the visual here is satisfying, even without it.

Answer 4

To show some of the more unexpected relationships, we begin with the formula \begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*} Observation:$\quad \text{inradius}=r=(2n -1)k$

Proof: We let inradius(r)=area(a)/semiperimeter(s) and we let $j=(2n-1)$

$a=\dfrac{AB}{2} =\dfrac{(j^2+2jk)(2jk+2k^2)}{2}\\ \quad=jk(j^2+3jk+2k^2)$

$s=\dfrac{A+B+C}{2} =\dfrac{(j^2+2jk)+(2jk+2k^2)+(j^2+2jk+2k^2)}{2}\\ \quad =j^2+3jk+2k^2$

$r=\dfrac{a}{s} =\dfrac{jk(j^2+3jk+2k^2)}{(j^2+3jk+2k^2)}=jk =(2n-1)k$

Observation: $\quad x^2=2(2n-1)^2k^2$

Proof:$\quad x^2=r^2+r^2=2r^2=2(2n-1)^2k^2$

Observation: $\quad y^2=2k^2C$

Proof: \begin{equation} \quad y^2=r^2+(B-r)^2 =j^2k^2 +(2jk+2k^2\space - \space jk)^2\\ =2 j^2 k^2 + 4 j k^3 + 4 k^4\\ =2k^2(j^2+ 2 j k + 2 k^2)\\ =2k^2C \end{equation}

Observation: $\quad z^2=(2n-1)^2C$

Proof: \begin{equation} \quad z^2=r^2+(A-r)^2 =j^2k^2 + (j^2 +2jk\space - \space jk)^2\\ =j^4 + 2 j^3 k + 2 j^2 k^2\\ =j^2(j^2+ 2 j k + 2 k^2)\\ =(2n-1)^2C \end{equation}

$\textbf{Update:}$ $$\text{Given:}\quad x^2=2 j^2 k^2\quad y^2=2k^2C\quad z^2=j^2C$$

\begin{align*} \dfrac{x^2y^2}{z^2} &=\dfrac{(2 j^2 k^2) (2 k^2 C)}{j^2C}=4k^4\\ \therefore\quad x^2y^2&=4k^4z^2\implies xy=2k^2z \end{align*}

Do keep in mind that side A and side C are always odd and that side B is always even as shown in this table of elements generated by the formula above. Also note how, in row$1,\space C-B=1\space$ and in column$1,\space C-A=2.$ To generalize, as can be seen by glancing at the formula, $\space C-A=2k^2\space$ and $\space C-B=(2n-1)^2.\quad$

\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 \\ \hline \end{array}