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

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 $$a<=b for a pythagorean triple, and let's write $$(x,y,z)$$ where $$x<=y 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.

• Just curious : are these properties 'well known'? If well known, perhaps a self contained proof (perhaps using other 'well known' properties with self contained proofs recursively) exists somewhere? I personally still fail to bridge the required gaps kindly offered as exercise by the experts who answered. Jul 12, 2022 at 23:37
• Primitive triples have $\{a,b\}=\{2mn,m^2-n^2\}$, $c=m^2+n^2$ where $m,n$ are of different parity and $m>n>0$. If $m^2-n^2<2mn$, then $c-b=m^2+n^2-2mn=(m-n)^2$ is the square of an odd number; if $2mn<m^2-n^2$, then $c-b=2n^2$ is twice a square; so the only possibilities for $c-b$ are $1,2,8,9,18,25,32,49,50,\dots$. If we allow nonprimitive triples, then of course every value for $c-b$ is possible; to get $c-b=t$, just use the triple $(a,b,c)=(3t,4t,5t)$. Jul 19, 2022 at 6:54
• Jul 19, 2022 at 7:35

Let $$(a,b,c)$$ be a primitive Pythagorean triple with $$a. Then $$\{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, then $$c-b=2n^2$$ is twice a square. Moreover, if $$m>3n$$, then $$2mn<(2/3)m^2<(8/9)m^2, so for every $$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 $$a=2mn, $$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.

• Thanks for this easy (for me) to follow and complete proof! Was this a known (to you) or well known property? Jul 21, 2022 at 9:56
• No. All I knew was the $2mn,m^2-n^2,m^2+n^2$ parametrization of Pythagorean triples, the rest I worked out or, in the case of the formulas for the distance from a vertex to the incenter, searched for on the web. Jul 21, 2022 at 12:23
• Big thanks and congratulations. I am happy to hear this is not so common knowledge which I rather incidentally found. Jul 21, 2022 at 12:38
• @FirstName LastName Yesterday, my reputation dropped $15$ points, because one of my answers to this question was no longer marked as correct. Did I do something to warrant this? Please let me know. I’m hoping that my answer with the drawings might be correct if the other was not. Jul 22, 2022 at 16:54
• @FirstName LastName Starting a bounty would cost you reputation for no good reason. I can manage without it. Glad I could help and it was a fascinating problem to solve. Jul 22, 2022 at 19:18

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

• Sorry for late reaction. I probably misunderstand, and, also, I am not expert at all of Pythagorean triples. Are you telling us you have a proof for the general case of any $n$? Also: $C-B$ does not need to be a square, does it? What about $(8,15,17)$ where $C-B=2$. I am sure I missed a point here and I apologise. Jul 10, 2022 at 21:32
• @FirstName LastName $\qquad$ Do not use $(8,15,17).\quad$ In primitive triples, half have $B>A$ like $\space(3,4,5), (5,12,13)\space$ and half have $A>B$ like $\space(15,8,17), (33,12,37).\quad$ We can see that $\space 5-4=13-12=1,\space$ $\space 17-8=9,\space$ and $\space37-12=25,\space$ and that $C-B$ is an odd square. For proof, start playing around with algebra using the functions of my formula for sides. Jul 10, 2022 at 22:24
• @FirstName LastName Euclid's formula reverses the order of $A$ and $B$ when generating imprimitive triples; mine does not. I developed my formula without even knowing that Euclid’s formula exists. Later I found that Euclid’s formula generates the same triples as mine if you let. \begin{align*} A= &(2n-1+k)^2-k^2\\ B=2&(2n-1+k)k\\ C=&(2n-1+k)^2+k^2 \end{align*} It will take me time to work on a proof but I will try. Jul 11, 2022 at 0:51
• @FirstName LastName Perhaps I got sloppy with my math or it only works for $n>1$. It will take me time to check because I have very limited time available for MSE. Jul 11, 2022 at 23:22
• @FirstName LastName Your example of $(15,8,17)$ has $C-A=2$ and $C-B=9=3^2=\big(2(2)-1\big)^2=(2n-1)^2.$ You seem to keep confusing A and B. For simplicity, try to remember that A is always odd. Jul 12, 2022 at 0:41

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.

• You are right, I will change picture to reduce confusion (note geogebra did not allow me to use lowercase x y z so I used uppercase). Jul 12, 2022 at 21:19
• @FirstName LastName I see nothing wrong with the picture, BTW, is this answer correct for your OP question? Jul 12, 2022 at 21:20
• Proof of math.stackexchange.com/questions/4488978 also left some exercise to the reader and I ticked it as answer, so I want to be fair and will tick yours, which also leaves some exercise to the reader as well. But, for both questions I am not able as reader to complete the exercise ... but then, I am not all that skilled :-) which is why I ask ... Jul 12, 2022 at 21:38
• @FirstName LastName I’ll see what I can do algebraically now that we know the line length relationships Jul 12, 2022 at 22:20

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 $$\space C-B=(2n-1)^2.\quad$$

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

• @FirstName LastName What other relationships would you like to prove? BTW $\text{circumradius}(c)=\dfrac{product}{4*area} =\dfrac{ABC}{4\bigg(\dfrac{AB}{2}\bigg)}=\dfrac{ABC}{2AB} =\dfrac{C}{2}$ Jul 16, 2022 at 23:34
• @FirstName LastName I have updated my answer to show $xy=2k^2z.$ I am concerned, however, that you may still sometimes mistake which side is A and which side is B. See the note in my update. Jul 17, 2022 at 15:41
• @FirstName LastName $\quad \text{for }\space(3,4,5),\space x=\sqrt{2}\quad y=\sqrt{10}\quad z=\sqrt{5}$ $$xy=\sqrt{2}\sqrt{10}=\sqrt{20}=2\sqrt{5}=2z$$ Jul 17, 2022 at 23:06
• @FirstName LastName $\quad y>z\space$ because $A=3,\space B=4.\quad$ I repeat that you must keep track of what is the horizontal and what is the vertical component of the triples. $\quad A<B\implies z<y.$ Jul 17, 2022 at 23:50
• @FirstName LastName I added a drawing to my answer so you can see, with A being horizontal, and B being vertical, why z is smaller than y for a 3,4,5 triangle. For triangles where $A>B$, like $(15,8,17)$, line-z would be the longest. Jul 19, 2022 at 3:54

Incidentally, I found a bit more general evidence.

I leave some details as exercise but, one finds and has

$$y^2=c(c-b)$$

and

$$z^2=c(c-a)$$

so

$$y^2=z^2(c-b)/(c-a)$$

Call $$r$$ radius of inscribed circle, then, one finds and has

$$r=(a+b-c)/2$$

and

$$x^2=2r^2$$

So one further calculates (by squaring $$r$$)

$$x^2=(c-a)(c-b)$$

But then

$$x^2y^2=(c-a)(c-b)z^2(c-b)/(c-a)=z^2(c-b)^2=z^2n^2$$

So eventually

$$xy=(c-b)z=nz$$

This holds for any rectangular triangle. It just so happens that for pythagorean triples n is a natural number making it look a bit more appealing :-).

So for example (no natural pythagorean triple) with

$$a=1$$

$$b=2$$

$$c=\sqrt 5$$

geogebra tells me (up to only 4 decimals as limited check)

x=0.5402

y=0.7265

z=1.6625


julia tells me

julia> a=1
1

julia> b=2
2

julia> c=sqrt(a^2+b^2)
2.23606797749979

julia> sqrt(5)
2.23606797749979

julia> x=0.5402
0.5402

julia> y=0.7265
0.7265

julia> z=1.6625
1.6625

julia> sqrt(c*(c-b))
0.726542528005361

julia> sqrt(c*(c-a))
1.6625077511098139

julia> c-b
0.2360679774997898

julia> x*y
0.3924553

julia> (c-b)*z
0.39246301259340055


So that seems to confirm (up to limited amount of decimals).