What percent of primitive Pythagorean triples have an even number as their smallest leg?

Question

What I'm trying to ask here is, if you take a larger and larger set of consecutive primitive Pythagorean triples, what percent of that set will have an even number as their smallest leg? Ex: 8,15,17. There's a way to generate a Pythagorean triple for every odd integer, $a^2+(\frac{a^2-1}{2})^2=(\frac{a^2-1}{2}+1)^2,$ but Pythagorean triples that have a even number as their smallest leg are not so easy. Can anybody help/give suggestions? Thanks!

Answer 1

This turns out to be a reasonably complicated question. To answer a question of the form "what proportion of an infinite set", one first has to decide on an ordering of that infinite set.

The most convenient ordering on Pythagorean triples $(a,b,c)$ comes from the classical parametrization $$ a = k(m^2-n^2), \quad b=k(2mn), \quad c=k(m^2+n^2), $$ where $m>n>0$ are relatively prime integers, not both odd, and $k$ is a positive integer. One can then count approximately how many Pythagorean triples there are with $1\le k,m,n\le x$, and how many of them have either $k$ even or $b$ as the smaller side. Those for which $b$ is the smaller side—that is, for which $2mn < m^2-n^2$, or $(\frac mn)^2 - 2\frac mn-1 > 0$—correspond to numbers $m,n$ with $m>(1+\sqrt2)n$. Out of all pairs with $m>n>0$, this corresponds to a proportion of $\frac1{1+\sqrt2} = \sqrt2-1$. Of course the even $k$ correspond to a proportion of $\frac12$. So the triples $(k,m,n)$ yielding an odd shorter side comprise a proportion $\big(1-(\sqrt2-1)\big)(1-\frac12) = 1-\frac1{\sqrt2}$, meaning that those yielding an even shorter side comprise a proportion $\frac1{\sqrt2}$.

There are some assumptions being swept under the rug—for example, that $k$ being even and $2mn$ being less than $m^2-n^2$ are asymptotically independent; and also that these proportions don't change when we restrict to relatively prime pairs $(m,n)$ that are not both odd. But I believe these assumptions can be verified with a lengthier argument.

So in conclusion: under this ordering, the percentage of Pythagorean triples with the shorter leg even seems to be $\frac1{\sqrt2} \approx 70.71\%$. (And if we restrict to primitive Pythagorean triples—those for which the three sides are relatively prime—then the $k$ variable disappears, and the percentage then becomes $\sqrt2-1 \approx 41.42\%$.)

The most natural ordering probably comes not from saying that $k,m,n\le x$, but rather that all three sides of the triangle are less than $y$, so that $k(m^2+n^2)\le y$. In this case, instead of the proportion of the triangle with vertices $(0,0)$, $(x,0)$, and $(x,x)$ that lies under the line $m=(\sqrt2+1)n$, I believe we should take the proportion of the circular wedge $\{m^2+n^2\le y,\, m>n\}$ that lies under that line—and that proportion turns out to be exactly $\frac12$! So under this ordering, the percentage of Pythagorean triples with the shorter leg even seems to be $\frac34$, and the percentage of primitive Pythagorean triples with the shorter leg even seems to be $\frac12$.

Answer 2

It depends on which formula you use. The one that shows the $trend$ most clearly is one I discovered which generates all triples where $GCD(A,B,C)=(2m-1)^2, m\in\mathbb{N}$. This includes all primitives. F(n,k) produces no trivial triples and uses all natural numbers.

$$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$

where $n,k\in\mathbb{N}$, where $n$ is a set number and $k$ is the member number or "count" within the set. It produces triples that look like the sample of sets shown here where side-$B$ is always even. Note also that, in $Set_1$, the values of $A$ include every odd integer $>1$. $$\begin{array}{c|c|c|c|c|c|c|} n & Triple_1 & Triple_2 & Triple_3 & Triple_4 & Triple_5 & Triple_6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline Set_{6} &143,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 & 253,204,325 \\ \hline \end{array}$$ As you can see in this sample (and in the formula) side-$B$ may begin smaller than side-$A$ in $Set_2$ and above, but the $k^2$ factor always makes it outgrow side-$A$. Offhand, I would say that $B<A$ much less than half the time. I suspect, if you wrote a program to test my conjecture, you would find, going through $n$ sets to the same depth $k$ in each, you would find that the percentage of $B<A$ would decrease the higher the numbers where $n=k$. The number of elements where $B<A$ per set increases with each increase set number forming a roughly $45^\circ$ diagonal through the sets/members but there are none in $Set_1$ and (not seen except for $27,36,45$) the percentage of non-primitives increases with set number and depth.

My guess is that, with the $all\text{-}primitives$ and no $B<A$ in $Set_1$ coupled with a non-primitive any time $k$ is a $1$-or-more multiple of any factor of $(2n-1)$ in $Set_2$ and above, the overall percentage of primitives where $B<A$ will hover around the low $40s\%$.

I did a check in a spreadsheet from $Set_1$ to $Set_{20}$ to a depth of 27 counting primitive where $B<A$. Here are the counts of the first $9$ sets:

$C_1=0\quad C_2=2\quad C_3=3\quad C_4=5\quad C_5=4\quad C_6=7\quad C_7=9\quad C_8=5\quad C_9=13\quad $

For these $9$ sets to a depth of $9$, there are $\frac{48}{81}=59.26\%$ In the $20$ sets to a depth $27$, the highest element in $Set_{20}$ where $B<A$, there are $\frac{230}{540}=42.59\%$. I believe the trend will converge to something near $41\%$ as indicated in Greg Martin's answer.