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!
What percent of primitive Pythagorean triples have an even number as their smallest leg
number theorypythagorean triplesstatistics
Related Solutions
Great question!
Looking at your data, it seems that you think that not every odd number appears in a primitive Pythagorean triangle. Is that right? And the first number you haven't found is $23$.
It may be a good idea to see if you can specifically find a triangle containing $23$. One thought may be to look at $(a, b, c) = (st, (s^2 - t^2)/2, (s^2 + s^2)/2)$, so try to write $a = st$.
For you, that means $23 = st$. This is a great example to try, since there are only two possible factorizations as $23$ is prime! Either $s = 23$ and $t = 1$, or $s = 1$ and $t = 23$.
Suppose we take $s = 23, t = 1$. Then $(s^2 - t^2)/2 = (529-1)/2 = 264$ and $(s^2 + t^2)/2 = 265$. Checking, we verify that $23^2 + 264^2 = 265^2$, and this is a primitive triple.
So it turns out that $23$ appears after all!
If you carry on with this sort of thinking, you should find that you can describe all odd $a$ appearing in a primitive Pythagorean triangle.
Let's turn this hint into a strategy. You collected data (very good!), formatted it for detecting patterns, and started to identify a pattern.
It can be very hard to identify patterns with incomplete data, so it's a good idea to try to push and prod at your data a little bit. Hopefully you'll find that it's full enough for the pattern to be there.
One particularly good way is to consider explicitly some of the edge cases in your data. In this case, I looked at the smallest odd number that doesn't appear in the data --- only to find that it actually was there. But maybe the next smallest odd number is interesting? Or maybe not!
Good luck!
The only thing I don't [understand] is the last part where it's given as "One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula"
It uses unique fractionization $\Rightarrow$ uniqueness of reduced fractions (with denominators $> 0),\,$ i.e.
$\qquad\qquad \begin{align}\gcd(\color{#c00}{c,b})=1\\ \gcd(j,k)= 1\end{align}$, $\ \ \dfrac{c}b = \dfrac{j}k\ \Rightarrow\ \begin{align} c&\,=\,j\\ b &\,=\, k\end{align},\ \ \ {\rm for}\ \ b,c,j,k\in \Bbb Z,\ b,k > 0$
Follow the link for a simple proof using Euclid's Lemma (hint: $\,j = ck/b\,\Rightarrow\,\color{#c00}{b\mid c}\,k\,\Rightarrow\,b\mid k)$
Remark $ $ A more conceptual way to derive this parametrization of Pythagorean triples is to employ arithmetic of Gaussian integers $\,\Bbb Z[i] = \{ a + b\,i\,: a,b\in\Bbb Z\}$. Like integers they enjoy (Euclidean) division (with smaller remainder) and this implies they too satisfy the analog of the Fundamental Theorem of Arithmetic = existence and uniqueness of factorization into primes (= irreducibles). This implies that coprime factors of a square must themselves be squares (up to unit factors $\,\pm1,\pm i)$
Thus if $\ z^2 = x^2 + y^2 = (x-y\,i) (x+ y\,i) $ and $\,x,y\,$ are coprime then one easily checks that $\,x-y\,i,\,x+y\,i\,$ are coprime, so being coprime factors of the square $\,z^2$ they must themselves be squares (up to a unit factor). Thus e.g. $\ x + y\ i\, =\, (m + n\ i)^2 =\ m^2 - n^2 + 2mn\, i,\,$ hence $\,x = m^2-n^2,\ y = 2mn\,$ (using the unit factor $1$; using the other unit factors $\, -1,\pm i\,$ merely changes signs or swaps $\,x,y\,$ values). Notice how very simple the solution is from this perspective.
This is a simple prototypical (arithmetical) example of the simplification that results from the transformation of nonlinear problems into linear problems by working in larger algebraic extension rings. See here for some further discussion of such.
Best Answer
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 $\frac12$.