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\title{$10/24/26$: Factoring a Pythagorean Date}

\author{Scott Duke Kominers}
\address{Harvard Business School; Department of Economics and Center of
  Mathematical Sciences and Applications, Harvard University; and a16z
  Frontier Research}
\email{kominers@fas.harvard.edu}

\thanks{I used LLMs to assist in preparing this recreational math note,
  particularly GPT-6 Astra and Claude Opus 5.5 (both accessed in part via
  Poe with the support of Quora, where I am an advisor).  The subject
  matter and $\sqrt{a^2+b^2}=c\text{elebration}$ are my own, and of
  course any errors remain my responsibility.}

\date{October 24, 2026}

\begin{document}

\begin{abstract}
Written numerically, the date October 24, 2026 corresponds to a
Pythagorean triple---the last one of the twenty-first century!  The
associated $10$--$24$--$26$ right triangle has area twice its perimeter;
a classical factorization shows that exactly three integer-sided right
triangles have this property and, more generally, counts those with any
prescribed integer inradius.  We also observe that $60$ is the smallest
perimeter shared by two noncongruent integer-sided right triangles, one
of which is the $10$--$24$--$26$ triangle.
\end{abstract}

\maketitle

On October 24, 2026, the date records a Pythagorean triple:
\[
  10^2+24^2=26^2.
\]

\begin{figure}[H]
\centering
\begin{tikzpicture}[x=.085cm,y=.085cm,
  every node/.style={font=\small},
  gridline/.style={draw=black!24,line width=.18pt,line cap=butt},
  squareedge/.style={draw=black!70,line width=.55pt},
  triangleedge/.style={draw=black,line width=1.05pt,line cap=butt}]
  \coordinate (A) at (0,0);
  \coordinate (B) at (24,0);
  \coordinate (C) at (0,10);
  \coordinate (D) at (0,-24);
  \coordinate (E) at (24,-24);
  \coordinate (F) at (-10,0);
  \coordinate (G) at (-10,10);
  \coordinate (H) at (34,24);
  \coordinate (I) at (10,34);

  \fill[black!3] (A)--(D)--(E)--(B)--cycle;
  \begin{scope}
    \clip (A)--(D)--(E)--(B)--cycle;
    \foreach \i in {1,...,23} {
      \draw[gridline] (\i,0)--(\i,-24);
      \draw[gridline] (0,-\i)--(24,-\i);
    }
  \end{scope}
  \draw[squareedge] (A)--(D)--(E)--(B)--cycle;

  \fill[black!6] (A)--(F)--(G)--(C)--cycle;
  \begin{scope}
    \clip (A)--(F)--(G)--(C)--cycle;
    \foreach \i in {1,...,9} {
      \draw[gridline] (-\i,0)--(-\i,10);
      \draw[gridline] (0,\i)--(-10,\i);
    }
  \end{scope}
  \draw[squareedge] (A)--(F)--(G)--(C)--cycle;

  \fill[black!9] (B)--(H)--(I)--(C)--cycle;
  \begin{scope}
    \clip (B)--(H)--(I)--(C)--cycle;
    \foreach \i in {1,...,25} {
      \pgfmathsetmacro{\t}{\i/26}
      \draw[gridline] ($(B)!\t!(C)$)--($(H)!\t!(I)$);
      \draw[gridline] ($(B)!\t!(H)$)--($(C)!\t!(I)$);
    }
  \end{scope}
  \draw[squareedge] (B)--(H)--(I)--(C)--cycle;

  \fill[white] (A)--(B)--(C)--cycle;
  % Draw the sides separately: butt caps stop the bold strokes exactly at
  % the vertices, rather than letting a joined path project past them.
  \draw[triangleedge] (A)--(B);
  \draw[triangleedge] (B)--(C);
  \draw[triangleedge] (C)--(A);
  \draw[line width=.55pt] (0,2)--(2,2)--(2,0);

  \node[fill=white,inner sep=1.5pt] at (12,-12) {$24^2$};
  \node[fill=white,inner sep=1.5pt] at (-5,5) {$10^2$};
  \node[fill=white,inner sep=1.5pt] at (17,17) {$26^2$};
\end{tikzpicture}
\caption{The date triangle and the three squares in its Pythagorean
identity $10^2+24^2=26^2$.}
\label{fig:date-triangle}
\end{figure}

Let $m$, $d$, and $y$ be the month, day, and two-digit year.  We call a
date \emph{Pythagorean} if $m^2+d^2=y^2$; thus the month and day are the
legs and the year is the hypotenuse.

Our particular Pythagorean date gives the triangle
$\Delta:=(10,24,26)=2(5,12,13)$.  The $5$--$12$--$13$ triangle is
equable: its area and perimeter both equal $30$.  Dilation by $2$
multiplies area by $4$ and perimeter by $2$, so our date triangle
satisfies the identity
\[
  \area(\Delta)=120=2\cdot 60=2\perim(\Delta).
\]
This corresponds to an inradius condition.  Indeed, every triangle $T$
of area $K$, semiperimeter $s$, and inradius $r$ satisfies $K=rs$; since
the perimeter is $2s$, it follows that
\begin{equation}\label{eq:area-perimeter}
  \frac{\area(T)}{\perim(T)}=\frac r2.
\end{equation}
Thus the displayed area--perimeter identity says precisely that our date
triangle $\Delta$ has inradius~$4$.

There are exactly two other integer-sided right triangles with the same
area-to-perimeter ratio.  Together, the three form something of a
``Pythagorean trio:''
\[
  (9,40,41),\qquad (10,24,26),\qquad (12,16,20).
\]
We derive this list as the $r=4$ case of a general elementary
classification below.  The factorization approach we use is the
classical argument for equable right triangles, run with an arbitrary
inradius; it appears explicitly in the work of Omland~\cite{Omland}.
Curry~\cite{Curry} instead derives the general classification by first
counting the primitive triangles and then recovering all integer-sided
right triangles by dilation; Robbins~\cite{Robbins} gives a short
independent count of the primitive ones. (Recall that an integer-sided
right triangle is \emph{primitive} if the greatest common divisor of
its three side lengths is $1$.)

Our purpose here is to use the factorization as an economical route to
the particular geometry and calendar consequences of $10/24/26$.  As a
final application, we observe that October 24, 2026 is the last
Pythagorean date of the twenty-first century.

\section*{A factorization}

Let $T$ be a right triangle with integer legs $a$ and $b$, hypotenuse
$c$, and inradius $r$.  The incircle, together with the radii to its
points of tangency on the two legs, cuts off an $r\times r$ square at
the right angle, so both tangent segments from the right-angle vertex
have length $r$.  Writing $x$ and~$y$ for the tangent lengths from the
acute vertices at the far ends of the legs $a$ and $b$, respectively,
equality of tangent lengths gives $a=r+x$, $b=r+y$, and $c=x+y$.  Hence
\begin{equation}\label{eq:inradius}
  2r=a+b-c.
\end{equation}
Moreover, $a+b-c$ is even, because
\[
  0=a^2+b^2-c^2\equiv a+b-c \pmod 2.
\]
Thus the inradius of an integer-sided right triangle is an integer.

\begin{theorem}\label{thm:classification}
For every positive integer $r$, the integer-sided right triangles of
inradius $r$ are in bijection with the unordered factorizations
\[
  \xi\eta=2r^2
\]
into positive integers.  The factorization $2r^2=\xi\eta$ corresponds to
the triangle
\begin{equation}\label{eq:parametrization}
  (a,b,c)=(2r+\xi,\,2r+\eta,\,2r+\xi+\eta).
\end{equation}
Consequently, the number $N(r)$ of such triangles is
\begin{equation}\label{eq:count}
  N(r)=\frac{\tau(2r^2)}{2},
\end{equation}
where $\tau(n)$ denotes the number of positive divisors of $n$.
\end{theorem}

\begin{proof}
By~\eqref{eq:inradius}, we have $c=a+b-2r$.  Substitution into
$a^2+b^2=c^2$ gives
\[
  a^2+b^2=(a+b-2r)^2,
\]
which rearranges to
\begin{equation}\label{eq:key-factorization}
  (a-2r)(b-2r)=2r^2.
\end{equation}
Both factors in~\eqref{eq:key-factorization} are positive; for example,
$2r=a+b-c<a$ because $b<c$.  Thus $\xi=a-2r$ and $\eta=b-2r$ give a
positive factorization of $2r^2$.

Conversely, if $\xi\eta=2r^2$ and $(a,b,c)$ is defined
by~\eqref{eq:parametrization}, then
\[
  a^2+b^2-c^2=4r^2-2\xi\eta=0
\]
and $(a+b-c)/2=r$.  Hence~\eqref{eq:parametrization} gives an
integer-sided right triangle of inradius $r$.  Interchanging $\xi$ and
$\eta$ only interchanges the legs, so unordered factorizations give
precisely the congruence classes of triangles.

Finally, $2r^2$ is not a square.  Its divisors therefore pair off
without a fixed point, giving~\eqref{eq:count}.
\end{proof}

If we have
\[
  r=2^{\alpha}\prod_{i=1}^{\ell} p_i^{\beta_i},
\]
where the $p_i$ are distinct odd primes, then
Theorem~\ref{thm:classification} immediately gives the more explicit
formula
\begin{equation}\label{eq:multiplicative-count}
  N(r)=(\alpha+1)\prod_{i=1}^{\ell}(2\beta_i+1).
\end{equation}

\section*{Power-of-$2$ inradii}

Specializing Theorem~\ref{thm:classification} to $r=2^k$ yields the
family containing both the equable $5$--$12$--$13$ triangle and the
$10$--$24$--$26$ date triangle.

\begin{corollary}\label{cor:power-of-two}
For every integer $k\geq 0$, there are exactly $k+1$ integer-sided right
triangles $T$ satisfying
\begin{equation}\label{eq:power-of-two}
  \frac{\area(T)}{\perim(T)}=2^{k-1}.
\end{equation}
They are indexed by $0\leq j\leq k$ and have sides
\begin{align*}
  a_{k,j}&=2^{k+1}+2^j,\\
  b_{k,j}&=2^{k+1}+2^{2k+1-j},\\
  c_{k,j}&=2^{k+1}+2^j+2^{2k+1-j},
\end{align*}
with $\gcd(a_{k,j},b_{k,j},c_{k,j})=2^j$.
\end{corollary}

\begin{proof}
Equation~\eqref{eq:area-perimeter} turns the
condition~\eqref{eq:power-of-two} into the requirement that the inradius
be $2^k$.  The unordered factorizations of
\[
  2(2^k)^2=2^{2k+1}
\]
are
\[
  2^j\cdot 2^{2k+1-j}\qquad(0\leq j\leq k);
\]
substitution into~\eqref{eq:parametrization} gives the displayed sides.
For the $\gcd$ assertion, it suffices to show that
$\gcd(a_{k,j},b_{k,j})=2^j$.  (For any integer-sided right triangle, the
greatest common divisor of the legs equals that of all three sides: if
$g$ divides both legs, then $g^2$ divides $c^2$, so $g$ divides $c$.)
Put $h=k-j$.  After division by $2^j$, the first two sides are
\[
  2^{h+1}+1
  \quad\text{and}\quad
  2^{h+1}(2^h+1).
\]
Their greatest common divisor is $1$, since $2^{h+1}+1$ is coprime to
$2^{h+1}$, because it is odd, and is coprime to $2^h+1$, because
$(2^{h+1}+1)-2(2^h+1)=-1$.
\end{proof}

By the $\gcd$ assertion, the $j=0$ triangle is the only primitive member
of each family, so for every $k\geq 0$ there is a unique primitive
integer-sided right triangle of inradius $2^k$.  Since
$(a_{k,j},b_{k,j},c_{k,j})=2^j(a_{k-j,0},b_{k-j,0},c_{k-j,0})$, the
$j$-th triangle in Corollary~\ref{cor:power-of-two} is $2^j$ times the
unique primitive triangle of inradius $2^{k-j}$.  Thus the family for
inradius $2^k$ is obtained by doubling every triangle of inradius
$2^{k-1}$ and adjoining one new primitive triangle.

For $k=1$, Corollary~\ref{cor:power-of-two} recovers the classical fact
that the only integer-sided right triangles with area equal to their
perimeter are
\[
  (6,8,10)\qquad\text{and}\qquad(5,12,13).
\]
For $k=2$, the three factorizations
\[
  32=1\cdot 32=2\cdot 16=4\cdot 8
\]
give
\begin{equation}\label{eq:trio}
  (9,40,41),\qquad(10,24,26),\qquad(12,16,20).
\end{equation}
Equivalently, the three triangles are
\[
  (9,40,41),\qquad 2(5,12,13),\qquad 4(3,4,5).
\]
Our Pythagorean date triangle is the middle member; $(12,16,20)$
also corresponds to a Pythagorean date (December 16, 2020).

\begin{figure}[ht]
\centering
\begin{tikzpicture}[x=.14cm,y=.14cm,
  every node/.style={font=\small},
  triangle/.style={line width=.65pt},
  incircle/.style={draw=black!65,fill=black!8,line width=.45pt}]
  \begin{scope}[shift={(0,28)}]
    \draw[triangle] (0,0)--(40,0) node[midway,below] {$40$}
      --(0,9) node[midway,above right=-1pt] {$41$}
      --cycle node[midway,left] {$9$};
    \draw[incircle] (4,4) circle (4);
    \draw[line width=.55pt] (0,1)--(1,1)--(1,0);
    \node[anchor=west] at (43,4.5) {$(9,40,41)$};
  \end{scope}
  \begin{scope}[shift={(0,14)}]
    \draw[triangle,line width=1pt] (0,0)--(24,0) node[midway,below] {$24$}
      --(0,10) node[midway,above right=-1pt] {$26$}
      --cycle node[midway,left] {$10$};
    \draw[incircle,line width=.7pt] (4,4) circle (4);
    \draw[line width=.55pt] (0,1)--(1,1)--(1,0);
    \node[anchor=west] at (43,5)
      {$\boldsymbol{(10,24,26)}$};
  \end{scope}
  \begin{scope}
    \draw[triangle] (0,0)--(16,0) node[midway,below] {$16$}
      --(0,12) node[midway,above right=-1pt] {$20$}
      --cycle node[midway,left] {$12$};
    \draw[incircle] (4,4) circle (4);
    \draw[line width=.55pt] (0,1)--(1,1)--(1,0);
    \node[anchor=west] at (43,6) {$(12,16,20)$};
  \end{scope}
\end{tikzpicture}
\caption{The Pythagorean trio: the three integer-sided right triangles
of inradius $4$, drawn to a common scale.  Their incircles are
congruent.}
\label{fig:inradius-four}
\end{figure}

\section*{A second coincidence}

The perimeter $60$ of the $10$--$24$--$26$ triangle has a further
distinguishing property.

\begin{proposition}\label{prop:perimeter}
The smallest positive integer that occurs as the perimeter of two
noncongruent integer-sided right triangles is $60$.  The two triangles
are
\[
  (10,24,26)=2(5,12,13)
  \quad\text{and}\quad
  (15,20,25)=5(3,4,5).
\]
\end{proposition}

\begin{proof}
Every integer-sided right triangle is an integer multiple of a primitive
one.  By Euclid's parametrization, after interchanging the legs if
necessary, every primitive Pythagorean triangle has a unique pair of
\emph{generators} $u>v>0$ with $\gcd(u,v)=1$ and $u\not\equiv v\pmod 2$,
and its sides are
\[
  u^2-v^2,\qquad 2uv,\qquad u^2+v^2,
\]
giving perimeter $2u(u+v)$.  The primitive perimeters at most $60$ are
thus
\[
  12,\quad 30,\quad 40,\quad 56,
\]
arising respectively from $(u,v)=(2,1),(3,2),(4,1),(4,3)$.  Indeed, when
$u\geq 5$, we have $2u(u+v)\geq 2\cdot 5\cdot 6=60$, with equality only
for $(u,v)=(5,1)$.  That pair is excluded because $5$ and $1$ are both
odd---and fittingly so, since Euclid's formula applied to $(5,1)$
produces the non-primitive triangle $(24,10,26)$, which is precisely our
date triangle.  Hence every admissible pair with $u\geq 5$ gives
perimeter greater than $60$.  In particular, no primitive triangle has
perimeter $60$, and each primitive perimeter in the displayed list comes
from a unique generator pair $(u,v)$.

If two noncongruent integer-sided right triangles have the same
perimeter $P\leq 60$, then their primitive reductions must therefore be
two distinct triangles from the displayed list: a fixed primitive
triangle has only one dilation with perimeter $P$.  Hence $P$ is
divisible by two distinct members of that list.  The smallest least
common multiple of a pair is
\[
  \operatorname{lcm}(12,30)=60.
\]
Thus no $P<60$ can occur twice.  At $P=60$, the primitive perimeters
$12$ and $30$ divide $P$, whereas $40$ and $56$ do not; hence they yield
exactly the two triangles in the statement.
\end{proof}

\begin{figure}[!ht]
\centering
\begin{tikzpicture}[x=.13cm,y=.13cm,
  every node/.style={font=\small},
  triangle/.style={line width=.7pt},
  incircle/.style={draw=black!65,fill=black!8,line width=.45pt}]
  \begin{scope}
    \draw[triangle,line width=1pt] (0,0)--(24,0) node[midway,below] {$24$}
      --(0,10) node[midway,above right=-1pt] {$26$}
      --cycle node[midway,left] {$10$};
    \draw[incircle] (4,4) circle (4);
    \draw[line width=.55pt] (0,1)--(1,1)--(1,0);
    \node[anchor=north] at (12,-4) {$\boldsymbol{2(5,12,13)}$};
  \end{scope}
  \begin{scope}[shift={(34,0)}]
    \draw[triangle] (0,0)--(20,0) node[midway,below] {$20$}
      --(0,15) node[midway,above right=-1pt] {$25$}
      --cycle node[midway,left] {$15$};
    \draw[incircle] (5,5) circle (5);
    \draw[line width=.55pt] (0,1)--(1,1)--(1,0);
    \node[anchor=north] at (10,-4) {$5(3,4,5)$};
  \end{scope}
\end{tikzpicture}
\caption{The smallest pair of noncongruent integer-sided right triangles
with a common perimeter.  Both perimeters are $60$; their inradii are
$4$ and $5$, respectively.}
\label{fig:perimeter-sixty}
\end{figure}

\section*{Coda: a calendar corollary}

\begin{corollary}\label{cor:calendar}
There are exactly twelve Pythagorean dates in each century.  Using the 
$m$/$d$/$y$ convention, they are:
\[
  \begin{array}{llll}
    3/4/05,  & 4/3/05,   & 6/8/10,  & 8/6/10,\\
    5/12/13, & 12/5/13,  & 9/12/15, & 12/9/15,\\
    8/15/17, & 12/16/20, & 7/24/25, & \boldsymbol{10/24/26}.
  \end{array}
\]
Moreover, October 24, 2026 is the last Pythagorean date of the
twenty-first century.
\end{corollary}

\begin{proof}
Let $r$ be the inradius of the triangle corresponding to a Pythagorean
date $(m,d,y)$.  Equation~\eqref{eq:inradius} gives
\[
  2r=m+d-y<m\leq 12,
\]
because $d<y$.  Hence $r\leq 5$.  Theorem~\ref{thm:classification} gives
the following complete lists for these five inradii:
\[
  \begin{array}{c|l}
    r & \text{integer-sided right triangles of inradius }r\\ \hline
    1 & (3,4,5)\\
    2 & (5,12,13),\ (6,8,10)\\
    3 & (7,24,25),\ (8,15,17),\ (9,12,15)\\
    4 & (9,40,41),\ (10,24,26),\ (12,16,20)\\
    5 & (11,60,61),\ (12,35,37),\ (15,20,25).
  \end{array}
\]
Imposing $m\leq 12$ and $d\leq 31$, and retaining both orders of the
legs when both give valid dates, yields exactly the twelve dates
displayed above.  Their largest year entry is $26$, and it occurs only
for $10/24/26$.
\end{proof}

Because the defining equation $m^2+d^2=y^2$ is symmetric in $m$ and $d$,
being Pythagorean is a property of the calendar day itself rather than
of the way it is written: in day/month/year notation, our date is
$24/10/26$, and $24^2+10^2=26^2$ still holds.

The table also shows that no Pythagorean date has inradius $5$, so our
date triangle has the largest inradius possible.  All three triangles of
inradius $3$ occur as dates (in 2015, 2017, and 2025), and two members
of the Pythagorean trio supply the century's last two Pythagorean dates,
$12/16/20$ and $10/24/26$.

Allowing the two-digit year to be a leg gives sixteen further dates per
century, the last two being $7/25/24$ and $10/26/24$.  Thus October 24,
2026 is the last date of the century whose three entries form a
Pythagorean triple in \emph{any} order.  After that, the calendar is
silent until April 5, 2103 ($4/5/03$), and there are no further
Pythagorean dates in our strict sense until March 4, 2105---barring some
truly substantial calendar reform, that is: the third member of the
inradius-$4$ trio, $(9,40,41)$, is still waiting for September 40, 2041.

\enlargethispage{5\baselineskip}

\begin{thebibliography}{9}

\bibitem{Curry}
S.~B. Curry,
\emph{Right triangles with inradius $r$},
Alabama J. Math., Spring/Fall 2005, 1--8.

\bibitem{Omland}
T.~Omland,
\emph{How many Pythagorean triples with a given inradius?},
J. Number Theory \textbf{170} (2017), 1--2.

\bibitem{Robbins}
N.~Robbins,
\emph{On the number of primitive Pythagorean triangles with a given
inradius},
Fibonacci Quart. \textbf{44} (2006), no.~4, 368--369.

\end{thebibliography}

\end{document}