This question was inspired by a tweet by Cliff Pickover.
I'm seeking a right triangle with integer sides, and whose area is a repdigit number greater than $666666$.
I know that Pythagorean Triples can be generated by choosing two integers with $m>n$, and then calculating the triple $(m^2-n^2,2mn,m^2+n^2)$
For the example from the tweet, choosing $m=37$ and $n=26$ gives the triple $(693, 1924, 2045)$.
The area of such a triangle would be $(m^2-n^2)\cdot mn$
I tried using Wolfram Alpha with the command: "Solve $(m^2-n^2)\cdot mn = N$ over the integers", replacing $N$ with $7$ and $8$ digit repdigit numbers. Wolfram timed out when I got to $66,666,666$ and found no examples for repdigits less than that.
Another method I thought about was to take a large repdigit number, then split its prime factors, along with an extra $2$ from the triangle area formula, between the two legs of the triangle and test to see if the hypotenuse was an integer. Another guess and check method that did not bear fruit. This method also reached the limit of online calculators.
So I'm stuck because I don't have a method better than guessing and checking, and I've reached the limit of online calculators.
It feels to me like there must be solutions. However, if a solution exists then finding it will likely involve number theory techniques that are beyond my capabilities. That said, I'm still curious about what responses Math SE will come up with.
Best Answer
I will check up almost all numbers up to 100 digits (except some numbers) that there is no other repdigit which is the area of a Pythagorean Triangle except $666666.$ I will use the CAS Maxima. Its documentation can be found here. I will also explain why I do not expect that there are more numbers with this property.
Let us summarize the facts already stated in your post and in comments: the cathetus of all Pythagorean Triangle can be generated by
$$f(m^2-n^2),\, 2fmn$$ where $m,n,f \in \mathbb{Z^+},$ the corresponding hypothenuse is $f(m^2+n^2)$, the corresponding area $A$ is $$A=f^2(m^2-n^2)mn$$ We are looking for numbers $m,n,f,t \in \mathbb{Z^+}, n<m, d \in \{1,2,3,4,5,6,7,8,9\}$ such that $$\frac{10^t-1}{9}d=f(m-n)(m+n)mn$$ At least one of $n$, $m$ of $m+n$ is even, so $2|A$ snf therefore $d \notin \{1,3,5,7,9\}.$ Either $n$ or one of $(m-n),m,(m+n)$ is divisible by $3$, so $3|A$. Therefore $d=6$ or $d \in \{2,4,8\}$ and $3|t$. $t$ is the number of places of the repdigit.
If we check the area $A$ we can see that $A \bmod 10$ is in $\{0,4,6\}$ That can easily be done by Maxima by iterating over all possible values for $m,n,f \mod {10}$
But repdigit numbers with digit $0$ we can ignore, so we have $d=6$ or $d=4$ and $3|t$
There was another property that I discovered by accident:
So $A \bmod 17$ is never in $\{3,5,12,14\}$
This has the consequence that for repdigit numbers with digit $4$ the numbers where the number of digits cannot be $9,13,14,15 \mod 16$ and for repdigit numbers with digit $6$ the number of digits cannot be $3,7,10 \bmod 16$. But we will not use this in the following investigations.
We can also estimate the size of the factors of $A$:
$$ 1 \le n \le cm \\ m <=m+n \le m(1+c) \\ m(1-c) \le m-n <=m \\ \implies \\ (1-c)m^3\le A \le c(1+c)m^4 $$
$$ cm \le n \le m \\ (1+c)m \le m+n \le 2m \\ 1 \le m-n \le (1-c)m \\\ \implies \\ c(1+c)m^3 \le A \le 2(1-c)m^4 $$
We have $$\min\{c(1+c),(1-c)\} m^3 \le A \le \max\{c(1+c),2(1-c)\} m^4$$ for arbitrary $c$. We choose $c=\sqrt 2-1$ on the left side and $c=\frac 3 2 $ on the right side to get
$$(2-\sqrt 2) m^3 \le A \le \frac 3 2 m^4$$
We also know that not all of the four factors $n,m-n,m,m+n$ of $A$ can be smaller than $\sqrt[4]A$ because otherwise their product would be less than $A$. The smallest of these for numbers ith either $n$ or $m-n$, the largest is $m+n$, the second largest is $m$. So $$\min\{n,m-n\}\le \sqrt[4]A$$
Now how can we check if a given repdigit number $N$ satisfies the equation $N=nm(m-n)(m+n)$? We can check for each factor $b$ of $N$ if there is an appropriate $m$. This means, we solve for each factor $b$ of $N$ the equation $$x^3b-xb^3=N$$ If this equation has an integer solution then we have found a Pythagorean Triangle generated by $(m, n)=(x, b). $ We have ignored that a factor $f^2$ can still be part of the area formula. So we have to repeat this procedure for all equations $$x^3n-xn^3=\frac N {f^2}$$ where $f^2$ is a square factor of $N$. But if we do not find an integer solutions for these equations than $N$ ist not the area of a Pythagorean Triangle.
Actually we do not have to test all $n$ that divide $N$. From the inequalities above we know that $n<frac 1 {2-\sqrt(2)}\sqrt[3]A$. But in my implementation I use the fact that $\min\{n,m-n\}\le \sqrt[4]A$ So we have to solve the equation $x^3b-xb^3=N$ for the divisor $b$ of $N$ or we have $b=m-n$, then we substitute $n$ by $m-b$ in $m^3n-mn^3=N$ and get the equation $$2bx^3-3b^2x^2+b^3x=N$$ Here a integer solution gives the tuple $(m, n)=(x, x-b)$
Here is a sketch of the algorithm in a pseudo programming language. The function
squarepart(n)retrieves the largest square number dividing an integerMaxima has the function
divisors(n)?that returns the list of divisors for a number. And it has the function $realroots(f,bound)$ that retrieves alle real roots of a polynomial. Ifboundis less than $1$ and the $root$ is an integer than Maxima returns this integer. One can check with the functionintegerp(number)if a number is an integerMy procedure in Maxima looks like this
The first for loop loops over all $f$ where $f^2$ is a divisor of the repdigit number $R$. The second one loops all divisors of $\frac R{f^2}$. The third loops constructs from d an N both equations wwe mentioned above. The fourth loop loops over the at most three roots of equations. The if-clause checks if the real solution which always exists, is an integer solution. If so, the result will be displayed.
we get the following result:
for all other numbers I tried, e.g. ((10^60-1)/9*6) I got
This last calculation took about 50 seconds on my notebook.
Finally I execureed the following function
This loops over all repdigit numbers of length less or equal $100$ except that of length $\{71,82,83,84,89,90,93,94,95,96,97,98\}$ and calls the function get_pythbase4 for that number. This gave me the following output
So only for $6666666$ is displayed a solution.
Note 1: The run_loop function that I actually executed still contained checks for M=2 and M=8, as I only realized later that this was not necessary. Therefore my output is a little bit "beautified".
Note2: The reason why I excluded some numbers for the loop was that Maxima was not able to factorize them in reasonable time or, for 84,90, 96, that the number of different prime factors was larger than 20 and therefore the numbers of divisors was larger than $10^6$ which would take a long time to process.
Note3: The procedure does factorization of a the same number B more fpr each f and for each M. That is not very performant, but this makes the source code simpler and I have no interest in optimizing this procedure.
This value converges to $0$ if $N$ goes to infinity. Note 4: The prime factorization of the missing number can be calculated at Dario Alpern's Web site. My algorithms can be modified that it uses the prime factorization supplied by this page to find the missing results.
I want to (over-)estimate the number of Pythagorean Triangles with an area less or equal $N$. So how many tuples (f,m,n)$ are there such that the area they define is smaller $N$. We have
$$\sum_{f=1}^{\sqrt N} \sum_{m=1}^{\sqrt[3]{\frac N {f^2}}}\sum_{n=1}^{m}1 \\ =\sum_{f=1}^{\sqrt N} \sum_{m=1}^{\sqrt[3]{\frac N {f^2}}} m \\ \le \sum_{f=1}^{\sqrt N} \frac12 (\sqrt[3]{\frac N {f^2}} )^2 \\ = \frac12 N^{\frac 23} \sum_{f=1}^{\sqrt N} f^{-\frac 43}\\ \le\frac12 N^{\frac 23} (1+\int_1^{\sqrt N}f^{-\frac 43}df)\\ \frac12 N^{\frac 23} ( 1-(\sqrt N)^{-\frac13}+1)\le N^{\frac 23}$$
So not more than $N^{\frac 23}$ number of the $N$ numbers les or equal $N$ are the area of a Pythagorean Triangle. $N$ has about $\log_{10}N$ digits. Now assume we draw randomly $\log_{10}N$ numbers from these $N$ numbers. Then we expect $$\left(\frac {N^{\frac 23}}N \cdot \frac{\log_{10}N}N \right)N= \frac{\log_{10}N}{N^{\frac 23}}$$ of the drawn numbers are equal to the area of a Pythagorean Triple. This value converges to $0$ if $N$ goes to infinity. So if the set of repdigit numbers less or equal $N$ behaves like a set of the same size of random numbers then I would not expect to find more repdigit numbers that are equal to the area of a Pythagorean Triangle. But of course it is possible that there is a relation between repdigit numbers and the Pythagorean Triangles and they do not bhave like a set of random number.
Bonus: Another digit pattern popular in recreational mathematics are palindromes. To find palindrome Pythagorean Trianles is much easier. Here is a small Python program that checks if Pythagorean Triangle with $1\le n<m\le 1000, f=1$ if its area is a palindrome number
Its output ist
$666666$ is in this list because it is a palindrome, too. An we see there are much more palindromes than repunit numbers that meet our requirement. The number of palindromes less than $N$ is $\sqrt N$, because we can choose the digits of the left halve of a digit an then get the right part my mirroring the left part. The expected number that the elements of a random set of size $\sqrt N$ are Pythagorean Triangle area is therefore $N^{\frac 16}$. So I expect there are Palindrome Numbers of size $N$ that are equal to the area of Pythagorean Triangles even for large $N$. But nevertheless they may be hard to find. But actually I only calculated an upper bound of the number of Pythagorean Triangles and therefore of the expected numbers.