Triangular repdigits

Question

I would like to know whether $55$, $66$ and $666$ are the only triangular numbers that are repdigits, i.e., numbers at least $10$ whose digits w.r.t. base 10 all agree.

In more sophisticated terms, I am asking about solutions of the equation $$\frac{n(n+1)}{2} = \frac{d}{9}(10^k-1)$$ where $k \geq 2$ and $n$ are positive integers and $d \in \{1,2, \dots, 9\}$.

Here are some quick ideas:

Considering the remainder of triangular numbers modulo 100 yields that this can only work for $d \in \{1,5,6\}$. I only looked into the case $d = 5$ in greater detail. Then, the equation above can be rephrased as $$(6n+3)^2 = 4\cdot10^{k+1}-31.$$ If $k$ is odd, there are no solutions as $31$ is no difference of two squares. (Alternatively, use that the RHS is $\equiv 6 \ (\text{mod. } 11)$, but $6$ is a quadratic nonresidue of $11$.)

For $k$ even, we would like to find all solutions to the Pell-like equation $$x^2 - 10y^2 = -31$$ where $y$ is twice a power of $10$. Two such solutions are given by $(x_0,y_0) = (3,2)$ and $(x_0', y_0') = (63,20)$. The theory of Pell's equation yields that all solutions of this equation are given by $$\begin{pmatrix}19 & 60 \\ 6 & 19\end{pmatrix}^m \ \textbf{v}$$ where $m \geq 0$ and $\textbf{v} = (x_0, y_0)^{T}$ or $\textbf{v} = (x_0', y_0')^{T}$. But does this really help to prove that none of these has the property that the second component is twice a power of ten unless $m=0$?

Answer 1

Following user523984's suggestion in the comments:

From triangle = repdigit $$\frac{k(k+1)}2 = \frac{d(10^j-1)}9$$ we get $$k = \frac{-1 \pm \frac13 \sqrt{9 + 8d(10^j-1)}}{2}$$ so we require $9 + 8d(10^j-1) = y^2$.

Case-splitting on $j = 3m + i$ we get $y^2 = 8d \cdot 10^i (10^m)^3 + (9 - 8d)$. Substitute $x = 10^m$, $c = 8d \cdot 10^i$ and put into Weierstrass form as $y'^2 = x'^3 + (9 - 8d)c^2$ with $y' = cy$, $x' = cx$. Then we can ask a CAS to find integral points and filter to relevant ones.

For example, in Sage:

# Elliptic curves to solve the triangular repdigit problem
for d in range(1, 10):
    for i in range(3):
        c = 8 * d * 10^i
        ec = EllipticCurve([0, 0, 0, 0, (9 - 8*d)*c^2])
        for xprime, yprime, _ in ec.integral_points():
            if xprime % c == 0 and yprime % c == 0:
                x, y = xprime // c, yprime // c
                # We require x to be a power of 10
                if x == 0:
                    continue

                m = int(log(x) / log(10) + 0.5)
                if x == 10^m:
                    k = (abs(y) / 3 - 1) / 2
                    rep = d * (10^(3*m+i) - 1) / 9
                    if rep > 0:
                        print(k, rep)

print("Search complete")

Run this online at sagecell.sagemath.org

It turns out that Ballew and Weger's result is correct, even though their argument does appear to have some calculation errors (including, but not limited to, an apparent belief that $0$ cannot be a digit in $y$).