Repdigit Triangular Numbers

Theorem
The only repdigit numbers which are also triangular are:
 * $55, 66, 666$

Proof
Let $\dfrac {d \paren {10^j - 1}} 9$ be a $j$-digit repdigit number.

Suppose it is triangular.

For $d = 1$, write:

Since $3 k + 1 \perp 3 k + 2$, one of them is equal to $2^{j + 1}$ while the other is equal to $5^j$.

This gives $2^{j + 1} \pm 1 = 5^j$.

Since $j \ge 1$, taking modulo $4$:
 * $0 \pm 1 \equiv 1^j \equiv 1 \pmod 4$

so we have:
 * $2^{j + 1} + 1 = 5^j$

By 1 plus Power of 2 is not Perfect Power except 9, the above equation has no solutions for $j > 1$.

For $j = 1$ we have equality.

Hence $1$ is the only triangular repunit.

By Odd Square is Eight Triangles Plus One:
 * $x = 1 + \dfrac {8 d \paren {10^j - 1}} 9$ is a perfect square.

For $d = 2, 4, 7, 9$, $x$ ends in $2$ or $3$, which is not a perfect square by Square Modulo 5/Corollary.

For $d = 3, 8$, $x$ ends in $5$.

Since it is a perfect square, $x$ must end in $25$.

This is satisfied only when $d = 3, j = 1$.

For $d = 3, j > 1$ and $d = 8, j = 1$, $x$ ends in $65$.

For $d = 8, j > 1$, $x$ ends in $05$.

This corresponds to $\dfrac {x - 1} 8 = 3$.

For $d = 5$, $x$ is in the form $44 \dots 41$.

We show that $x$ is square only when $x = 441$.

This corresponds to $\dfrac {x - 1} 8 = 55$.

For $d = 6$, $x$ is either:
 * $49$ for $j = 1$
 * $53 \dots 329$ for $j > 1$, with $j - 1$ $3$'s

we show that $x$ is square only when $x = 49, 529, 5329$.

These correspond to $\dfrac {x - 1} 8 = 6$, $66$ and $666$.

We have hence demonstrated that the only triangular repdigit numbers are:
 * $1, 3, 6, 55, 66, 666$