Each digit of $\dfrac{n(n+1)}{2}$ equals $a$ 
For which digits $a$ does there exist an integer $n \geq 4$ such that each digit of $\dfrac{n(n+1)}{2}$ equals $a$? 

I was first thinking of looking at $\dfrac{n(n+1)}{2} \pmod{100}$, but it doesn't look like it has a small period so it may be hard to arrive at a conclusion. Is there an easier way to solve this question?

Note that $n(n+1)/2 = z$  has an integer solution iff $1+8z$ is a square.
So the question is whether $1 + \dfrac{8 a}{9} (10^k-1)$ can be a square, or equivalently whether $9 - 8 a + 8 a \cdot 10^k$ can be a square.
You can rule out $a = 2, 3, 4, 7, 8, 9$ this way because $9-8a$ is not a square $\mod (8 a \cdot 10^k)$ for $k=1$ or $2$.
That leaves $a=1$.
EDIT: For $a = 1$, you want $1 + 8 \cdot 10^k$ to be a square with $k \ge 2$.  Suppose it is $x^2$.  Since $x^2 \equiv 1 \mod 5^k$, we may assume  $x \equiv 1 \mod 5^k$.
So let $x = 1 + 5^k y$.  Then we have $1 + 2 y \cdot 5^k + y^2 \cdot 5^{2k} = 1 + 8 \cdot 10^k$, or $2 y + y^2 \cdot 5^k = 2^{k+3}$.  In particular, $y \mid 2^{k+3}$, so $y$ is a power of $2$.  If $y = 2^j$, the equation becomes
$2^{j+1} + 2^{2j} \cdot 5^k = 2^{k+3}$.  Considering factors of $2$, two of $j+1, 2j, k+3$ must be equal.  But it's easy to rule out all these possibilities.
See also OEIS sequence A045914.
EDIT: It's also interesting to consider other bases than $10$.  Here there are some infinite families of solutions, as well as some (such as $52 \dfrac{98^4-1}{98-1} = \dfrac{9944 \cdot 9945}{2}$ or $105 \dfrac{159^6-1}{159-1} = \dfrac{4634175 \cdot 4634176}{2}$) for which I don't see an obvious pattern.
In base $8j+1$, where $j$ is an integer, $j \dfrac{(8j+1)^{2k}-1}{8j+1-1} = \dfrac{n(n+1)}{2}$ where $n = \dfrac{(8j+1)^k-1}2$.
