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?
 A: 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 $n = \dfrac{(8j+1)^k-1}2$.
A: Inspired by Robert Israel's answer, here is an alternative proof that $a=1$ cannot occur as the sole digit of a multi-digit triangular number.
If it did, then
$${n(n+1)\over2}={10^k-1\over9}$$
which can be rewritten as
$$(3n+1)(3n+2)=2\cdot10^k=2^{k+1}\cdot5^k$$
On the left hand side we have a product of two consecutive numbers, which are necessarily relatively prime.  Thus one must be the $2^{k+1}$ and the other the $5^k$.  It's easy to see that $2^{k+1}$ and $5^k$ differ by $1$ only for $k=0$ and $k=1$, neither of which gives a multi-digit triangular number.
A: Use induction to show that $\sum_{i=0}^n i = \frac {n(n+1)}{2} $
Then the number with digit $a$, $aaa...a = \sum_{i=0}^n i$ for some $n $
These are the triangular numbers
Now to start the problem we can look at lists of Triangular numbers to see that both 55 and 66 are candidates.
Bellew and Weger showed that 666 is the largest such number.
Then the complete list of repdigit triangular numbers is 55, 66, and 666.
