Are there finitely many triangular powers of $10$? Because strange conjectures have the tendency to enter my mind, I've become convinced that the only powers of $10$ that take the form $\frac{k(k+1)}{2}$ for $k\in \mathbb{Z}^{+}$ (i.e. is triangular) are $10^{0}$ and $10^{1}$. However, I'm having a difficult time making progress towards how to prove or disprove this.  Any thoughts on how we might be ale to show this?  Possible approaches?  Similar problems whose solutions might utilize similar methods?  
 A: $k(k+1)=2^{n+1}\cdot 5^n$
In $k(k+1)$ one factor is odd and one is even.
So necessarily one of them is $2^{n+1}$ and the other one is $5^n$.
If $n\ge2$:
$5^n>4^n=2^{2n}>2^{n+1}+1$
So $2^{n+1}$ is much smaller than $5^n$ for $n\ge2$, thus they cannot be consecutive numbers.
So the only solutions are for $n=0$ and $n=1$.
A: Because $k$ and $k+1$ are coprime, one of them must be the power of $2$ and the other must be the power of $5$.  The exponent on the $2$ has to be one larger than the exponent on the $5$ because of the two in the denominator.  But once $k \gt 1, 5^k \gt 2^{k+1}+1$, so no other powers of $10$ are triangular.
A: Suppose that $k(k+1) = 2 \times 10^n$. Then, we can solve for $k$ in terms of $n$ via the quadratic formula:
$$
k = \frac{1}{2} (\sqrt{8 \times 10^n + 1}-1)
$$
Thus, $k$ will be an integer, whenever $8 \times 10^n + 1$ is a perfect square (the square root will be odd, so the two at the denominator will be cancelled as  the numerator will be even).
Suppose that $8 \times 10^n + 1 = x^2$, then $(x+1)(x-1) = 2^{n+3}5^n$. For all $n \geq 3$, we have $5^n \geq 2^{n+3} + 2$.
Since only one of $x+1$ and $x-1$ can be a multiple of $5$ and hence $5^n$, we must have either:
$x+1 \geq 5^n$, $x-1 \leq 2^{n-3}$: You can see this does not happen at all.
Or:
$x-1 \geq 5^n$, $x+1 \leq 2^{n-3}$: You can see this too does not happen at all.
Hence, only $n=1,0$ remain, giving the required values of $k$. Hence, your conjecture is now a full fledged result. Congratulations.
