Q:

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 $n=1$.

