Prove that $4k(k+1) =5^n - 1$ has infinitely many solutions in positive integers

Question

Similarly, I believe $k(k+1) = 3^n -1$ has only the trivial solution $k = 1, n = 1$

ps: Sorry, I mis-typed the equation. Well, the solution now is fairly trivial. So no need to prove the title equation.

Answer 1

$k(k+1) = 3^n - 1$.

If $n>1$, then $RHS \equiv -1$ (mod 9). But we have $k(k+1) \equiv 2$ (mod 9).

Answer 2

Hint: $(2k+1)^2=4k(k+1)+1$. Substitute.

Answer 3

With the new version of the first equation, you are correct about both equations.

First equation: $4k(k+1) = 5^n - 1$. Rewriting by completing the square (see Thomas Andrews' hint) gives $$ (2k+1)^2 = 5^n $$ which we can see has a solution whenever $n$ is even.

Second equation: Consider it modulo $9$. For $n \ge 2$, $3^n - 1 \equiv 8 \pmod{9}$. On the other hand, $k(k+1)$ is either divisible by $3$, or $\equiv 2 \pmod{9}$. So $n = 1$, and thus $k(k+1) = 2$, implying $k = 1$.