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Are there any integer solutions to the Diophantine equation $$ 10^k + 332 = 3n^2 $$ other than the solutions given by $(k, n) = (2, \pm 12)$?

One approach would be to consider that if $k$ is even, then solutions correspond to solutions to the Pell-like equation $$ x^2 - 3y^2 = -332 $$ in which $x$ is a power of $10$, and if $x$ is odd, then solutions correspond to solutions to the Pell-like equation $$ x^2 - 30y^2 = -3320 $$ in which $x$ is a power of $10$. (Alternatively, they also correspond to solutions to the Pell-like equation $$ x^2 - 30y^2 = 996 $$ where $y$ is a power of $10$)

We can then solve these Pell equations, and check when the solution we get for $x$ is a power of $10$. One possible approach for this is to use various lower bounds for linear forms in logarithms (one which I am familiar with is due to Matveev) to find bounds on the power of $10$, and then do a brute force search for solutions below this bound. The bound which I obtained in the cases which I worked out explicitly is very large (for one of the cases that arises when $k$ is even, I get that $k$ is at most a number which is approximately $10^{15}$), however, but I am at least confident that the number of solutions is finite.

Is there any simpler approach? (Using more elementary methods?) Is there an approach for bounding the solutions for the Pell equations which give reasonable bounds which can be checked fairly quickly?

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Simpler? Probably not, but there are 3 cases: If $k$ is a multiple of 3, the the equation can be written $y^3+332=3n^2$. If $k$ is of the shape $3m+1$, then you can multiply the equation by $100$ and write it as $y^3 +33200=300n^2$. If $k$ is of the shape $3m+2$, multiply by $10$ and write $y^3 + 3320 =30n^2$. Then fire up Pari or Sage and find all the integer points on these three elliptic curves. If any of the $y$ values are a power of $10$, you might have a solution. (Yeah, definitely not simpler. But maybe interesting anyway.)

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