Considering all +ve int n>=0, such that 5*3^m + 4 = n^2 holds for some corresponding int m>=0. What is the sum of all such n?

Question

Considering all +ve int n>=0, such that 5*3^m + 4 = n^2 holds for some corresponding int m>=0. What is the sum of all such n ? After trying many ways, I got two solutions of m,0 and 2. But I do not know how to prove it. And one thing, please do not take it as a homework question.

Answer 1

Suppose we had a solution $5\times 3^m+4=n^2$ We'd deduce that $n^2-4=5\times 3^m$ hence $$(n-2)(n+2)=5\times 3^m$$

Now, the gcd of $n-2$ and $n+2$ is a divisor of $4$ (as the two numbers differ by $4$). In particular, that gcd is prime to both $3$ and $5$. Indeed, as $n$ is clearly odd we can deduce that $\gcd(n-2,n+2)=1$. Thus one of the two factors must be divisible by $3^m$ and the other must be $5$ or $1$.

Since we know that $n>2$ (as $n^2>4$ and $n>0$) we have $n+2≥5$. If $n+2=5$ then $n=3$ and we get a solution (with $m=0$). Otherwise $n+2>5$ so we must have $n-2=5$ or $n=7$. That too gives a solution, with $m=2$ Thus, these are the only two solutions and the answer is $3+7=10$.