Solve over positive integers:
$$\lfloor \sqrt{9n^2+1}\rfloor+\lfloor\sqrt{9n^2+2}\rfloor+...+\lfloor\sqrt{9n^2+24n}\rfloor=349 $$
I started with $x-1<\lfloor x\rfloor \le x$:
$$\lfloor \sqrt{9n^2+1}\rfloor > \sqrt{9n^2+1}-1 > 3n-1$$
Using this for each term and adding: $349 > 24n(3n-1)$. This is true for $n=2$, but not for $n=3$. So $n\le 2$. Also $\sqrt{9n^2+24n} < \sqrt{9n^2+24n+16}=3n+4$. So
$$\lfloor \sqrt{9n^2+24n}\rfloor< \sqrt{9n^2+24n} < 3n+4$$
Using this for each term and adding: $349 < 24n(3n+4)$. This is true for $n=2$, but not for $n=1$. So $n\ge 2$. Combining $n=2$. Is this reasoning correct? Can this equation be solved in a easier way?
I would start by noting that there are $24n$ terms, each of which is at least $3n$. We need $72n^2\lt 349$ so $n$ can only be $1$ or $2$. If $n=2$ there are $48$ terms with the range being from $\lfloor \sqrt {37}\rfloor =6$ to $\lfloor \sqrt {84}\rfloor =9$ which is clearly in the ballpark. If $n=1$ there are $24$ terms and the largest is $\lfloor \sqrt {33}\rfloor =5$, which is too small, so the only solution can be $n=2$.
Finally we note that for $n=2$ there are $12$ terms with a floor of $6,\ 15$ with a floor of $7,\ 17$ with a floor of $8,\ $ and $4$ with a floor of $9$. Indeed $$12\cdot 6+15\cdot 7+17\cdot 8+4\cdot 9=349$$