Very easy problem. Answer is all square numbers except 0,1,9.

Just set m=tn and calculate.

Here is my solution(a little lengthy): Assume $m=\frac{a}{x},n=\frac{b}{x} \text{ where} (a,b,x)=1,abx \neq 0$ If $(m,n)$ is a solution, so is $(-m,-n)$, therefore we can assume that $b,x>0$ Then the original equation becomes $$a^3=(27b^2+x^2)(a+2b)$$If $a>0$, then since $a+2b \mid 8b$, $a+2b=\frac{8}{3}b,4b,8b \Longrightarrow$ all cases lead to contradiction. If $a<0$, plugging in $-a$ instead of $a$, we will just solve $$a^3=(27b^2+x^2)(a-2b)$$$\Longrightarrow a>2b \Longrightarrow a=2b+c (c>0)$ $\Longrightarrow (2b+c)^3=(27b^2+x^2)c \text{ and } c \mid 8b (=sc)$ $\Longrightarrow (s+4)^3c^2=(27s^2c^2+64x^2)$ $\Longrightarrow (s+4)^3=27s^2+(\frac{8x}{c})^2$ $\Longrightarrow (s+1)(s-8)^2=(\frac{8x}{c})^2$ $\Longrightarrow s+1=k^2$ If $s=-1,8$ then $x=0$ and if $s=0$ then $b=0$ Therefore, the answer is $k^2 (k\neq0,1,3)$ (finding solutions for $k\neq0,1,3$ is easy.) I didn't know that $\frac{m-6n}{m+2n}=\frac{1}{(m+3n)^2}$. It could make the solution much shorter..

MNJ2357 wrote: Here is my solution(a little lengthy): Assume $m=\frac{a}{x},n=\frac{b}{x} \text{ where} (a,b,x)=1,abx \neq 0$ If $(m,n)$ is a solution, so is $(-m,-n)$, therefore we can assume that $b,x>0$ Then the original equation becomes $$a^3=(27b^2+x^2)(a+2b)$$If $a>0$, then since $a+2b \mid 8b$, $a+2b=\frac{8}{3}b,4b,8b \Longrightarrow$ all cases lead to contradiction. If $a<0$, plugging in $-a$ instead of $a$, we will just solve $$a^3=(27b^2+x^2)(a-2b)$$$\Longrightarrow a>2b \Longrightarrow a=2b+c (c>0)$ $\Longrightarrow (2b+c)^3=(27b^2+x^2)c \text{ and } c \mid 8b (=sc)$ $\Longrightarrow (s+4)^3c^2=(27s^2c^2+64x^2)$ $\Longrightarrow (s+4)^3=27s^2+(\frac{8x}{c})^2$ $\Longrightarrow (s+1)(s-8)^2=(\frac{8x}{c})^2$ $\Longrightarrow s+1=k^2$ If $s=-1,8$ then $x=0$ and if $s=0$ then $b=0$ Therefore, the answer is $k^2 (k\neq0,1,3)$ (finding solutions for $k\neq0,1,3$ is easy.) I didn't know that $\frac{m-6n}{m+2n}=\frac{1}{(m+3n)^2}$. It could make the solution much shorter.. MNJ2357 how did you prove that $m$ and $n$ have the same denominator?

Echoes12 wrote: MNJ2357 how did you prove that $m$ and $n$ have the same denominator? I didn't assume that $\gcd (a,x)=\gcd(b,x)=1$. It's just a simple construction. For example: $$m=\frac{1}{2}, n=\frac{3}{5}\Longrightarrow a=5, b=6, x=10$$.

Have $m^3=27n^2m+54n^3+m+2n=27n^2m+(6n)^3-6\cdot27n^3+m+2n$ $\Longrightarrow$ $(m-6n)(m^2+6mn+36n^2)=27n^2(m-6n)+(m+2n)$ $\Longrightarrow$ $(m-6n)(m+3n)^2=m+2n$ $\Longrightarrow$ $\frac{m-6n}{m+2n}=(\frac{1}{(m+3n)})^2$