อ้างอิง:
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Problem. Determine the number of pairs $(m,n)$ of positive integers such that $$m^3-n^3=728\,.$$
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I am not quite sure if this is a quick solution, but here it goes. Note that $$728=m^3-n^3=(m-n)(m^2+mn+n^2)\,,$$
where
$$m^2+mn+n^2=(m-n)^2+3mn>(m-n)^2\,.$$
This means
$$(m-n)^3<728<729=9^3\,.$$
Hence, $m-n<9$. Therefore, if $m=n+k$ for some integer $k$, then $k\in\{1,2,\ldots,8\}$.
Now, $k$ must also divide $728=2^3\cdot 7\cdot 13$, and it has to be an even number (because $m-n=k$ and $m^3-n^3=728$ must have the same parity). This means $k=2$, $k=4$, or $k=8$.
If $k=8$, then $m-n=8$ and $m^2+mn+n^2=\dfrac{728}{8}=91$. That is, $$m+n=\sqrt{\dfrac{4(m^2+mn+n^2)-(m-n)^2}{3}}=\sqrt{\frac{364-64}{3}}=10\,.$$ This implies $(m,n)=(9,1)$.
If $k=4$, then $m-n=4$ and $m^2+mn+n^2=\dfrac{728}{4}=182$. That is, $$m+n=\sqrt{\dfrac{4(m^2+mn+n^2)-(m-n)^2}{3}}=\sqrt{\frac{728-64}{3}}=\sqrt{\frac{664}{3}}\,,$$ which is not an integer. Therefore, there are no solutions in this case.
If $k=2$, then $m-n=2$ and $m^2+mn+n^2=\dfrac{728}{2}=364$. $$m+n=\sqrt{\dfrac{4(m^2+mn+n^2)-(m-n)^2}{3}}=\sqrt{\frac{1456-64}{3}}=22\,.$$ This implies $(m,n)=(12,10)$.