A number theory problem by Samina Siamwalla

Consider the cubes of consecutive integers $n$ and $n + 1$ . The gap between their cubes is

$\begin{aligned} (n + 1)^3 - n^3 &= (n^3 + 3n^2 + 3n + 1) - n^3 \\ &= 3n^2 + 3n + 1 \end{aligned}$

which is clearly an increasing function for all positive $n$ , i.e. the gap between consecutive cubes will grow as $n$ grows.

The gaps between the first few consecutive cubes is

• $2^3 - 1^3 = 8 - 1 = 7$

• $3^3 - 2^3 = 27 - 8 = 19$

• $4^3 - 3^3 = 64 - 27 = 37$

and the gaps will continue to grow so there are no consecutive cubes with a difference of $21$ .

The only other difference of cubes which is small enough that it might be $21$ is $3^3 - 1^3$ but that equals $26$ .

Thus the given equation has no solution over the positive integers.

From the relation $m^3 - n^3 = 21$ we have that $(m - n)(m^2+mn+n^2) = 3\cdot7$

If $m$ and $n$ are both integers then $(m - n)$ and $(m^2+mn+n^2)$ are also integers. So by the FTA there are only four possibilities to check:

• $m-n=1$ and $m^2+mn+n^2=21$ , this gives $3 n^2 + 3 n - 20 = 0$ that has no integer solution ( $\Delta =249$ ).
• $m-n=3$ and $m^2+mn+n^2=7$ , this gives $3 n^2 + 9 n + 2 = 0$ that has no integer solution ( $\Delta =57$ ).
• $m-n=7$ and $m^2+mn+n^2=3$ , this gives $3 n^2 + 21 n + 46 = 0$ that has no integer solution ( $\Delta =-111$ ).
• $m-n=21$ and $m^2+mn+n^2=1$ , this gives $3 n^2 + 63 n + 440 = 0$ that has no integer solution ( $\Delta =-1311$ ).

Remember that $an^2+bn+c=0$ may have integer solutions for $n$ only if the discriminant $\Delta=b^2-4ac$ is a square number.