Some Cubes
The sum of the cubes of two numbers is 7163. What are these
numbers?
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The sum of the cubes of two numbers is 7163.
What are these numbers?

Factors could be useful.
An excellent crop of solutions here! Syed Farhan Iskander of Foxford School School and Community College listed the cubes of every whole number until he got to a cube that was larger than 7163. Then he systematically examined the differences between 7163 and these cubes to find cases where the difference was a cube number. In this way he showed that the only two numbers $x$ and $y$ such that $x^3 + y^3 = 7163$ are 11 and 18. Ling Xiang Ning of Tao Nan School, Singapore and Elizabeth and Ella of Madras College also used this method.
We give two further solutions using different methods, one by Koopa Koo from Boston College which uses the prime factorisation of 7163 and another by Adam of King Jame's School, Knaresborough, based on the parity of the numbers.
Firstly Adam's solution: Let the numbers be $x$ and $y$. Assume they are integers. Then $$x^3 + y^3 = 7163 *.$$ This implies that one of the numbers must be odd and the other even. Let $x$ be the odd number. Then, $x = 2n + 1$ and $y = 2m$ where $n$ and $m$ are integers. Substituting these in the equation marked $*$ gives :
\begin{eqnarray} \\ (2n + 1)^3 + (2m)^3
&=& 7163 \\ 8n^3 + 12n^2 + 6n + 1 + 8m^3 &=& 7163
\\ 8n^3 + 12n^2 + 6n + 8m^3 &=& 7162 \\ 4n^3 + 6n^2 + 3n +
4m^3 &=& 3581 \\ n(4n^2 + 6n + 3) + 4m^3 &=& 3581.
\end{eqnarray}
Now $4m^3$ must be even, so that $n(4n^2 + 6n + 3)$ must be odd.
This means that $n$ must be odd. I can therefore write $n$ as $n =
2z + 1$ where $z$ is an integer, i.e. $x = 4z + 3$. This gives
\begin{eqnarray} (4z + 3)^3 + (2m)^3 &=&
7163 \\ 64z^3 + 144z^2 + 108z + 27 + 8m^3 &=& 7163 \\ 64z^3
+ 144z^2 + 108z + 8m^3 &=& 7136 \\ 16z^3 + 36z + 27z + 2m^3
&=& 1784. \end{eqnarray}
Now $2m^3$ is even so that $z(16z^2 + 36z +27)$ must be even. This
means that $z$ must be even. I have now greatly simplified the problem because there are only three even values of $z$ for which $z(16z^2 + 36z + 27)$ is less than 1784. I will try each of these in turn to see which gives an integer solution for $m$.
If $z=0$, then $z(16z^2 + 36z + 27) = 0$ and hence $2m^3 = 1784$. This does not have an integer solution for $m$.
If $z=2$ then $z(16z^2 + 36z + 27) = 326$, then
\begin{eqnarray} 2m^3 &=& 1784 326 =
1458 \\ m^3 &=& 729 \\ m &=& 9.
\end{eqnarray}
Because $z = 2$ gives an integer value for $m$ we have $z=2$, $m=9$
gives the solution $x = 4z + 3 = 11$ and $y = 2m = 18$. So the
numbers are 11 and 18.Now Koopa's solution: We have $7163 = 13 \times 19 \times 29$ by prime factorisation. Hence, $$x^3 + y^3 = (x + y)(x^2  xy + y^2) = 13 \times 19 \times 29.$$Now it falls into the following cases:
(1) $x + y = 1$ which is rejected as $x, y \geq 1$,
(2) $x + y = 13$, $x^2  xy + y^2 = 19 \times 29$,
(3) $x + y = 19$, $x^2  xy + y^2 = 13 \times 29$,
(4) $x + y = 7163$, $x^2  xy + y^2 = 1$,
(5) $x + y = 29$, $x^2  xy + y^2 = 13 \times 19.$
Consider (2), $x^2  xy + y^2 = (x + y)^2  3xy = 19 \times 29$ implying $169  19 \times 29 = 3xy = 382$ which is impossible as $x$ and $y$ are positive.
Consider (3), $361  13\times 29 = 3xy =  16$ which is another contradiction.
Consider (4), $51308569  1 = 3xy$ implying that $xy = 17102856$ and $x + y = 7163$. Hence
\begin{eqnarray} \\ (7163  y)y &=&
17102856 \\ y^2  7163y + 17102856 &=& 0.
\end{eqnarray}
The discriminant of this quadratic is negative implying that no
real solutions exist and hence this case is rejected.Lastly (5),
\begin{eqnarray} 841  247 &=& 3xy \\ xy
= 198, \ (x + y) &=& 29 \\ y^2  29 y + 198 &=& 0
\\ (y  18)(y  11) &=& 0. \end{eqnarray}
Hence the two numbers are 11 and 18.