The sum of the cubes of two different prime numbers is 5256.

Question

Is an simple way to solve the problem?

The sum of the cubes of two different prime numbers is 5256. What are the two primes?

Here is what I did: assume the two numbers are $x$ and $y$, then I have $x^3+y^3=5256$. I can try x and y, but I don't think it is the most effective method. Can anyone help me?

Answer 1

Maybe the easiest way is notice that $$ 18^3 = 5832 > 5256, $$ So if $x^3 + y^3 = 5256$, then $x$ and $y$ are between $1$ and $17$. But you also know they are prime, so you have that $$ x, y \in \{2, 3, 5, 7, 11, 13, 17\}. $$ We can then cube them to get $$ x^3, y^3 \in \{8, 27, 125, 343, 1331, 2197, 4913\}. \tag{1} $$ Now we subtract each element from $5256$: $$ 5256 - x^3, 5256 - y^3 \in \{5248, 5229, 5131, 4913, 3925, 3059, 343\}. \tag{2} $$ Since $x^3 = 5256 - y^3$, $x^3$ is in BOTH the set (1) and the set (2). The intersection of these two sets is $\{343, 4913\} = \{7^3, 17^3\}$. So $x = 7$ or $x = 17$, and correspondingly, $y = 17$ or $y = 7$, respectively.

Answer 2

We might as well take $x \ge y$. Then $18 \gt \sqrt[3]{5256} \gt 17$, so $x$ cannot be greater than $17$. $x$ has to be at least $\sqrt[3]{5256/2} \gt 13$ so we know $x=17$ as it is the only prime in the range. Now $y=\sqrt[3]{5256-17^3}=7$ and we are done without checking any cases.