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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?

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    $\begingroup$ Yes, try the biggest one first ($17$), since that will tell you immediately that $2, 3, 5$ are too small anyways. Then you're down to four. But the answer below is more constructive, so go for that. $\endgroup$ – Arthur Sep 19 '16 at 18:39
  • $\begingroup$ Curiously enough, the only solution to the equation $x^3 + y^3 = 5256$ to use negative numbers involves two composite even numbers, e.g., $x = -14$. $\endgroup$ – Mr. Brooks Sep 19 '16 at 21:04
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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.

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  • $\begingroup$ (+1) this was my idea as well, but even more concise. $\endgroup$ – 6005 Sep 19 '16 at 19:51
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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.

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