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Problem:

Let $x^3 + y^3 = z!$, where $z > 12$.

1) Prove that the largest prime less than $z$ does not divide x.

2) Prove that $x + y$ is a multiple of $330$.

I noticed that since

$$z > 12,$$

$x$ and $y$ have the same parity.

Then $x^3 + y^3$ could be factored as

$$(x+y)(x^2 - xy + y^2),$$

where both factors had to be either even or odd. Here is where I got stuck.

Any insights on this problem? Any help is appreciated!

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Hint: by Bertrand, the largest prime $p$ less than $z$ is greater than $z/2$. If it divided $x$, then it also divides $y$, and $p^3$ would divide $z!$.

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Note the prime factorization of $330$ is $2 \cdot 3 \cdot 5 \cdot 11$ show that all these must occur on the right hand side for $z!$ which is trivial for $z$ greater then $12$. now we $x^3+y^3=0$ mod $a < z $ so as you noted $(x+y)(x^2−xy+y^2)=0$ mod $a<z$ you can verify for yourself that if ($x$ mod a not equal to $0$ or $y$ mod $a$ not equal to $0$) then $(x^2−xy+y^2)$ mod $a$ not equal to $0$ for $a$ in $(2,3,5,11)$. so either ($x$ mod $a = 0$ and $y$ mod $a = 0$) or $(x+y) \mod a = 0$. in either case ($(x+y)=0 \mod a$) for $a = (2,3,5,11)$. so $x+y \mod 2 \cdot 3 \cdot 5 \cdot 11 = 0$. with the previous answer by Robert Isreal your question is answered.

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