1. ## congruence

Show that for any positive integer $n$, there exist a prime number $p$ and another positive integer $m$ such that:
(1) $p$ is a prime numer of form $6k+5$;
(2) $n$ is not a multiple of $p$;
(3) $n-m^3$ is a multiple of $p$.

2. Edit: nevermind, didn't think it through.

Edit 2: Here are some thoughts.

p being congruent to 5 mod 6 is the same as saying p is an odd prime congruent to 2 mod 3.

Here is an experiment: Note that

$1^3\equiv1\pmod{11}$

$2^3\equiv8\pmod{11}$

$\dots$

$10^3\equiv10\pmod{11}$

produces all equivalence classes mod 11 besides zero.

If this is true for all odd primes congruent to 2 mod 3 and we can prove it, then the only other piece we'd need is that there exist an infinitude of such primes.

3. We can rephrase this in the language of congruences:

(1) $p\equiv 5\pmod{6}$
(2) $n\not \equiv 0\pmod{p}$
(3) $n\equiv m^3\pmod{p}$ has a solution

With this in mind, it suffices to show that the map $f:\mathbb{Z}_p\to \mathbb{Z}_p$ defined by $f(x)=x^3$ is a bijection. Can you do this?