congruence

**Shanks** · September 10th 2010, 02:11 AM

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 $p$ .

**undefined** · September 10th 2010, 05:18 AM

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.

**roninpro** · September 10th 2010, 07:00 AM

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?

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