When $\,a-2 \mid a^3+4,\,$ or $\,a-n\mid f(a)$ for a polynomial $f(x)$ with integer coef's.

Question

I need to find for which values of $a\in\mathbb{N}$ the following happens:

$$a-2\ |\ a^3+4$$

that is, for which values of $k\in\mathbb{Z}$ the following holds:

$$a^3+4 = k(a-2)$$

I could not find anything useful from that.

Usually for a proof involving division, for example, proving $8|3^{2n}+7$ I'd suppose $3^{2k}+7 = 8k$ and prove by induction that it also holds for $k+1$, but for this case is diferent.

Answer 1

Hint : $$a^3+4=(a-2)(a^2+2a+4)+12$$

Answer 2

Hint: $a-2\mid a^3-8$. When can $a-2\mid a^3+4$?

Answer 3

Hint $ $ Apply divisibility mod reduction to simplify the dividend,

namely, recall that if $\bmod d\!:\ n\equiv \bar n\ $ then $\,\ \bbox[5px,border:1px solid #c00]{d\mid n\!\iff\! d\mid \bar n}\,\ $ Applied here we get

$\bmod\, \underbrace{a-2}_{\color{#c00}{\textstyle a\equiv 2}}\!:\ $ $\,\underbrace{\color{#c00}a^3\!+\!4\equiv \color{#c00}2^3\!+\!4}_{\textstyle\color{#0a0}{f(a)\equiv f(2)}}\equiv 12,\,\ $ so $\ \ \bbox[5px,border:1px solid #c00]{a\!-\!2\mid a^3\!+\!4\!\iff\! a\!-\!2\mid 12}$

$\bmod\, \underbrace{a-2}_{\color{#c00}{\textstyle a\equiv 2}}\!:\ $ $\ \ \color{#0a0}{f(a)\equiv f(2)},\ \ $ therefore $\,\ \ \bbox[5px,border:1px solid #c00]{a\!-\!2\mid f(a)\!\iff\! a\!-\!2\mid f(2)}$

where we used: $\ \color{#0a0}{f(a)\equiv f(2)}\pmod{\!a\!-\!2}\,$ for any polynomial $\,f(x)\,$ with integer coefficients, by the Polynomial Congruence Rule (or the Polynomial Remainder Theorem). Using congruences as above is generally much simpler than using long-division, since here we don't need the quotient $\,f(a)\div a\!-\!2,\,$ just the remainder $\,f(a)\bmod a\!-\!2\,$ [$ = f(2),\,$ a simple polynomial evaluation].

Answer 4

A sensible way to approach problems like this is with a change of variables: let $a-2=d$, so that the question becomes for what values of $d$ does $d$ divide $a^3+4=(d+2)^3+4=d^3+6d^2+12d+12$. It's clear you can ignore the $d^3+6d^2+12d$ portion. In fact, you really don't need to compute those terms at all; you could just skip to the $2^3+4=12$.

Now the divisors of $12$ are $\pm1$, $\pm2$, $\pm3$, $\pm4$, $\pm6$, and $\pm12$, from which the integer values of a are obtained by adding $2$ to each: $a\in\{1,3,0,4,-1,5,-2,6,-4,8,-10,14\}$. If you just want values in $\mathbb{N}$, you can toss out the negative values, leaving (in order) $a\in\{0,1,3,4,5,6,8,14\}$.

Remark: This is really just a lowbrow version of Bill Dubuque's congruence-based approach (in particular in the skip-to-the $2^3+4$ part). Its chief advantage, if any, is that changing variables in order to simplify the statement of a problem is, in many cases, a straightforward thing to do.