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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.

Bill Dubuque
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Poperton
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4 Answers4

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

Peter
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  • I can see that when $a-2$ is a multiple of $12$ it will work, that is, when $a$ is in the form $12k+2$. Is it right? But I'd also have to check for when $(a^2+2a+4)$ is a multiple of $12$, which is a harder task – Poperton Apr 03 '17 at 22:27
  • No, we just need $a-2|12$. You need not care about the cofactor $a^2+2a+4$ – Peter Apr 03 '17 at 22:31
  • So the answer would be $a = 12k+2$ only? What about the other term? – Poperton Apr 03 '17 at 22:34
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    No again, take the positive divisors of $12$ and add $2$ to get all positive solutions. – Peter Apr 03 '17 at 22:38
  • @Peter, if you just take the positive divisors of $12$, you'll miss the natural numbers $a=0$ and $1$. (Sorry, I know this is four years late, but I just came across the question.) – Barry Cipra Jun 08 '21 at 21:27
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Hint: $a-2\mid a^3-8$. When can $a-2\mid a^3+4$?

Thomas Andrews
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We can simplify $\,d\mid n\,$ by replacing $\,n\,$ by $\,\bar n = n\bmod{d}\,$ (divisibility mod reduction),

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:

$\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 its remainder $f(a)\bmod (a\!-\!2)\,$ [$ = f(2),\,$ a simple polynomial evaluation].

Generally this easily extends to the nonmomic primitive case: for $\,f(x)\in\Bbb Z[x]\,$ of degree $\,d$

$${\rm if}\,\ \gcd(j,k)\!=\!1\,\ {\rm then}\,\ \ \bbox[1px,border:1px solid #c00]{\bbox[6px,border:1px solid #c00]{k\:\!\:\!a\!-\!j\mid f(a) \!\iff\! \, k\:\!\:\!a\!-\!j\mid k^d f \left({\small \frac{j}k}\right)\in\Bbb Z}}\qquad\qquad\quad$$

where the scaling by $\,k^d\,$ serves to keep the arithmetic fraction-free.

Bill Dubuque
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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.

Barry Cipra
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