# Simple number theory problem

I found this question in a textbook on number theory:

For which integer c will $\;\displaystyle{\frac{c^6 - 3}{c^2 + 2}}\;$ also be an integer?

I wonder if there is a solution which is not based on trial and error.

If $(c^6 - 3)/(c^2 + 2)$ is an integer, then so is $$\frac{c^6 - 3}{c^2 + 2} - (c^4 - 2c^2 + 4) = \frac{c^6 - 3}{c^2 + 2} - \frac{c^6 + 8}{c^2 + 2} = \frac{-11}{c^2 + 2},$$ that is, $c^2 + 2$ divides $11$. The only way this can happen is if $c^2 = 9$, so $c = \pm 3$.

• For the record, this is not magic: this is simply long division of polynomials.
– lhf
Jun 3, 2013 at 20:25
• Quick question, probably painfully obvious but I can't see it. Where is $c^4 - 2c^2 + 4$ coming from? Jun 3, 2013 at 20:29
• Long division. e.g. wolframalpha.com/input/?i=%28x%5E6-3%29%2F%28x%5E2%2B2%29 Jun 3, 2013 at 20:31
• @TylerHilton As lhf says, this is long division of polynomials. You need the $c^4$ to get from $c^2 + 2$ to something of degree $6$; then subtract $c^4(c^2 + 2)$ from $c^6 - 3$ to get a polynomial of lesser degree and repeat. Jun 3, 2013 at 20:31
• @TylerHilton, as $(a^3+b^3)=(a+b)(a^2-ab+b^2),$ putting $a=c^2,b=2$ we get $c^6+8=(c^2+2)(c^4-2c^2+4)$ Jun 4, 2013 at 3:17

$\,c^2\!+2\mid \color{#0a0}{c^6\!-3}\,\Rightarrow\,$ mod $\,c^2\!+2\!:\,\ c^2\equiv \color{#c00}{ -2},\ \ \color{}0\equiv \color{#0a0}{(c^2)^3\!-3}\equiv (\color{#c00}{-2})^3\!-\!3 \equiv -11,\$ so $\ c^2\!+2\mid 11.$

• Nice solution, thank you!
– user80827
Jun 3, 2013 at 20:54
• Nice use of congruence. Welcome back. Jun 3, 2013 at 21:19