# Proof that $3n(n^2+6)$ can't be an odd cube

How can I prove that $3n(n^2+6)$ can't be an odd cube?

• You mean an odd cube? – joriki Oct 14 '12 at 15:00

If $n$ is not a multiple of 3, $3n(n^2+6)$ is a multiple of 3 but not a multiple of 9, so it cannot be a cube. So let $n = 3m$, and we want to show $9m(9m^2+6) = 27(m(3m^2+2))$ cannot be an odd cube, or equivalently, that $m(3m^2+2)$ cannot be an odd cube.

Now, if it is an odd cube, $m$ must be odd, and thus $\gcd(m,3m^2+2) = 1.$ Therefore, both $m$ and $3m^2+2$ are cubes; let $m = k^3.$ Now, $3m^2+2 = 3k^6+2$ is a cube. But $3k^6+2$ is either 5 or 2 mod 7, while cubes are 1,0, or -1 mod 7. This can be calculated explicitly, or you can note that Fermat's little theorem tells us that $k^6$ is 1 mod 7 if $k$ is not a multiple of 7.

• I got everything except the cubes are 1,0 or -1 mod 7 – Jorge Fernández Hidalgo Oct 14 '12 at 15:28
• By Fermat's little theorem, either $x \equiv 0 \pmod{7}$ or $x^6 \equiv 1 \pmod{7}$. In the second case, we can factor that as $(x^3-1)(x^3+1) \equiv 0 \pmod{7}$, so if $x$ is not a multiple of 7, $x^3$ is 1 or -1 mod 7. – only Oct 14 '12 at 15:30

It is worth emphasis that the idea in only's answer works much more generally. Namely

Theorem $\$ If $\rm\:a,b,j,k,m\:$ are naturals with $\rm\:(m,b) = 1,\:$ and $\rm\:jk+1 = p\:$ is prime then

$\rm\qquad\qquad\quad m\,(a\,m^{\,j} + b)\:$ is a $\rm\:k$-th power of a natural

$\rm\qquad\qquad\!\Rightarrow\ a+b\,\ \ or\,\ \ b\$ is a $\rm\:k$-th power $\rm(mod\ p)$

Proof $\rm\,\ (m,\,a\,m^{\,j}+b) = (m,b) = 1,\:$ thus $\rm\:m\:$ and $\rm\:a\,m^{\,j}+b\:$ are coprime factors of a $\rm\:k$-th power hence, by unique factorization, they are themselves $\rm\:k$-th powers, say $\rm\:m = c^k,\:$ and $\rm\:d^k\! = a\,m^{\,j} + b = a\, c^{\,j\,k}\!+ b = a\, c^{\,p-1}\!+b\equiv a+b\ \ or\ \ b\pmod p\:$ by little Fermat. $\ \$ QED

Remark $\$ This consequence of unique factorization often proves very useful in number theory. Essentially this was exploited by Fermat in his method of infinite descent. Indeed, Weil wrote

Infinite descent a la Fermat depends ordinarily upon no more than the following simple observation: if the product $ab$ of two ordinary integers (resp. two integers in an algebraic number-ﬁeld) is equal to an $n$-th power, and if the gcd of $a$ and $b$ can take its values only in a given ﬁnite set of integers (resp. of ideals), then both $a$ and $b$ are $n$-th powers, up to factors which can take their values only in some assignable ﬁnite set. For ordinary integers this is obvious; it is so for algebraic number-ﬁelds provided one takes for granted the ﬁniteness of the number of ideal-classes and Dirichlet’s theorem about units.