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Claim: $\forall$x$\in$Z, $[16|x^3]$ $\Rightarrow$ $[2|x]$

Also provide a counterexample which disproves the converse.

My attempt

Proof. Suppose $16|x^3$. Then $x^3=16q$, where $q\in$$Z$. Taking the cube root of both sides of the eqn. $x^3=16q$ gives $x$ $=$ $\sqrt[3]{16q}$ $=$$2$$\sqrt[3]{2q}$. Because $x$ $=$ $2$$\sqrt[3]{2q}$, it follows that, by the Quotient Remainder Theorem, the quotient $q$ $=$ $\left\lfloor\frac{x}{2}\right\rfloor$ $=$ $\sqrt[3]{2q}$ and the remainder $r$ $=$ $(x-2$$\left\lfloor\frac{x}{2}\right\rfloor$) $=$ $0$. Thus $2|x$, as claimed.

Counterexample for converse: Let $x=2$. Then by the QRT, we have $2=2q$ $\Rightarrow$ $q=1$ and $r=0$. Thus $2|x$. However, $16\nmid 8$ since $8$ $=$ 16$\left\lfloor\frac{8}{16}\right\rfloor$$+$$(8-16\left\lfloor\frac{8}{16}\right\rfloor$) whereby $(8-16\left\lfloor\frac{8}{16}\right\rfloor$)$\neq$$0$.

Karam
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    Another perhaps easier method: write $x$ as a unique product of prime factors, and consider how many times $2$ appears in both $x$ and $x^3$. – James Jun 05 '21 at 12:38
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    Try proving the contrapositive: "if 2 does not divide $x$ then 16 does not divide $x^3$". You can translate "2 does not divide $x$" by "x is odd" or again "There is a $k$ such that $x=2k+1$ and then showing that $x^3$ is also odd. – MasB Jun 05 '21 at 13:15

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