Question

Show that cube of any positive integer is of the form $4\mathrm{m}$, $4\mathrm{m}+1$ and $4\mathrm{m}+3$ for some integer m.

Answer 1

We need to prove that the cube of any positive integer is of the form $4\mathrm{m}$,$4\mathrm{m}+1$ or $4\mathrm{m}+3$ , for some integer m.

Step$1$: Use Euclid’s division lemma

Let $\mathrm{a}$ be any positive integer and $\mathrm{b}=4$.

According to Euclid’s division lemma, we know that

$\mathrm{a}=\mathrm{bq}+\mathrm{r}[0\le \mathrm{r}<\mathrm{b}]$

$\mathrm{a}=3\mathrm{q}+\mathrm{r}[0\le \mathrm{r}<4]$

According to the given details, the possible values of r are,

$\mathrm{r}=0$,$\mathrm{r}=1$, $\mathrm{r}=2$,$\mathrm{r}=3$

Step $2$: Find $\mathrm{a}³$ for different value of $\text{r}$

Case 1: When $\mathrm{r}=0$,

$\mathrm{a}=4\mathrm{q}+0$

$\mathrm{a}=4\mathrm{q}$

On taking cubes on both sides LHS and RHS, we get,

$\mathrm{a}³=\left(4\mathrm{q}\right)³$

$\mathrm{a}³=64\mathrm{q}³$

$\mathrm{a}³=4(16\mathrm{q}³)$

$\mathrm{a}³=4\mathrm{m}$ [where m is an integer = 16q³]

Case 2: When $\mathrm{r}=1$,

$\mathrm{a}=4\mathrm{q}+1$

On taking cubes on both sides LHS and RHS, we get,

$\mathrm{a}³=(4\mathrm{q}+1)³$

$\mathrm{a}³=64\mathrm{q}³+1³+3\times 4\mathrm{q}\times 1(4\mathrm{q}+1)$

$$\begin{gathered} \mathrm{a}³=64\mathrm{q}³+1+48\mathrm{q}²+12\mathrm{q}\mathrm{a}³ \\ =4(16\mathrm{q}³+12\mathrm{q}²+3\mathrm{q})+1 \end{gathered}$$

$\mathrm{a}³=4\mathrm{m}+1$ [where$\mathrm{m}$ is an integer and $\mathrm{m}$ $=16\mathrm{q}³+12\mathrm{q}²+3\mathrm{q}$]

Case 3: When$\mathrm{r}=2$,

$\mathrm{a}=4\mathrm{q}+2$

On taking cubes on both sides LHS and RHS, we get,

$$\begin{gathered} \mathrm{a}³=(4\mathrm{q}+2)³\mathrm{a}³ \\ =64\mathrm{q}³+2³+3\times 4\mathrm{q}\times 2(4\mathrm{q}+2)\mathrm{a}³ \\ =64\mathrm{q}³+8+96\mathrm{q}²+48\mathrm{q}\mathrm{a}³ \\ =4(16\mathrm{q}³+2+24\mathrm{q}²+12\mathrm{q}) \end{gathered}$$

$\mathrm{a}³=4\mathrm{m}$ [where$\mathrm{m}$ is an integer $\mathrm{m}$$=16\mathrm{q}³+2+24\mathrm{q}²+12\mathrm{q}$]

Case 4: When $\mathrm{r}=3$,

$\mathrm{a}=4\mathrm{q}+3$

Step $3$: On taking cubes on both sides of LHS and RHS, we get,

$$\begin{gathered} \mathrm{a}³=(4\mathrm{q}+3)³\mathrm{a}³ \\ =64\mathrm{q}³+27+3\times 4\mathrm{q}\times 3(4\mathrm{q}+3)\mathrm{a}³ \\ =64\mathrm{q}³+24+3+144\mathrm{q}²+108\mathrm{q}\mathrm{a}³ \\ =4(16\mathrm{q}³+36\mathrm{q}²+27\mathrm{q}+6)+3 \end{gathered}$$

$\mathrm{a}³=4\mathrm{m}+3$ [where$\mathrm{m}$ is an integer and $\mathrm{m}$$=16\mathrm{q}³+36\mathrm{q}²+27\mathrm{q}+6$]

Thus, the cube of any positive integer is in the form of $4\mathrm{m}$,$4\mathrm{m}+1$or $4\mathrm{m}+3$.

Hence Proved.