# Show that cube of any positive integer is of the form $4m, 4m + 1$ or $4m + 3$, for some integer $m$.

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Given:

Positive integer $m$.

To do:

We have to prove that cube of any positive integer is of the form $4m, 4m + 1$ or $4m + 3$, for some integer $m$.

Solution:

By Euclid\'s division algorithm,

If $a$ and $b$ are two positive integers,

$a\ =\ bq\ +\ r$, where  $0\ \underline{< }\ r\ <\ b$

Let $a$ be the positive integer and $b$ is equal to 4,

$a\ =\ 4q\ +\ r$, where $0\ \underline{< }\ r\ <\ 4$,

So, $r\ =\ 0,\ 1,\ 2,\ 3$

$a^{3} =(4 q+r)^{3}$

$=64 q^{3}+r^{3}+12 q r^{2}+48 q^{2} r$

$a^{3}=(64 q^{3}+48 q^{2} r+12 q r^{2})+r^{3}$ where, $0 \leq r<4$

When $r=0$,

$a^{3}=64 q^{3}$

$=4(16 q^{3})$

$a^{3}=4 m$ where, $m=16 q^{3}$ is an integer.

When $r=1$,

$a^{3} =64 q^{3}+48 q^{2}+12 q+1$

$=4(16 q^{3}+12 q^{2}+3 q)+1$

$=4 m+1$ where, $m=(16 q^{2}+12 q^{2}+3 q)$ is an integer.

When $r=2$,

$a^{3} =64 q^{3}+144 q^{2}+108 q+27$

$=64 q^{3}+144 q^{2}+108 q+24+3$

$=4(16 q^{3}+36 q^{2}+27 q+6)+3$

$=4 m+3$ where, $m=(16 q^{3}+36 q^{2}+27 q+6)$ is an integer.

Hence, the cube of any positive integer is of the form $4m, 4m + 1$ or $4m + 3$, for some integer $m$.

Updated on 10-Oct-2022 13:27:08