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