Question

Question 2
Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

Answer 1

Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers a and 4, there exist non-negative integers q and r such that
$a=4q+r,where0\le r<4\Rightarrow a^3=(4q+r{)}^3\Rightarrow 64q^3+r^3+12q{r}^2+48q^{2}r\Rightarrow a^3=(64q^3+48q^{2}r+12q{r}^2)+r^3\dots (i)$
Where, $0\le r<4$

Case I
When r =0,
Putting r = 0 in Equation (i), we get
$a^3=64q^3=4\left(16q^3\right)=4m$
Where $m=16q^3$ is an integer.

Case II
When r =1,
Putting r = 1 in Equation(1), we get
$a^3=64q^3+48q^2+12q+1=4(16q^3+12q^2+3q)+1=4m+1$
Whre $m=(16q^3+12q^2+3q)$ is an integer

Case III
When r = 2
Putting r = 2 Equation(1), we get
$a^3=64q^3+96q^2+48q+8=4(16q^3+24q^2+12q+2)=4m$
Where $m=(16q^3+24q^2+12q+2)$ is an integer.

Case IV:

When r =3.

Putting r = 3 Equation(1), we get
$a^3=64q^3+144q^2+108q+27=64q^3+144q^2+108q+24+3=4(16q^3+36q^2+27q+6)+3=4m+3$
Where $m=(16q^3+36q^2+27q+6)$ is an integer.
Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m +3 for some integer m.