Question

Question 1
Show that the cube of a positive integer of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Answer 1

Let a be an arbitrary positive integer. Then, by Euclid's division algorithm, corresponding to the positive integers a and 6, there exist non - negative integers q and r such that;
$a=6q+r,where,0\le r<6$
$\Rightarrow a^3=(6q+r{)}^3=216q^3+r^3+3\times 6qr(6q+r)$
$[\therefore (a+b{)}^3=a^3+b^3+3ab(a+b)]$
$\Rightarrow a^3=(216q^3+108q^{2}r+18q{r}^2+r^3)$ ...(i)
where, $0\le r<6$.
Case I
When r = 0, putting r = 0 in eq (i), we get,
$a^3=216q^3=6\left(36q^3\right)=6m$
where, $m=36q^3$ is an integer.
Case II
When r =1 , putting r = 1 in eq. (i), we get,
$a^3=(216q^3+108q^2+18q)+1$
$=6(36q^3+18q^2+3q)+1$
$=6m+1$
where, $m=(36q^3+18q^2+3q)$ is an integer.
Case III
When r = 2, putting r = 2 in eq. (i), we get,
$a^3=(216q^3+216q^2+72q)+8$
$a^3=(216q^3+216q^2+72q+6)+2$
$\Rightarrow a^3=6(36q^3+36q^2+12q+1)+2=6m+2$
where $m=(36q^3+36q^2+12q+1)$ is an integer.
Case IV
When r =3 , putting r = 3 in eq. (i), we get,
$a^3=(216q^3+324q^2+162q)+27$
$=6(36q^3+54q^2+27q)+24+3$
$=6(36q^3+54q^2+27q+4)+3=6m+3$
where, $m=(36q^3+54q^2+27q+4)$ is an integer.
Case V
When r = 4, putting r = 4 in eq. (i), we get,
$a^3=(216q^3+432q^2+288q)+64$
$=6(36q^3+72q^2+48q)+60+4$
$=6(36q^3+72q^2+48q+10)+4=6m+4$
Where, $m=(36q^3+72q^2+48q+10)$ is an integer.
Case VI
When r = 5, putting r = 5 in eq. (i), we get,
$a^3=(216q^3+540q^2+450q)+125$
$=6(36q^3+90q^2+75q)+120+5$
$=6(36q^3+90q^2+75q+20)+5=6m+5$
Where, $m=(36q^3+90q^2+75q+20)$ is an integer.
Hence, the cube of a positive integer of the form 6q + r, where q is an integer and r = 0 1, 2, 3, 4, 5 is also of the forms 6m , 6m + 1, 6m + 2, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.