Question 23

Show that the square of any positive integer cannot be of the form 5q+2 or 5q+3 for some integer q.

Solution:

Let a be an arbitrary positive integer. Then by Euclid’s divison Algorithm, corresponding to the positive integers a and 5, there exist non-negative integers m and r such that
$a=5m+r,\text{where}0\le r<5$
$\Rightarrow a^2={(5m+r)2}^{}=25{\mathrm{m2}}^{}+{\mathrm{r2}}^{}+10mr$ $[\therefore {(a+b)2}^{}={\mathrm{a2}}^{}+2ab+{\mathrm{b2}}^{}]$ ….(i)
$\Rightarrow {\mathrm{a2}}^{}=5(5{\mathrm{m2}}^{}+2mr)+{\mathrm{r2}}^{}$
Where , $0\le r<5$
CASE I When r=0, then putting r=0 in Eq. (i), we get
${\mathrm{a2}}^{}=5\left(5{\mathrm{m2}}^{}\right)=5q$
where, $q=5{\mathrm{m2}}^{}$ is an integer
CASE II When r=1, then putting r=1, is Eq. (i) we get ${\mathrm{a2}}^{}=5(5{\mathrm{m2}}^{}+2m)+1$
$\Rightarrow q=5q+1$
where, $q=(5{\mathrm{m2}}^{}+2m)$ is an integer
CASE III When r=3, then putting r=3 in Eq. (i), we get
${\mathrm{a2}}^{}=5(5{\mathrm{m2}}^{}+4m)+4=5q+4$
where, $q=(5{\mathrm{m2}}^{}+4m)$ is an integer.
CASE IV When r=3, then putting r=3, in Eq. (i), we get
${\mathrm{a2}}^{}=5(5{\mathrm{m2}}^{}+6m)+9=5(5{\mathrm{m2}}^{}+6m)+5+4$
$=5(5{\mathrm{m2}}^{}+6m+1)+5=5q+4$
where, $q=(5{\mathrm{m2}}^{}+6m+1)$ is an integer.
CASE V When r=4 , then putting r=4, in Eq.(i) we get
${\mathrm{a2}}^{}=5(5{\mathrm{m2}}^{}+8m)+16=5(5{\mathrm{m2}}^{}+8m)+15+1$
$\Rightarrow {\mathrm{a2}}^{}=5(5{\mathrm{m2}}^{}+8m+3)+1=5q+1$
where, $q=(5{\mathrm{m2}}^{}+8m+3)$ is an integer
Hence the square of an y positive integer cannot be of the form $5q+2\text{or}5q+3$ for any integer q.