Question

Question 1
Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

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 m and r, such that,
$a=4m+r$, where $0\le r<4$
$\Rightarrow a^2=(4m+r{)}^2=16m^2+r^2+8mr\dots (i)$
Where, $0\le r<4$

Case I
When r = 0, then putting r= 0 in EQuation (i), we get
$a^2=16m^2=4(4m{)}^2=4q$
Where, $q=4m^2$ is an integer.

Case II
When r =1, then putting r = 1 in Equation (i), we get
$a^2={(4m+1)}^2=16m^2+8m+1=4(4m^2+2m)+1=4q+1$
Where, $q=4(4m^2+2m)$ is an integer.

Case III
When r = 2, then putting r = 2 in Equation (i) we get
$a^2=16m^2+4+16m=4(4m^2+4m+1)=4q$
Where, $q=(4m^2+4m+1)$ is an integer.

Case IV
When r =3 , then putting r = 3 in Equation (i), we get
$a^2=16m^2+924m=16m^2+24m+8+1=4(4m^2+6m+2)+1=4q+1$
Where, $q=(4m^2+6m+2)$ is an integer.
Hence, the square of any positive integer is either of the form 4q or 4q +1 from some integer q.