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

Explanation

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, "where" 0 le r lt 5`
`Rightarrow a^(2)=(5m+r)^(2)=25m^(2)+r^(2)+10mr` `[therefore (a+b)^(2)=a^(2)+2ab+b^(2)]` ....(i)
`Rightarrow a^(2)=5(5m^(2)+2mr)+r^(2)`
Where , `0 le r lt 5`
CASE I When r=0, then putting r=0 in Eq. (i), we get
`a^(2)=5(5m^(2))=5q`
where, `q=5m^(2)` is an integer
CASE II When r=1, then putting r=1, is Eq. (i) we get `a^(2)=5(5m^(2)+2m)+1`
`Rightarrow q=5q+1`
where, `q=(5m^(2)+2m)` is an integer
CASE III When r=3, then putting r=3 in Eq. (i), we get
`a^(2)=5(5m^(2)+4m)+4=5q+4`
where, `q=(5m^(2)+4m)` is an integer.
CASE IV When r=3, then putting r=3, in Eq. (i), we get
`a^(2)=5(5m^(2)+6m)+9=5(5m^(2)+6m)+5+4`
`=5(5m^(2)+6m+1)+5=5q+4`
where, `q=(5m^(2)+6m+1)` is an integer.
CASE V When r=4 , then putting r=4, in Eq.(i) we get
`a^(2)=5(5m^(2)+8m)+16=5(5m^(2)+8m)+15+1`
`Rightarrow a^(2)=5(5m^(2)+8m+3) +1=5q+1`
where, `q=(5m^(2)+8m+3)` is an integer
Hence the square of an y positive integer cannot be of the form `5q+2 or 5q+3` for any integer q.