Question

Prove that any square of positive integer is of the form of 4m of 4m +1 for some integer m

Answer 1

hello,
let us assume that there are 2 positive integers a and b,then there exists unique integers q and r which satisfies a=bq+r         (0≤r<b)
take b=2
then a=2q+r            (0≤r<2) i.e r=0 and 1
case 1,when r=0
a=2q
a²=(2q)²
a²=4q²
a²=4m           where m=q² for some integer m
case 2,when r=1
a=2q+1
a²=(2q+1)²
a²=4q²+1+4q             (a+b)²=a²+b²+2ab
a²=4q²+4q+(1)
a²=4(q²+q)+1
a²+4m+1              where q²+q=m for some integer m
hence proved that sqaure of any positive integer is in the form of 4m or 4m+1 for some integer m