Show that any square no cannot be put in the form 4q+2
Answers
Let a be an arbitrary positive integer. Then, by, Euclid's division algorithm, corresponding to the positive intgers a and 4, there exist non - negative integers m and r, such that
a = 4m + r, where 0 ≤ < 4
a2 = 16m2 + r2 + 8mr ...(i)
where, 0 ≤ < 4 [∵(a+b)2 = a2 + 2ab + b2]
Case I when r = 0, then putting r = 0 in Eq.(i), we get
a2 = 16m2 = 4(4m2) = 4q
where, q = 4m2 is an integer.
Case II When r = 1, putting r = 1 in Eq.(i), we get
a2 = 16m2 + 1 + 8m
= 4(4m2 + 2m) + 1 = 4q + 1
where, q = (4m2 + 4m + 1) is an integer.
Case III When r = 2, then putting r = 2 in Eq.(i), we get
a2 = 16m2 + 4 + 16m
= 4(4m2 + 4m + 1) = 4q
where, q = (4m2 + 6m + 2) is an integer.
Case IV When r = 3, then putting r = 3 in Eq.(i), we get
a2 = 16m2 + 9 + 24m = 16m2 + 24m + 8 + 1
= 4(4m2 + 6m + 2) + 1 = 4q + 1
where, q = (4m2 + 6m + 2) is an integer.
Hence, the square of any positive integer is either of the form 4 q + 1 for some integer q.
Solution:
Let a be any positive integer and b=2.
Every positive integer is of the form 2m or 2m+1 or 2m+2.
Case 1:
When a=2m
a^2=2m(2m)
=4q,where q=m^2
Case 2:
When a=2m+1
a^2=(2m+1)^2
=4m^2+4m+1
=4m(m+1)+1
=4q+1,where q=m(m+1)
Case 3:
When a=2m+2
a^2=(2m+2)^2
=4m^2+8m+4
=4m(m+2)+4
=4q+4,where q=m(m+2)
Hence proved that square of any positive integer doesn't occur in the form 4q+2