Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.
By Euclid’s Lemma, b = a × q + r, 0 ≤ r < a
Here, b is any positive integer.
Let a be an arbitrary positive integer. Then corresponding to the positive integers a and 4, there exist non-negative integers m and r, such that
a = 4m + r, where 0 ≤ r < 4
Squaring both the sides using (a + b)2 = a2 + 2ab + b2
⇒ a2 = (4m + r)2 = 16m2 + 8mr + r2
Case I When r = 0 we get
a2 = 16m2 which is an integer.
Case II When r = 1 we get
a2 = 16m2 + 1 + 8m
⇒ a2 = 4(4m2 + 2m) + 1 = 4q + 1
where q = (4m2 + 2m) in an integer.
Case III When r = 2 we get
a2 = 16m2 + 4 + 16m
⇒ a2 = 4 (4m2 + 4m + 1) = 4q
where q = (4m2 + 4m + 1) is an integer.
Case IV When r = 3 we get
a2 = 16m2 + 9 + 24m = 16m2 + 24m + 8 + 1
⇒ a2 = 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 4q or 4q + 1 for some integer q.