#### Question

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

#### Solution

By Euclid’s division Algorithm

a = bm + r, where 0 ≤ r ≤ b

Put b = 4

a = 4m + r, where 0 ≤ r ≤ 4

If r = 0, then a = 4m

If r = 1, then a = 4m + 1

If r = 2, then a = 4m + 2

If r = 3, then a = 4m + 3

Now, (4m)^{2} = 16m^{2}

= 4 × 4m^{2}

= 4q where q is some integer

(4m + 1)^{2} = (4m)^{2} + 2(4m)(1) + (1)2

= 16m^{2} + 8m + 1

= 4(4m^{2} + 2m) + 1

= 4q + 1 where q is some integer

(4m + 2)^{2} = (4m)^{2} + 2(4m)(2)+(2)^{2}

= 16m^{2} + 24m + 9

= 16m^{2} + 24m + 8 + 1

= 4(4m^{2} + 6m + 2) + 1

= 4q + 1, where q is some integer

Hence, the square of any positive integer is of the form 4q or 4q + 1 for some integer m

Is there an error in this question or solution?

Solution Prove that the Square of Any Positive Integer is of the Form 4q Or 4q + 1 for Some Integer Q. Concept: Euclid’s Division Lemma.