. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Dear Student

Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore,
x = 3q, 3q+1 and 3q+2

Now as per the question given, by squaring both the sides,
we get,
x^2=(3q)^2
=9q^2
=3×3q^2
Let 3q^2= m
Therefore
x^2= 3m ..........................(1)

x^2= (3q + 1)^2
= (3q)^2+1^2+2×3q×1
= 9q^2+ 1 +6q
= 3(3q^2+2q) +1
Substitute,
3q2+2q = m, to get,
x^2= 3m + 1 .................................. (2)


x^2=(3q+2)^2
=(3q)^2+2^2+2×3q×2
=9q^2+4+12q
=3(3q^2+4q+1)+1
Again, substitute, 3q^2+4q+1 = m, to get,

x^2= 3m + 1................................. (3)

Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Thanks