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, x2=(3q)^2=9q^2=3×3q^2
Let 3q^2= m
Therefore, x^2= 3m ..........................(1)

x^2= (3q + 1)2= (3q)^2+12+2×3q×1 = 9q^2+ 1 +6q = 3(3q^2+2q) +1 Substitute, 3q^2+2q = m, to get,
x^2= 3m + 1 .................................. (2)
x^2=(3q+2)^2=(3q)^2+22+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