Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
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 = 9q2 = 3 × 3q2
Let 3q2 = m
Therefore, x2 = 3m ……………………..(1)
x2 = (3q + 1)2 = (3q)2+ 12 + 2 × 3q × 1 = 9q2 + 1 +6q = 3(3q2+2q) + 1
Substitute, 3q2 +2q = m, to get,
x2 = 3m + 1 ……………………………. (2)
x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1
Again, substitute, 3q2+4q+1 = m, to get,
x2 = 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.