Question

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

Answer 1

Step-by-step explanation:

Let x=3q

squaring both sides

x²=9q²

x²=3(3q²)

x=3m (m=3q²)

Let x=3q+1

squaring both sides

x²=9q²+6q+1

x²=3(3q²+2q)+1

x=3m+1 (m=3q²+2q)

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Answer 2

Answer:

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.