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.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Let x be any positive integer and b = 3.
Then, by euclid’s algorithm
x = 3q + r, where q ≥ 0 and r = 0, 1, 2 [0 ≤ r ≤ b]
Case (i) : For r = 0, x = 3q, = x2 = 9q2, taking 3 as common,
x2 = 9q2 = 3 (3q2), which is of the form 3m, where m = 3q2.
Case (ii) : For r = 1, x = 3q + 1
x2 = 9q2 + 1 + 6q, taking 3 as common,
= 3 (3q2 + 2q) + 1, which is of the form 3m + 1, where m = 3q2 + 2q
Case (iii) : For r = 2, 3q + 2
x2 = 9q2 + 4 + 12q = (9q2 + 12q + 3) + 1, taking 3 as common,
= 3 (3q2 + 4q + 1) + 1, which is of the form 3m +1, where m = 3q2 + 4q + 1
Hence, x2 is either of the form 3m, 3m + 1 for some integer m.
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(iii) 2x2– 6x + 3 = 0
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