why 3q,3q+1,3q+2 has square 3m,3m+1, 3m+2
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Sia ? 6 years, 6 months ago
By Euclid’s division algorithm
a = bq + r, where 0 ≤ r ≤ b
Put b = 3
a = 3q + r, where 0 ≤ r ≤ 3
If r = 0, then a = 3q
If r = 1, then a = 3q + 1
If r = 2, then a = 3q + 2
Now, (3q)2 = 9q2
= 3 × 3q2
= 3m, where m is some integer
(3q + 1)2 = (3q)2 + 2(3q)(1) + (1)2
= 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1, where m is some integer
(3q + 2)2 = (3q)2 + 2(3q)(2) + (2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 4
= 3(3q2 + 4q + 1) + 1
= 3m + 1, where m is some integer
Hence the square of any positive integer is of the form 3m, or 3m +1
But not of the form 3m + 2
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