Show that the square of an integer is of the form 3q and 3q + 1
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Answer:
3k1 or 3k2 + 1 or 3k3 + 1
Step-by-step explanation:
Let a be any positive integer and b = 3 .
Then a = 3q + r for some integer q 2
And r = 0, 1, 2 because 0 ≤r<3 Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2 a2 = (9q)2 or 9q2 + 6q+1 or 9q2 + 12q + 4
= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
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