Use Euclid's Division Algorithm to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let a and b are two positive integers such that a is greater than b; then:
a = bq + r; where q and r are also positive integers and 0 ≤ r < b
Taking b = 3, we get:
a = 3q + r; where 0 ≤ r < 3
⇒ The value of positive integer a will be
3q + 0, 3q + 1 or 3q + 2
i.e., 3q, 3q + 1 or 3q + 2.
Now we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m, or 3m + 1 for some integer m.
Square of 3q = (3q)2
= 9q2 = 3(3q2) = 3m; 3 where m is some integer.
Square of 3q + 1 = (3q + 1)2
= 9q2 + 6q + 1
= 3(3q2 + 2q) + 1 = 3m + 1 for some integer m.
Square of 3q + 2 = (3q + 2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1 = 3m + 1 for some integer m.
The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Hence the required result.
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