why 3q,3q+1,3q+2 has square 3m,3m+1, 3m+2

1.	why 3q,3q+1,3q+2 has square 3m,3m+1, 3m+2
Answer» By Euclid’s division algorithma = bq + r, where 0 ≤ r ≤ bPut b = 3a = 3q + r, where 0 ≤ r ≤ 3If r = 0, then a = 3qIf r = 1, then a = 3q + 1If r = 2, then a = 3q + 2Now, (3q)2\xa0= 9q2= 3 × 3q2= 3m, where m is some integer(3q + 1)2\xa0= (3q)2\xa0+ 2(3q)(1) + (1)2= 9q2\xa0+ 6q + 1= 3(3q2\xa0+ 2q) + 1= 3m + 1, where m is some integer(3q + 2)2\xa0= (3q)2\xa0+ 2(3q)(2) + (2)2= 9q2\xa0+ 12q + 4= 9q2\xa0+ 12q + 4= 3(3q2\xa0+ 4q + 1) + 1= 3m + 1, where m is some integerHence the square of any positive integer is of the form 3m, or 3m +1But not of the form 3m + 2

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