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We know that, We can express a number which is not a multiple of 3 as ( 3m + 1) or ( 3m +2).

( 3m+1)2 = (3m)2 + 2×3m×1 + 12 = 9m2 + 6m + 1 = 3 ( 3m2 + 2m) + 1 ………..(1)

…………..using, (x+y)2 = x2 + 2xy + y2

( 3m+2)2 = (3m)2 + 2×3m×2 + 22 = 9m2 + 12m + 4 = 9m2 + 12m + 3 +1

= 3( 3m 2 + 4m +1) +1 ………..(2)

…………..using, (x+y)2 = x2 + 2xy + y2

Now, observe (1) and (2) carefully, clearly on division by 3 both these terms will leave remainder 1.

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