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(Note: the solution does not help much if you do not know what the problem is. OK?)

Problem: Demonstrate that 121 does not divide f (n) = n ^ 2 + 3n + 5

Solution:
f (n) = n ^ 2 + 3n +5 +28 - 28 = (n + 7) (n - 4) +33

is clear then that 11 divides f (n) if and only if n is of the form 4 + 11k. But in that case, f (4 + 11k) = 121k (k + 1) + 33. Consequently 121

not divide f (n) for any n. Tip

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