Show that n^2+n+1 is not divisible by 5

Given

n^2+n+1 is not divisible by 5 \forall n \in N

Proof

We will prove that the given statement is true by contradiction.

Let n \in N. Assume n^2+n+1 is divisible by 5. Let,

n^2+n+1 = 5k where k \in N

n^2+n+1-5k = 0

The above equation is a quadratic equation. The roots of a quadratic equation are given by

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

n = \dfrac{-1\pm\sqrt{1^2-4(1)(1-5k)}}{2(1)}

n = \dfrac{-1\pm\sqrt{20k-3}}{2}

We know that n \in N. If n \in N then \sqrt{20k-3} must also \in N. This means that 20k-3 must be a perfect square. But this is not possible. 20k-3 always have 7 in ones place. A perfect square cannot have 7 in ones place. Therefore, n cannot be an integer. But this is contradictory since n \in N. This implies that our assumption that n^2+n+1 is divisible by 5 is incorrect.

Therefore, by contradiction we can say that n^2+n+1 is not divisible by 5 \forall n \in N.

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