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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