# How can we prove using mathematical induction that n^2+n+1 is an odd number if n is a natural number?

Mar 31, 2017

See Explanation.
or try again yourself keeping in mind that : -
Any even number can be represented as 2t and any odd number can be represented as 2t+1 where t is also a natural number

#### Explanation:

Assuming you know the general algorithm of Principle of mathematical induction;

1 . Checking if the statement is true for $n = 1$.
${1}^{2} + 1 + 1 = 3$ which is odd

2 . Assuming statement is true for some natural number $k$.
i.e. ${k}^{2} + k + 1$ is odd.
i.e. ${k}^{2} + k + 1 = 2 l + 1$ where $l$ is another natural number.

[any even number can be represented as 2t and any odd number can be represented as 2t+1 where t is also a natural number]

3 . To prove that statement is true for natural number next to $k$ i.e. $k + 1$.

$\implies$ to prove that ${\left(k + 1\right)}^{2} + \left(k + 1\right) + 1$ is also odd.
${\left(k + 1\right)}^{2} + \left(k + 1\right) + 1 = {k}^{2} + 1 + 2 k + k + 1 + 1$
$= \left({k}^{2} + k + 1\right) + 2 k + 2$
$= \left(2 l + 1\right) + 2 \left(k + 1\right)$ (from point 2.)
$= 2 l + 1 + 2 \left(k + 1\right)$
$= 2 \left(l + k + 1\right) + 1$
$= 2 p + 1$ $\equiv$ ODD (where p is a natural number)

now since all three, $l , k , 1$ are natural numbers, their sum i.e. $l + k + 1$ is also a natural number and can be represented by another natural number.