An easy True/False problem

Let $n$ be the positive integer so $(n+1)$ is $n$ 's consecutive number. Thus $(n+1)$ is an positive integer too.

$n < n + 1$

$\Rightarrow n^2 < (n + 1)^2$

$\Rightarrow n^2 < n^2 + 2\cdot n +1$

$\Rightarrow n^2 < n^2 + n + 1 < n^2 + 2\cdot n +1$ $\Rightarrow n^2 < n^2 + n +1 < (n + 1)^2$ Since the value of $(n^2 + n +1)$ is between the values of two perfect squares of two consecutive integers, then $\sqrt{(n^2 + n + 1)}$ cannot be an integer.

Counterexample, let n=2:

then $2^2 + 2 + 1 = 7 \neq x^2$ , $\forall x \in N$