Need help with proof by contradiction of implication.

Question

Question reads:

$$\forall n \in \mathbb{N}, \forall m \in \mathbb{N}, (n^4 + n^2 + 1 \ne m^2)$$

In english, this reads: for every n and m in the set of natural numbers, $n^4 + n^2 + 1$ does not equal a perfect square.

I would like to prove by contradiction because it seems like the most obvious route, but have gotten stuck and need help. My goal is to negate the implication above and either prove or disprove: $$n^4 + n^2 + 1 = m^2$$ As follows:

$$\sim(\forall n \in \mathbb{N}, \forall m \in \mathbb{N} ,(n^4 + n^2 + 1 \ne m^2)) = \exists n \in \mathbb{N}, \sim(\forall m \in \mathbb{N}, (n^4 + n^2 + 1 \ne m^2))$$ $$\exists n \in \mathbb{N}, \sim(\forall m \in \mathbb{N}, (n^4 + n^2 + 1 \ne m^2)) = \exists n \in \mathbb{N}, \exists m \in \mathbb{N}, \sim(n^4 + n^2 + 1 \ne m^2)$$ $$\exists n \in \mathbb{N}, \exists m \in \mathbb{N}, \sim(n^4 + n^2 + 1 \ne m^2) = \exists n \in \mathbb{N}, \exists m \in \mathbb{N}, (n^4 + n^2 + 1 = m^2)$$ My goal is now to prove or disprove: $$n^4 + n^2 + 1 = m^2$$ By factoring: $$n^2(n^2 + 1) + 1 = m^2$$ $$n^2(n^2 + 1) = m^2 - 1$$ Clearly, both of the the items on the left of the equation can divide the right hand side, but I'm not sure where to go from here. Any help would be appreciated.

Answer 1

$$(n^2)^2<n^4+n^2+1<(n^2+1)^2$$ so that $n^4+n^2+1$ is between two consecutive squares.

Answer 2

Just do it:

$n^4 + n^2 + 1 = m^2$

$n^4 + n^2 + (1-m^2) = 0$

$n^2 =\frac {-1 \pm \sqrt{1 -4(1-m)^2}}2$

$n^2 = -\frac 12 \pm \frac {\sqrt{4m^2 -3}}2$

Then $4m^2 -3$ must be a perfect square (or else this is irrational). And $4m^2 = (2m)^2$ is also a perfect square.

Are there any two perfect squares that are exactly $3$ apart?

Well, the difference between any two consecutive squares, $(k+1)^2, k^2$ is $(k+1)^2 - k^2 = 2k + 1$.

So if $(2m)^2 - k^2 =3$ then $2m > k$ and $3= (2m)^2 - k^2\ge (k+1)^2 - k^2 = 2k + 1$.

So $k \le 1$.

If $k = 0$ then $(2m)^2 = 3$ which is impossible.

So $k =1$ and $(2m)^2 -3 =1$ so $m =1$.

and therefore

....

$n^2 =-\frac 12 \pm \frac {4m^2 - 3}2 = $

$-\frac 12 \pm \frac {4*1^2 -3}2 = 0, -1$

$n^2 = -1 $ is impossible so $n = 0$.

Which, if you accept that $0$ is a natural number is perfectly acceptable. But as you probably don't accept that $0$ is a natural number there are no solution.