$ \ ( \forall p \in \mathbb{Z})( \exists q \in \mathbb{N})[(p^2 = 2q^2) \implies (p^2 + q^2 > 4)]$

Question

Question:

Is the following statement true or false:

$ \ ( \forall p \in \mathbb{Z})( \exists q \in \mathbb{N})[(p^2 = 2q^2) \implies (p^2 + q^2 > 4)]$

My attempt:

The statement is true because the premise of the "if then$ statement is false.

Answer 1

If $0\in\mathbb{N}$, then it's false with $p=q=0$.

Otherwise, suppose $p\in\mathbb{Z}$ such that $q\in\mathbb{N}$ satisfies $p^2=2q^2$. Then $q\geq 1$ implies that $p^2\geq 4$ and $q^2>0$. Consequently $p^2+q^2 > 4$. This proves the statement.

As you pointed out, instead of doing the short argument above we could have noticed that it's impossible to have $p^2=2q^2$ for $p\in\mathbb{Z}$ and $q\in\mathbb{N}$. Indeed, if $p^2=2q^2$ for $p,q\in\mathbb{Z}$, $q\ne0$, then $$ 2 = \frac{p^2}{q^2} $$ implies that $\sqrt{2} = p/q$ is rational.