If the square of an irrational number r is irrational, can it be equal to a + br, where both a and b are rational

Question

I am trying to evaluate such a statement:

$$\forall r \in \mathbb{R} \setminus \mathbb{Q}: \ ({r^{2}} \notin \mathbb{Q} \implies \forall a \in \mathbb{Q} \ \forall b \in \mathbb{Q}: {r^{2}} \neq a + br)$$

but it seems to exceed my skills. Could you please help?

Answer 1

The following is a counterexample:

Set $r = \sqrt{2} +1 \in \mathbb{I}$.

Then $r^{2} = (\sqrt{2} +1)^2=1+2\sqrt{2}+2=1+2(\sqrt{2}+1) = 1+2r.$