Given the binary operation $x*y=x^2+4xy+y^2$ show that $a*1 \in \mathbb{N}$ has infinitely many solutions, where $a$ is irrational.

Question

Consider the binary operation:

$$x * y = x^2 + 4xy + y^2$$

defined on $\mathbb{R}$. I have to show that there are infinitely many irrational numbers $a$ such that $a * 1$ is a natural number. This means we need:

$$a * 1 = a^2 + 4a + 1$$

to be natural with $a$ being irrational. So we need to have:

$$(a+2)^2-3 \in \mathbb{N}$$

with infinitely many irrational $a$'s. I don't see any irrational $a$ that would result in the above being natural, let alone infinitely many $a$'s. So how should I proceed?

Answer 1

Well if we set $(a + 2)^2 - 3 = n$ we can solve for $a$ to obtain:

$$a = \pm \sqrt{n + 3} - 2$$

All the numbers you are looking for are of this form but not all of them irrational - $n + 3$ has to be a non-square. There are of course infinitely many such values (all except $n = 1, 6, 13, 22, \dots$). So, for example, $\sqrt{5} - 2$, $-\sqrt{5} - 2$, $\sqrt{6} - 2$, $-\sqrt{6} - 2$ satify your requirement.

Answer 2

You are very close. Now think of taking irrational numbers of the form $\sqrt{n} - 2$ where $n$ is a big enough positive integer (and shouldn't be square, of course).

Answer 3

You're almost there.. just a few more steps should do it:

Since we're saying that the expression: $(a+2)^2-3$ should be a natural number, we can pick any $a$ such that $(a+2)^2-3$ is positive. Let's turn to the graph for some help. In the graph we can see that it is positive for infinitely many (irrational) $a$.

Answer 4

You’ve made great progress so far, so I’ll pick up from where you left off:

Choose some $n\in\mathbb{N},$ and let $a$ be the algebraic number such that

$\left( a+2 \right)^2-3=n.$

Then we can rearrange for

$a=-2\pm \sqrt{n+3}.$

Now clearly

$$n\in\mathbb{N}\implies n+3>0\implies \sqrt{n+3}\in\mathbb{R},$$

This means that $a\in\mathbb{R}\ \forall\ n\in\mathbb{N}$. Now it just remains to show that there exist infinitely many $n$ such that $a\in\mathbb{R}\backslash\mathbb{Q}$.

I assume if you have encountered this problem that you would be comfortable to recall the/a standard proof by contradiction (Just assume $2=\dfrac{p}{q}$ and rearrange) to prove $\sqrt{2}$ is irrational. Then it is also not difficult to show that $x\sqrt{2},\ x\in\mathbb{N}$ is irrational (because otherwise $\sqrt{2}$ rational). Similarly adding an integer to an irrational number remains irrational. So just choose the infinitely many $n$ such that $n+3$ is twice a square, say $$n+3=2x^2,\ x\in\mathbb{N}.$$

Well then by the above $$a=-2\pm \sqrt{n+3}=-2\pm\sqrt{2x^2}=-2\pm x\sqrt{2}\in\mathbb{R}\backslash\mathbb{Q}.$$

I hope this helps, stay safe!