# Solutions to $x^2 = x + 1$ are irrational

I'm trying to prove that all the solutions to the equation $x^2 = x + 1$ are irrational. This statement is equivalent to: If $x^2 = x + 1$, then $x$ is irrational.

I want to prove this using contraposition. The contrapositive statement is: If $x$ is rational, then $x^2 \neq x + 1$.

A rational number is one that can be described as a ratio. So, $r = \frac{k_1}{k_2}$, where $k_1$ and $k_2$ are integers and $k_2 \neq 0$.

Let $x$ be a rational number.

Using the above definition of a rational number we simplify $x^2 = x + 1$ to $(\frac{k_1}{k_2})^2 = \frac{k_1}{k_2} + 1$, where $k_1$ and $k_2$ are integers and $k_2 \neq 0$. All is left is to show that this equation cannot be satisfied for any combination of $k_1$ and $k_2$ (considering they are integers).

How can we do this? Can someone point me in the right direction?

The polynomial $x^2-x-1$ is irreducible over $\mathbb{F}_2$, hence it is irreducible over $\mathbb{Q}$ and cannot have any rational root. (Not really) Alternative way: assuming that $\frac{p}{q}$ is a root of $x^2-x-1$, we have $$p^2 = pq + q^2 \tag{1}$$ but since $p$ and $q$ are not both even (due to $\gcd(p,q)=1$), the RHS and LHS of $(1)$ have opposite parity, contradiction. (Really) Alternative way: since the roots of $x^2-x-1$ add to one by Vieta's theorem, it is enough to prove that one of them (say, the greatest) is irrational to have that both of them are irrational. But if a number $x>1$ fulfills $x^2=x+1$, it also fulfills $x=1+\frac{1}{x}$, hence its ordinary continued fraction is given by $x=[1;1,1,1,1,1,\ldots]$. That is not a finite ordinary continued fraction, hence $x\not\in\mathbb{Q}$.

• Explain your downvote. – Jack D'Aurizio Oct 2 '16 at 20:41

HINT: if I multiply through by $k_2$ I get $${(k_1)^2\over k_2}=k_1+k_2.$$ This means $(k_1)^2\over k_2$ is an integer. This isn't obviously a contradiction . . .

. . . but what if we assumed at the beginning that $k_1\over k_2$ is in lowest terms?

Note that to get a contradiction from the assumption that $k_1\over k_2$ was in lowest terms, we need to know that $k_2\not=1$ - that is, that there is no integer satisfying $x^2=x+1$. But this has a nice proof! HINT: is $x$ even or odd?

• This was a lot of help, thanks! However, if both k1 and k2 are odd, how can I prove that the left side of the expression is a non-integer? I understand the right side of the expression is always an integer. I updated my question with my work so far. – Georan Sep 29 '16 at 3:45
• @Georan I think you're misunderstanding my hint re: even/oddness: my point is that $x$ can't be an integer, because for any integer $x$, $x+1$ and $x^2$ have different parities (either odd/even or even/odd). The $k_i$ don't enter into this part; we only look at them once we've determined $x$ isn't an integer, which lets us conclude that $k_2\not=1$ (and it's this that really sets us up for a contradiction as hinted at in the first part of my answer). – Noah Schweber Sep 29 '16 at 3:47
• I don't think I'm getting it (brain's running a little slow at this time :P). I understand why 'x' can't be an integer and why k2 /= 1, but what contradiction are we trying to set up? – Georan Sep 29 '16 at 4:22

By the rational roots theorem...

I just want to make a note that this is not the easiest way to do this problem. The best way (in my opinion) would be to actually find the solutions and prove their irrationality.

A hint for your proposed solution method: look at the denominator of each side. What equality would $k_2$ have to satisfy for these rational numbers to be equal?

I'd just rewrite it as $x^2 - x - 1 = 0$, run the quadratic formula, and then see that the roots are irrational, i.e. that $\sqrt{5}$ is irrational.

$4+1=4(x^2-x)+1=(2x-1)^2$ but $5$ isn't a perfect square.