Prove by contradiction that if $a$ is rational, then $a^2 \neq a + 1$ 
Prove by contradiction that if $a$ is rational, then $a^2 \neq a + 1$.

I can see how can I solve it for integers (dividing into to cases: even and odd). However, since a is rational, I'm expressing it as c/d, and I can't use the even and odd logic anymore since the numbers might be fractions.
Would love to get a direction.
Thanks!
 A: Hint
If $a^2=a+1,a\in \Bbb Q$ then we have$$a^2-a+{1\over 4}={5\over 4}\implies \left|a-{1\over 2}\right|={\sqrt{5}\over 2}\in\Bbb Q$$which is impossible.
A: If $a^2=a+1$ with $a\in\Bbb R$, then $a=\dfrac{1\pm\sqrt5}2$.  
If $a$ were rational, then $2a-1=\pm\sqrt5$ would be rational, which is a contradiction.
A: If $a=\frac cd$ were a reduced rational root, then
$$
c^2-cd-d^2=0
$$
implies both $c|d^2$ and $d|c^2$, which is only possible in the trivial case, so that $a=\pm 1$. Which obviously is not a root with either sign.
A: hint
$a$ cannot be an integer  since
$$a(a-1)=1 \implies$$
$$(a=1 \wedge a-1=1 )\vee (a=-1\wedge a-1=-1$$
which is not possible.
If $a=\frac cd$ with $c \wedge d=1$ and $a^2=a+1$, then
$$c^2=d(c+d)$$ 
now use Gauss Theorem :
If $A|BC$ and $(A\wedge B)=1$ then $A|C$.
A: Of course this problem may be readily solved via the observation that
$a^2 = a + 1 \Longrightarrow a^2 - a - 1 = 0, \tag 1$
and applying the quadratic formula to obtain
$a = \dfrac{1 \pm \sqrt 5}{2}, \tag 2$
clearly not a rational number.  This line of reasoning provides us with the shortest possible proof I can think of.
Nevertheless, I also feel it is far more engaging to pursue the desired result using only the methods and techniques of truly elementary number theory, so here we go:
Suppose then that the equation
$a^2 = a + 1 \tag 3$
is possessed of a rational solution; that is, we have
$a = \dfrac{p}{q}, \; p, q \in \Bbb Z; \tag 4$
we may of course assume that
$\gcd(p, q) = 1; \tag 5$
then the equation
$a^2 = a + 1 \tag 6$
becomes
$\dfrac{p^2}{q^2} = \dfrac{p}{q} + 1, \tag 7$
whence upon multiplication by $q^2$,
$p^2 = pq + q^2; \tag 8$
one conclusion we may draw from this equation is
$p^2 = pq + q^2 = q(p + q) \Longrightarrow q \mid p^2. \tag 9$
We pause to note that
$a \notin \Bbb Z,  \tag{10}$
lest, as pointed out by our colleague hamam_Abdallah in his answer, we have
$a(a - 1) = 1, \tag{11}$ 
which implies either
$a, a - 1 = 1 \Longrightarrow a = 1 + 1 = 2 \Rightarrow \Leftarrow a = 1, \tag{12}$
or
$a, a - 1 = -1 \Longrightarrow a = -1 + 1 = 0$
$\Rightarrow \Leftarrow a = -1; \tag{13}$
it is manifestly evident that (12) and (13) are fallacious; thus (10) binds, and from this we see that
$q \ne 1; \tag{14}$
otherwise (10) doesn't apply.
Now
$\gcd(p, q) = 1 \Longrightarrow \exists a, b \in \Bbb Z, \; ap + bq = 1; \tag{15}$
multiplying by $p$ we obtain
$ap^2 + bqp = p, \tag{16}$
whence
$q \mid p^2 \Longrightarrow q \mid p \Rightarrow \Leftarrow \gcd(p, q) = 1 \tag{17}$
in light of (14); this final contradiction implies it is impossible to find $p$ and $q$ satisfying (4); we conclude that
$a \notin \Bbb Q, \tag{18}$
$OE\Delta$.
