Intersections of Tangent Lines to Parabola Different Proof Let $f_a: \mathbb{R} \to \mathbb{R}$ with $f_a(x) = 2ax - a^2$ for $x \in \mathbb{R}$ and $a \in \mathbb{Z}^+$. It is easy to verify, that $f_a$ describes the tangent line to the parabola $x \mapsto x^2$ at $x = a$.

Prove: There are no three different values $a, b, c \in \mathbb{Z}^+$ such that $f_a(x) = f_b(x) = f_c(x)$ for some $x \in \mathbb{R}$.

The statement means there are no three tangent lines intersecting at the same point.
Approach. I present my approach as following. Let's intersect $f_a$ and $f_b$ for arbitrary $a, b \in \mathbb{Z}^+$ with $a \neq b$. We have $\displaystyle f_a(x^*) = f_b(x^*) \Leftrightarrow 2ax^* - a^2 = 2bx^* - b^2 \Leftrightarrow x^* = \frac{a + b}{2}$ and $f_a(x^*) = ab$. Assume there is a third tangent line $f_c$ with $c \neq a$ and $c \neq b$ which also passes through the point $(x^* \mid ab)$:
$f_c(x^*) = ab \Leftrightarrow 2cx^* - c^2 = ab \Leftrightarrow c^2 - 2cx^* + ab = 0 \Leftrightarrow c^2 - 2c(a + b) + ab = 0$
We've got a quadratic equation and should somehow reason, that there is no $c \in \mathbb{Z}^+$ satisfying it, unless $a = b$, $c = a$ or $c = b$. The solution for $c$ is $\displaystyle c = a + b \pm \sqrt{a^2 + ab + b^2}$. I don't know how to proceed from here, i.e. arrive at contradiction.
Any kind of help is appreciated.
 A: $$\begin{align}
\begin{cases}
2 a x - a^2 = 2 b x - b^2  \\
2 a x - a^2 = 2 c x - c^2  
\end{cases}&\quad\to\quad
\begin{cases}
2 x ( a - b ) = a^2-b^2 = ( a - b )( a + b ) \\
2 x ( a - c ) = a^2-c^2 = ( a - c )( a + c )
\end{cases} \\[6pt]
&\quad\to\quad\begin{cases}
2 x ( a - b )\cdot( a - c ) = ( a - b )( a + b )\cdot(a-c) \\
2 x ( a - c )\cdot( a - b ) = ( a - c )( a + c )\cdot(a-b)
\end{cases} \\
\\[0pt]
\text{(subtracting equations)}&\quad\to\quad
0 = ( a - b )( a - c )( b - c)
\end{align}
$$
A: If the tangent lines $$y=2ax-a^2; y=2bx-b^2$$intersect at the point $P=(p,q)$ then $$q=2ap-a^2$$ $$q=2bp-b^2$$
Subtract to obtain $$0=2(a-b)p-(a^2-b^2)$$
Now either $a=b$ (the tangents are identical) or we can cancel a factor $a-b$ to obtain $$p=\frac {a+b}2$$ If we choose $a$ as our first point and $P$ as the point of intersection, we are then looking for values of $b$ which satisfy $$b=2p-a$$and this has just one solution. So whichever values $a$ and points $P$ we choose, there are only ever (at most) two tangents which meet at $P$. (If a point is inside the parabola, it doesn't lie on a tangent, and if it is on the parabola, there is just one tangent).
