Interesting Olympiad Style Problem about Invariance

Problem: The following operations are permitted with the quadratic polynomial $ax^2 +bx +c:$ (a) switch $a$ and $c$, (b) replace $x$ by $x + t$ where $t$ is any real. By repeating these operations, can you transform $x^2 − x − 2$ into $x^2 − x − 1?$

My Attempt: Notice that the sum of coefficients $S\equiv a+b+c\pmod{t}$ is invariant. This is clear if we switch $a$ and $c.$ If we replace $x$ with $x+t$ then we have $ax^2+(2at+b)x+(at^2+bt+c)$ and so $S\equiv a+2at+b+at^2+bt+c\equiv a+b+c\pmod{t}.$ Now for $x^2-x-2$ we have $S\equiv -2\pmod{t}$ and at the end we want $S\equiv -1\pmod{t}$, which is impossible. I am not sure whether this is correct because $t\in \mathbb{R}.$ So any inputs will be much appreciated.

Solution 1:

The two operations preserve the discriminant $b^2-4ac$, now the discriminant of $x^2-x-2$ is $9$ while the discriminant of $x^2-x-1$ is $7$. So ...

Solution 2:

I don't understand. Is it $S = a+b+c\pmod{t}$ or $S= a+b+c$. Because I don't understand how you get $S\equiv a+b+c\pmod{t}$ in the second case.

Anyway, just calculate the discriminant and show that it doesn't change:

Mark new polynomial with $a'x^2+b'x+c'$

Case 1. If we change only we get from $ax^2+bx+c$ this $cx^2+bx+a$ so $a'=c$, $b'=b$ and $c'=a$ so $$D' = b'^2 -4a'c' = b^2-4ac = D$$

Case 2. If we replace $x$ with $x+t$ we get $$a(x+t)^2+b(x+t)+c = ax^2+(2at+b)x +at^2+bt+c$$ so $a' = a$, $b' = 2at+b$ and $c'=at^2+bt+c$ so $$D' = (2at+b)^2-4a(at^2+bt+c) = 4a^2t^2+4abt+b^2 -4at^2-4abt-4ac = D$$

So since the discriminant at begining is different from the end it is impossible.