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.

• You can use different values of $t$ on different occasions, so you have to prove the choices are compatible. Dec 30 '17 at 13:48

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 ...

• This is interesting, for the same answer I din't get any up vote. And I even answer before.
– Aqua
Dec 30 '17 at 13:55
• @JohnWatson 1) You answered before but you deleted your answer for some time. 2) I provide more details than you do 3) Your first paragraph is confusing, I find Mark Benett's comment much clearer 4) You did get some upvotes (4 as of now). Dec 30 '17 at 15:04
• I don't know why are you reacting so nervously. My comment (obviously) wasn't directed at you. And, yes I did delete it because my answer (as yours!) is not an answer on his question.
– Aqua
Dec 30 '17 at 15:29
• -1: this answer doesn't even address the OP's question. Dec 31 '17 at 2:33
• @MartinArgerami It is not clear whether the OP's question is more about the problem itself or the attempted solution, and the OP also says "any input appreciated" Dec 31 '17 at 7:30

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.