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. – Mark Bennet 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). – Ewan Delanoy 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. – Martin Argerami 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" – Ewan Delanoy 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.