Interesting Olympiad Style Problem about Invariance

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 ...

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.

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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.