putnam_like_set6_a1 - mistake in grading_scheme.md

[danio2010]

There are two mistakes in the solution: the incorrect signs in lines 24 and 28. The correct file content should be as below:

This step is worth 2 points.
Observe that $x^3 + a x^2 + b x + c = (x-\alpha)(x-\beta)(x-\gamma)$. Hence
$$
\begin{cases}
\alpha+\beta+\gamma = -a,\
\alpha\beta+\beta\gamma+\alpha\gamma = b,\
\alpha\beta\gamma = -c,
\end{cases}
$$

This step is worth 2 points.
Assume that $x^3 + A x^2 + B x + C$ is a desired polynomial. Then
$$
\begin{cases}
\alpha^2+\beta^2+\gamma^2 = -A,\
\alpha^2\beta^2+\beta^2\gamma^2+\alpha^2\gamma^2 = B,\
\alpha^2\beta^2\gamma^2 = -C.
\end{cases}
$$

This step is worth 5 points.
Doing standard calculations we see that:

• $-C = (\alpha\beta\gamma)^2=c^2$,
• $-A = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\alpha\gamma) = a^2-2b$,
• $B = (\alpha\beta+\beta\gamma+\alpha\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma) = b^2-2ac$.

This step is worth 1 point.
Answer: $x^3-(a^2-2b)x^2 + (b^2-2ac)x -c^2 = 0$.

[Ashutosh0x]

Hey @danio2010, nice catch! You're absolutely right the signs were off. I've put up a fix in PR #4 that corrects both lines 24 and 28: a^2+2b a^2-2b. It's a straightforward Vieta's expansion issue, since (-a)^2 - 2b = a^2 - 2b, not a^2 + 2b. Thanks for flagging this!

[Ashutosh0x]

Hey @danio2010, nice catch! You're absolutely right the signs were off. I've put up a fix in PR #4 that corrects both lines 24 and 28: a^2+2b → a^2-2b. It's a straightforward Vieta's expansion issue since (-a)^2 - 2b = a^2 - 2b, not a^2 + 2b. Thanks for flagging this!