2021 JMPSC Sprint Problems/Problem 1
Solving the right side gives . Distributing into the left side gives , so the answer is .
and , so the answer is .
Therefore, the answer is
- kante314 -
Solution by pog
Recall that, for any nonzero , , , we have that . By Vieta's, we see that , , and are the roots of the monic cubic polynomial where is an arbitrary constant equal to .
In the given equation, we have that , , and , so and , for (Finding the former of the latter two coefficients would lead to bashy casework, which we avoid in this extremely elegant solution.)
Since our cubic has roots , , and , the discriminant of our monic cubic polynomial (where , , and are the latter three coefficients of , respectively--do not confuse them for the roots of our polynomial equal to , , and , respectively) must not be equal to , or our cubic will have a double root, which it clearly does not (proof here is left to the reader).
Thus, substitution of the coefficients into our discriminant gives that must not be equal to . Letting , expanding gives that is not equal to . Thus must not be a root of our cubic, so from here we apply the cubic formula to find that is not equal to (approximately) , , or .
Recall that, from earlier, our answer is equal to . By bounding, is between the aforementioned values of and , so since it must be a multiple of (all answers in the Junior Mathematicians' Problem Solving Competition are integers), it is equal to and the requested answer is equal to . This corroborates with jasperE3's strict bounding.
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