SOLUTION: <pre> Suppose x is a positive number such that {{{x^2=1-x}}}. There is a unique choice of whole numbers p and q so that {{{x^8=p-qx}}}. Find p+q. My attempt: I know that if

Question 1196201:
Suppose x is a positive number such that . There is a unique choice of whole numbers p and q so that . Find p+q. My attempt: I know that if , then , which simplifies to . Then what?

Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

            It is VERY nice problem, admitting BEATIFUL solution.
            See below.

If x^2 = 1-x, then x^8 = (1-x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4. The idea is to replace in this expression, and in all expressions that follow, x^2 by (1-x) everywhere, where it is possible, and as many times as possible, until you get the desired expression of degree 1 (one). See how it works x^8 = 1 - 4x + [6(1-x)] - [4x*(1-x)] + [(1-x)*(1-x)] = I continue = = 1 - 4x + 6 - 6x - 4x + 4x^2 + 1 - 2x + x^2 = 8 - 14x + 5x^2 = I replace x^2 by (1-x) again = = 8 - 14x + 5*(1-x) = 8 - 16x + 5 - 5x = 13 - 21x. So, p = 13, q = 21 and p + q = 13 + 21 = 34. ANSWER

Solved.

This method is called " the lowering of a degree " method.

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It can be solved by different methods, but this one is a " true delight ".

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Suppose x is a positive number such that . There is a unique choice of whole numbers p and q so that . Find p+q. My attempt: I know that if , then , which simplifies to . Then what?
I got p + q = 34 You already know that: So, ---- Substituting ----- Substituting 1 - x for x² Finally, , so p = 13, and q = 21. Therefore, p + q = 13 + 21 = 34

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