2006 Alabama ARML TST Problems/Problem 14

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Problem

Find the real solution $(x, y)$ to the system of equations

$x^3-3xy^2=-610,$ $3x^2y-y^3=182.$

Solution

Note that $(x+yi)^3 = x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = (x^3-3xy^2) + (3x^2y-y^3)i = -610 + 182i$

So, we need to find $(x+yi)$ that satisfies $(x+yi)^3 = -610 + 182i$ . Even though there are three solutions, for simplicity, lets assume that one exists in the first quadrant in the complex plane, and try to find that one.

Note that $|(x+yi)^3|^2 = (-610)^2 + (182)^2 = 372100 + 33124 = 405224 = 8 \cdot 50653 = 2^3 \cdot 37^3$ .

Thus, $|x+yi|^2 = \sqrt[3]{2^3 \cdot 37^3} = 2\cdot37 = 74$ .

Therefore, $x^2 + y^2 = 74$ . Note that $(x,y) = ( 5,7)$ and $(7,5)$ are the only positive integer solutions to $x^2 + y^2 = 74$ . (Even though the solution may be non-integral, this is a good place to start.)

However, letting $r \cdot cis(\theta) = x+yi$ , [ $(x+yi)^3 = r^3 \cdot cis(3\theta)$ ] and bounding $\theta$ yields:

$-1 < \tan{(3\theta)} = -\dfrac{182}{610} < 0$ .

$135^\circ < 3\theta < 180^\circ$ . (Since $3\theta$ is obviously in the 2nd quadrant).

$45^\circ < \theta < 60^\circ$ .

$\tan{(\theta)} > 1$ .

Thus, $y>x$ . So, the only remaining positive integer solution is $(x,y) = (5,7)$ .

As a quick check: $(5+7i)^3 = (5)^3 + 3(5)^2(7)i + 3(5)(7)^2i^2 + (7)^3i^3 = -610 + 182i$

Thus, the solution is $(x,y) = \boxed{(5,7)}$ .

Note that $(5+7i)cis(120^\circ) = \left(\dfrac{-5-7\sqrt{3}}{2}\right) + \left(\dfrac{-7+5\sqrt{3}}{2}\right)i$ and $(5+7i)cis(240^\circ) = \left(\dfrac{-5+7\sqrt{3}}{2}\right) + \left(\dfrac{-7-5\sqrt{3}}{2}\right)i$

Thus, $(x,y) = \left(\dfrac{-5-7\sqrt{3}}{2},\dfrac{-7+5\sqrt{3}}{2} \right)$ and $\left(\dfrac{-5+7\sqrt{3}}{2},\dfrac{-7-5\sqrt{3}}{2} \right)$ are also solutions.

2006 Alabama ARML TST (Problems)
Preceded by: Problem 13	Followed by: 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2006 Alabama ARML TST Problems/Problem 14

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Problem

Solution

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