1983 AIME Problems/Problem 5

Problem

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solutions

Solution 1

One way to solve this problem is by substitution. We have

$x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$

Hence observe that we can write $w=x+y$ and $z=xy$ .

This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$ .

Because we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$ .

$w^2-7=2z \implies z=\frac{w^2-7}{2}$ .

Substituting, $w^3-21w+20=0$ , which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division).

The largest possible solution is therefore $x+y=w=\boxed{004}$ .

Solution 2

An alternate way to solve this is to let $x=a+bi$ and $y=c+di$ .

Because we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$ .

Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.

$(a+bi)^2+(c-bi)^2=7+0i$

$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$

Looking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates.

Looking at the real part of the equation and plugging in $a=c$ , $2a^2-2b^2=7$ , or $2b^2=2a^2-7$ .

Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ):

$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$

$2a^3-6ab^2=10$

Since we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield:

$2a^3-3a(2a^2-7)=10$

$-4a^3+21a=10$

$4a^3-21a+10=0$

Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=\frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$ , and their sum is $\boxed{004}$ .

Solution 3

Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$ , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$ . Substituting $c=\frac{b^2-7}{2}$ , we deduce that $b^3-21b-20=0$ , whose roots are $-4$ , $-1$ , and $5$ . Since $-b$ is the sum of the roots and is maximized when $b=-4$ , the answer is $-(-4)=\boxed{004}$ .

Solution 4

$x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ . Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ . Testing $p = 1$ , we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}$

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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