1979 AHSME Problems/Problem 19

Problem 19

Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$ .

$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$

Solution

Solution by e_power_pi_times_i

Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$ . However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$ , and dilating by $2$ gives $\pm2$ . The sum of the squares is $(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}$

Solution 2

Notice that we can change the equation to be $x^{256}=256^{32}$ .

The RHS can be simplified to $x^{256}=2^{256}$

Hence, the only real solutions are $x=-2,2$

Hence, $(-2)^2+(2)^2=8=\boxed{A}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1979 AHSME Problems/Problem 19

Contents

Problem 19

Solution

Solution 2

See also