Mock AIME 1 2010 Problems/Problem 6

Problem

Find the number of Gaussian integers $z$ with magnitude less than 10000 such that there exists a different Gaussian integer $w$ such that $z = w^4$. (The magnitude of a complex $a+bi$, where $a$ and $b$ are reals, is defined to be $\sqrt{a^2+b^2}$. A Gaussian integer is defined to be a complex number whose real and imaginary parts are both integers.)

Solution

Because $|z|<10000$ and $z=w^4$, we know that $|w|<\sqrt[4]{10000}=10$. Thus, the number of Gaussian integers within the open disk that satisfies $|w|<10$ is the number of possible $w$. This is equal to the number of lattice points in the open disk $x^2+y^2<100$ on the $xy$ plane. However, we have to be careful, because for every $z\neq0$, there are $4$ different values of $w$ which satisfy $z=w^4$. These $4$ different values are all $i$-multiples of each other, so they are all Gaussian integers, and none of them lie in the same quadrant. Furthermore, because, from the problem, $w\neq z$, we know that $w$ cannot be $0$ or $1$. However, $z$ can still equal $1$ if $w=-1$, so $w\neq 1$ will not affect our final answer. Thus, we will count the number of lattice points in the open disk $x^2+y^2<100$ on the positive $x$-axis and the first quadrant.

When $x=9$, $0\leq y\leq4$, so we have $5$ points here.

When $x=8$, $0\leq y\leq5$, so we have $6$ points here (remember we are dealing with a strict inequality, so $y=6$ does not work).

When $x=7$, $0\leq y\leq7$, so we have $8$ points here.

When $x=6$, $0\leq y\leq7$, so we have $8$ points here ($y=8$ does not work).

When $x=5$, $0\leq y\leq8$, so we have $9$ points here.

When $x\in\{1,2,3,4\}$, $0\leq y\leq9$, so we have $10$ points for each of these four values for a total of $40$ points.

We are not counting the points where $x=0$, because those are on the $y$-axis. Thus, our final answer is $5+6+8+8+9+40=\boxed{076}$.

See Also

Mock AIME 1 2010 (Problems, Source)
Preceded by
Problem 5
Followed by
Problem 7
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