2023 AIME I Problems/Problem 15

Problem 15

Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying

the real and imaginary parts of $z$ are integers;
$|z|=\sqrt{p}$ , and
there exists a triangle with side lengths $p$ , the real part of $z^{3}$ , and the imaginary part of $z^{3}$ .

Answer: 349

Suppose $z=a+bi$ ; notice that $\arg(z^{3})\approx 45^{\circ}$ , so by De Moivre’s theorem $\arg(z)\approx 15^{\circ}$ and $\tfrac{b}{a}\approx tan(15^{\circ})=2-\sqrt{3}$ . Now just try pairs $(a, b)=(t, \left(2-\sqrt{3}\right)t)$ going down from $t=\lfloor\sqrt{1000}\rfloor$ , writing down the value of $a^{2}+b^{2}$ on the right; and eventually we arrive at $(a, b)=(18, 5)$ the first time $a^{2}+b^{2}$ is prime. Therefore, $p=\boxed{349}$ .

Solution

Denote $z = a + i b$ . Thus, $a^2 + b^2 = p$ .

Thus, $z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) .$

Because $p$ , ${\rm Re} \left( z^3 \right)$ , ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have ${\rm Re} \left( z^3 \right) > 0$ and ${\rm Im} \left( z^3 \right) > 0$ . Thus, $\begin{align*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{align*}$

Because $p$ , ${\rm Re} \left( z^3 \right)$ , ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have the following triangle inequalities: $\begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}$

We notice that $| z^3 | = p^{3/2}$ , and ${\rm Re} \left( z^3 \right)$ , ${\rm Im} \left( z^3 \right)$ , and $| z^3 |$ form a right triangle. Thus, ${\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}$ . Because $p > 1$ , $p^{3/2} > p$ . Therefore, (3) holds.

Conditions (4) and (5) can be written in the joint form as $\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4)$

We have $\begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*}$ and $p = a^2 + b^2$ .

Thus, (5) can be written as $\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| < a^2 + b^2 . \hspace{1cm} (6)$

Therefore, we need to jointly solve (1), (2), (6). From (1) and (2), we have either $a, b >0$ , or $a, b < 0$ . In (6), by symmetry, without loss of generality, we assume $a, b > 0$ .

Thus, (1) and (2) are reduced to $a > \sqrt{3} b . \hspace{1cm} (7)$

Let $a = \lambda b$ . Plugging this into (6), we get $\begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) \end{align*}$

Because $p= a^2 + b^2$ is a prime, $a$ and $b$ are relatively prime.

Therefore, we can use (7), (8), $a^2 + b^2 <1000$ , and $a$ and $b$ are relatively prime to solve the problem.

To facilitate efficient search, we apply the following criteria:

\begin{enumerate} \item To satisfy (7) and $a^2 + b^2 < 1000$ , we have $1 \leq b \leq 15$ . In the outer layer, we search for $b$ in a decreasing order. In the inner layer, for each given $b$ , we search for $a$ . \item Given $b$ , we search for $a$ in the range $\sqrt{3} b < a < \sqrt{1000 - b^2}$ . \item We can prove that for $b \geq 9$ , there is no feasible $a$ . The proof is as follows.

For $b \geq 9$ , to satisfy $a^2 + b^2 < 1000$ , we have $a \leq 30$ . Thus, $\sqrt{3} < \lambda \leq \frac{30}{9}$ . Thus, the R.H.S. of (8) has the following upper bound $\begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}$

Hence, to satisfy (8), a necessary condition is $\begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{10}{27} . \end{align*}$

However, this cannot be satisfied for $\sqrt{3} < \lambda \leq \frac{30}{9}$ . Therefore, there is no feasible solution for $b \geq 9$ . Therefore, we only need to consider $b \leq 8$ .

\item We eliminate $a$ that are not relatively prime to $b$ .

\item We use the following criteria to quickly eliminate $a$ that make $a^2 + b^2$ a composite number. \begin{enumerate} \item For $b \equiv 1 \pmod{2}$ , we eliminate $a$ satisfying $a \equiv 1 \pmod{2}$ . \item For $b \equiv \pm 1 \pmod{5}$ (resp. $b \equiv \pm 2 \pmod{5}$ ), we eliminate $a$ satisfying $a \equiv \pm 2 \pmod{5}$ (resp. $a \equiv \pm 1 \pmod{5}$ ). \end{enumerate}

\item For the remaining $\left( b, a \right)$ , check whether (8) and the condition that $a^2 + b^2$ is prime are both satisfied.

The first feasible solution is $b = 5$ and $a = 18$ . Thus, $a^2 + b^2 = 349$ .

\item For the remaining search, given $b$ , we only search for $a \geq \sqrt{349 - b^2}$ . \end{enumerate} $$ (Error compiling LaTeX. Unknown error_msg)

Following the above search criteria, we find the final answer as $b = 5$ and $a = 18$ . Thus, the largest prime $p$ is $p = a^2 + b^2 = \boxed{\textbf{(349) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/V0KFMIXmp08

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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2023 AIME I Problems/Problem 15

Contents

Problem 15

Answer: 349

Solution

Video Solution