Do there exist an infinite number of complex solutions of $3^z+4^z=5^z$?

Solution 1:

It is a consequence of Hadamard's factorization theorem that any entire function of order $1$ with finitely many zeros is of the form $e^{az}p(z)$, where $a \in \mathbb{C}$ and $p(z)$ is a polynomial. The function $f(z) = 3^z + 4^z - 5^z$ is entire and of order $1$ but is not of this form, so it must have infinitely many zeros.

Solution 2:

Summary

For the equation $3^z + 4^z = 5^z$,

1. $z = 2$ is the only root when $\Re z \ge 2$.
2. There are infinitely many complex roots within the strip: $\;\;2 - \varepsilon < \Re z < 2$.

Details

Let $u, v$ be two positive numbers such that $u^2 + v^2 = 1$. Let $\mu = \frac{|\log u|}{2\pi}, \nu = \frac{|\log v|}{2\pi}$ and consider following Dirichlet polynomial equation:

$$\varphi(z) = u^z + v^z = 1\tag{*1}$$

1. By construction, $z = 2$ is a root of $(*1)$.

2. For $\Re z > 2$, we have: $$|u^z+v^z| \le |u^z| + |v^z| = u^{\Re z} + v^{\Re z} < u^2 + v^2 = 1.$$ This means $(*1)$ doesn't have any root with $\Re z > 2$.

3. For $\Re z = 2$, let $z = 2 + it$, we have: $$\begin{align} u^z+v^z = 1 \iff & u^2 e^{-2\pi\mu t i} + v^2 e^{-2\pi\nu t i} = 1\\ \iff & e^{-2\pi\mu t i} = e^{-2\pi\nu t i} = 1\\ \iff & \mu t, \nu t \in \mathbb{Z} \end{align}$$ For $t \ne 0$, it is clear the last condition $\mu t, \nu t \in \mathbb{Z}$ is possible when and only when $\frac{\mu}{\nu}$ is a rational number. This means:

   • if $\frac{\mu}{\nu} \in \mathbb{Q}$, $(*1)$ has infinitely many roots on the line $\Re z = 2$.
   • if $\frac{\mu}{\nu }\notin \mathbb{Q}$, $z = 2$ is the only root of $(*1)$ with $\Re z \ge 2$.

Let us now concentrate on the more interesting case where $\frac{\mu}{\nu} \notin \mathbb{Q}$. We know there are infinitely many pairs of positive integers $p,q$ such that $$\left|\frac{\mu}{\nu} - \frac{p}{q}\right| < \frac{1}{q^2}$$

For any such pair of integers $p, q$, let $t_{\mu} = \frac{p}{\mu}$, $t_{\nu} = \frac{q}{\nu}$, $t_{<} = \min(t_\mu,t_\nu)$, and $t_{>} = \max(t_\mu,t_\nu)$. When $t$ varies between $t_{<}$ and $t_{>}$, we have

$$\begin{align} & |\mu t - p | \le |\mu t_{\nu} - p| = | \frac{\mu}{\nu} q - p | < \frac{1}{q}\\ & |\nu t - q | \le |\nu t_{\mu} - q| = | \frac{\nu}{\mu} p - q | < \frac{\nu}{\mu q}\tag{*2} \end{align}$$

Let $K = \max(\frac{\nu}{\mu},1)$, one consequence of $(*2)$ is when $q > 4K$, the angles of following two circular arcs on unit circle

$$ C_\mu = \Big\{ \omega = e^{-2\pi\mu t i} : t \in [t_<, t_>]\Big\} \quad\text{ and }\quad C_\nu = \Big\{ \omega = e^{-2\pi\nu t i} : t \in [t_<, t_>]\Big\} $$ are both smaller than $\frac{\pi}{2}$.

Since $C_{\mu}$ touches $1$ at and only at $t = t_\mu$ and $C_{\nu}$ touches $1$ at and only at $t = t_\nu$, One of $C_{\mu}$ or $C_{\nu}$ ( $C_\mu$ if $t_\mu > t_\nu$, $C_\nu$ otherwise ) lies completely in the quadrant $\Re \omega > 0, \Im \omega \ge 0$ while the other one lies completely in the quadrant $\Re \omega > 0, \Im \omega \le 0$.

Since $C_\mu$ and $C_\nu$ touches $1$ at different $t$, we can deduce $0 < \Re \varphi(2+it) < 1$ for $t \in (t_<, t_>)$. Furthermore, it is not hard to see $\Im \varphi(2 + it_<) > 0$ and $\Im \varphi(2 + it_>) < 0$.

Let $z_< = 2 + i t_<$, $z_> = 2 + i t_>$ and $m = \min(\mu,\nu)$. In addition to $q > 4K$, let us assume we have chosen a $q$ so large such that $$e^{2\pi m\varepsilon}\cos(\frac{2\pi K}{q}) > 1.$$ Let $\mathscr{R}$ be the rectangular contour joining $z_<$, $z_>$, $z_>\!-\varepsilon$, $z_<\!-\varepsilon$ and then back to $z$. We will look at what happens to $\omega = \varphi(z) - 1$ when $z$ walks around $\mathscr{R}$ once.

1. Along the line segment $[z_<,z_>]$, above discussion implies $\omega$ moves from the quadrant $\Re \omega < 0, \Im \omega > 0$ to the quadrant $\Re \omega < 0, \Im \omega < 0$.

2. Along the line segment $[z_>, z_>\!-\varepsilon]$, it is easy to see $\Im\omega$ doesn't change sign and remains $< 0$ all the time.

3. Along the line segment $[z_>\!-\varepsilon, z_<\!-\varepsilon]$, we have $$\Re\omega \ge u^2 e^{2\pi\mu\varepsilon}\cos(\frac{2\pi}{q}) + v^2 e^{2\pi\nu\varepsilon}\cos(\frac{2\pi\nu}{\mu q}) - 1 \ge e^{2\pi m\varepsilon}\cos(\frac{2\pi K}{q}) - 1 > 0 $$

4. Along the line segment $[z_<\!-\varepsilon, z_<]$, $\Im\omega$ doesn't change sign again and remains $> 0$ all the time.

Combine these 4 observations, we can conclude when $z$ walks around $\mathscr{R}$ once, $\omega$ walks around the origin counter-clockwisely once. From this, we can conclude $\varphi(z) = 1$ has a root inside $\mathscr{R}$. Since there are infinitely many $q$ one can chose from, we conclude $\varphi(z) = 1$ has infinitely many roots in the strip $2 - \varepsilon < \Re z < 2$.

To answer the original question, let $u = (3/5)^2$ and $v = (4/5)^2$. It is known that $\log 2$, $\log 3$ and $\log 5$ are linearly independent over $\mathbb{Q}$. This means $\frac{\mu}{\nu} = \frac{\log u}{\log v} = \frac{\log 5 - \log 3}{\log 5 - 2\log 2} \notin \mathbb{Q}$. By above discussions, the conclusions in the summary follow immediately.