Question

The number of pairs of integer $(x, y)$ that satisfy the following two equations $\begin{cases} \cos(xy) = x \\ \tan(xy) = y \end{cases}$

Answer 1

1

Answer 2

The given system of equations is:

1. $\cos(xy) = x$
2. $\tan(xy) = y$

Let $A = xy$. The equations can be rewritten as:

1. $\cos(A) = x$
2. $\tan(A) = y$

From equation (1), $\cos(A) = x$. We know that the range of the cosine function is $[-1, 1]$. Since $x$ is an integer, $x$ must be one of the values $\{-1, 0, 1\}$.

We will analyze each possible value for $x$:

Case 1: $x = 0$

Substitute $x=0$ into equation (1): $\cos(0 \cdot y) = 0$ $\cos(0) = 0$ $1 = 0$

This is a contradiction. Therefore, $x$ cannot be $0$.

Case 2: $x = 1$

Substitute $x=1$ into equation (1): $\cos(1 \cdot y) = 1$ $\cos(y) = 1$

For $\cos(y)=1$, $y$ must be of the form $2n\pi$ for some integer $n$. Since $y$ is an integer, $y \in \mathbb{Z}$. The only way $2n\pi$ can be an integer is if $n=0$, because $\pi$ is an irrational number. If $n \ne 0$, then $2n\pi$ is irrational. So, $n=0 \implies y=0$. Now we have the pair $(x, y) = (1, 0)$. Let's check this pair in equation (2): $\tan(xy) = y$ $\tan(1 \cdot 0) = 0$ $\tan(0) = 0$ This is true. Thus, $(1, 0)$ is a solution.

Case 3: $x = -1$

Substitute $x=-1$ into equation (1): $\cos(-1 \cdot y) = -1$ $\cos(-y) = -1$

Since $\cos(-\theta) = \cos(\theta)$, we have $\cos(y) = -1$. For $\cos(y)=-1$, $y$ must be of the form $(2n+1)\pi$ for some integer $n$. Since $y$ is an integer, $y \in \mathbb{Z}$. The expression $(2n+1)\pi$ can only be an integer if $2n+1=0$ (which is not possible for an integer $n$) or if $\pi$ were rational (which is not true). Since $2n+1$ is a non-zero integer for any integer $n$, $(2n+1)\pi$ is an irrational number. An irrational number cannot be an integer. Therefore, there are no integer values for $y$ in this case.

Combining all cases, the only pair of integers $(x, y)$ that satisfies the given equations is $(1, 0)$. The number of such pairs is 1.