Solving a Diophantine equation: $y^x=x^{2007}$, $x$ and $y$ integers.

Question

I found this Diophantine equation and to solve it I used the definition of logarithm but the solution doesn't require the use of logarithmic rules. I solved it in this way: $$y^x=x^{2007}$$ $$\log_xy^x=2007$$ $$\log_xy=\frac {2007} {x}$$ Therefore $x$ is a divisor of $2007=3^2\cdot 223$, and the only possible values of $x$ are $1,3,9,223,669,2007$. How can I solve this equation with other methods?

Answer 1

Let $x=\prod_i {p_i}^{m_i}$, where $m_i\ge 1$ and $p_i$ prime, then $y=\prod_i {p_i}^{n_i}$ for $n_i\ge 0$. Plugging this in the equation have $2007 m_i = n_i x = n_i \prod_j {p_j}^{m_j}$.

Suppose that $p_j\nmid 2007$, then ${p_j}^{m_j} \mid m_j$ which implies $2^{m_j} \le m_j$, contradiction, hence $p_j \mid 2007$.

Also ${p_i}^{m_i}\mid 2007 m_i$. Note that $m_i$ is upper bounded. By your factor $2007=3^2\cdot 223$, for $p_i=3$ you have $m_i=1,2,3$, and for $p_i=223$ you have $m_i=1$.

You can now enumerate the 8 possibilities by hand using the above relation we derived.

If $x\mid 2007$ then everything works out. This includes $1, 3, 9, 223, 669, 2007$. The other two we need to check are $27$ and $6021$. Of these only $27$ works.