Xin đưa lên lời giải đó
--------------Rearranging we have $x^y + 19 = xy^2 + y^x$, so $y|(x^y + 19)$
Similarly we get $-y^x + 19 = xy^2 - x^y$, so $x|(y^x - 19)$.
Obviously $x \not = y$, so as $x,y$ are prime we have by FLT $x^y \equiv x \pmod{y}$.
Similarly $y^x \equiv y \pmod{x}$.
But this means $y|(x+19)$ and $x|(y - 19)$ so $y \le x + 19$ and $x \le y - 19$ if $y - 19$ is positive, but this implies $x = y - 19$, a clearly contradiction so there are no solutions.
Now consider when $y - 19$ is negative, so $y \le 19$. If $y = 2$, it must be that $x = 17$, which is obviously false. If $y = 3$, we must have $x = 2$. If you go through the other cases of $y \le 19$, it is obvious the only working cases are $\boxed{(x,y) = (2,3),(2,7)}$.
FLT: Fermat Little Theorem = Định lý nhỏ Fermat
Bài viết đã được chỉnh sửa nội dung bởi hxthanh: 26-10-2011 - 17:12