论文标题
关于总和的整数值和三个正理性数的乘积
On integer values of sum and product of three positive rational numbers
论文作者
论文摘要
在1997年，我们证明，如果$ n $是$ 4K的形式，\ quad 8k-1 \ quad {\ rm或} \ quad 2^{2m+1}（2k-1}（2k-1）+3，$ k，m \ in \ in \ mathbb n，$ in \ mathbb n，$，那么没有正值的$ x，y，y，y，y，y，z $ y+y+y+ $$最近，N。X. tho证明了以下语句：让$ a \ in \ mathbb n $奇怪，让$ n \ equiv 0 \ equiv 0 \ pmod 4 $或$ n \ equiv 7 \ equiv 7 \ pmod 8 $。那么方程式$$ xyz = a，\ Quad x+y+z = an。 $$在正理性数字中没有解决方案$ x，y，z。$ 我们结果的代表性示例是以下陈述：假设$ a，n \ in \ mathbb n $如此至少以下条件之一： $ \ bullet $ $ n \ equiv 0 \ pmod 4 $ $ \ bullet $ $ n \ equiv 7 \ pmod 8 $ $ \ bullet $ $ a \ equiv 0 \ pmod 4 $ $ \ bullet $ $ a \ equiv 0 \ pmod 2 $和$ n \ equiv 3 \ pmod 4 $ 美元 那么方程式$$ xyz = a，\ Quad x+y+z = an。 $$在正理性数字中没有解决方案$ x，y，z。$
In 1997 we proved that if $n$ is of the form $$ 4k, \quad 8k-1\quad {\rm or} \quad 2^{2m+1}(2k-1)+3, $$ where $k,m\in \mathbb N,$ then there are no positive rational numbers $x,y,z$ satisfying $$ xyz = 1, \quad x+y+z = n. $$ Recently, N. X. Tho proved the following statement: let $a\in\mathbb N$ be odd and let either $n\equiv 0\pmod 4$ or $n\equiv 7\pmod 8$. Then the system of equations $$ xyz = a, \quad x+y+z = an. $$ has no solutions in positive rational numbers $x,y,z.$ A representative example of our result is the following statement: assume that $a,n\in\mathbb N$ are such that at least one of the following conditions hold: $\bullet$ $n\equiv 0\pmod 4$ $\bullet$ $n\equiv 7\pmod 8 $ $\bullet$ $a\equiv 0\pmod 4$ $\bullet$ $a\equiv 0\pmod 2$ and $n\equiv 3\pmod 4$ $\bullet$ $a^2n^3=2^{2m+1}(2k-1)+27$ for some $k,m\in \mathbb N.$ Then the system of equations $$ xyz = a, \quad x+y+z = an. $$ has no solutions in positive rational numbers $x,y,z.$
