解不定方程a,b,c为整数

realnumber

$a,b,c\in Z,\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=1$,求a,b,c

青青子衿

回复 1# realnumber
似乎是无整数解的
\begin{align*}
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) &=n\\
\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} &= N =n-3
\end{align*}
Melvyn J. Knight's problem
asahi-net.or.jp/~KC2H-MSM/mathland/math03/xyz00.htm

A085514   Integers n representable as the product of the sum of three nonzero integers
with the sum of their reciprocals: n=(x+y+z)*(1/x+1/y+1/z).

根据OEIS上的数据，n=1, 9, 10, 11, 14, 15, 18, 26, 29, 30, ...
由于n取不到4，那么N取不到1。
oeis.org/A085514

realnumber

谢谢，看来被那2个初中，坑了

青青子衿

谢谢，看来被那2个初中，坑了
realnumber 发表于 2021-3-10 16:11

什么叫“二个初中”？