Solutions of symmetric three variable equations.

[alyafey22]

Inspired by this http://mathhelpboards.com/challenge-questions-puzzles-28/solving-system-equations-8521.htmlit is interesting to look at the solutions of

\(\displaystyle xz+y=a\)

\(\displaystyle xy+z=a\)

\(\displaystyle zy+x=a\)

where we look at the case \(\displaystyle a>0\)

Assume the following \(\displaystyle a=bc\) where \(\displaystyle a\neq 0\)

Then we have

If we assume that \(\displaystyle c\) is a solution we have

\(\displaystyle c^2+c=bc\) so \(\displaystyle c+1=b\) so

\(\displaystyle a=c(c+1)\). Hence if $a$ can be factorized to the multiplications of two consecutive numbers we have the least number as a solution.

It is also immediate to see that the permutations of \(\displaystyle (1,1,a-1)\) is always a solution.

Also it is immediate to see that \(\displaystyle -(c+1)\) is a solution.
Hence we have five solutions \(\displaystyle x=y=z=c,x=y=z=-(c+1),(1,1,a-1),(1,a-1,1),(a-1,1,1)\).

 

[jvicens]

Thank you for sharing this interesting observation! It is indeed fascinating to see how the solutions of these equations can be related to the factorization of a. It also highlights the importance of considering all possible solutions, including negative solutions and permutations. This can also be extended to other values of a, such as a<0 or a=0, and exploring how the solutions change in those cases. This forum post is a great example of how mathematical thinking and problem-solving skills can be applied to real-world problems.