# Solutions of symmetric three variable equations.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Inspired by this thread it 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 $$\displaystyle x=y=z=c,x=y=z=-(c+1),(1,1,a-1),(1,a-1,1),(a-1,1,1)$$.

Last edited: