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- #1

- Jan 17, 2013

- 1,667

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 \(\displaystyle x=y=z=c,x=y=z=-(c+1),(1,1,a-1),(1,a-1,1),(a-1,1,1)\).

\(\displaystyle xz+y=a\)

\(\displaystyle xy+z=a\)

\(\displaystyle zy+x=a\)

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 \(\displaystyle x=y=z=c,x=y=z=-(c+1),(1,1,a-1),(1,a-1,1),(a-1,1,1)\).

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