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Consider the diophantine equation $Q(x,y,z)=0$, where $x$, $y$ and $z$ are nonnegative integer unknowns and

$ Q(x,y,z)=x^3 + (-2y + 2)x^2 + ((z - 6)y + (2z + 1))x + ((2z - 4)y + 3z) $

Since the degree of $Q$ in $y$ and $z$ is $1$, the equation seems tractable. If $t\geq 0$, then $(x,y,z)=(t,t,t)$ is a solution. Are there any others ?

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Modulo mistakes in my algebra, the equation can be rewritten as $((x+2)y+2x+3)(2(x+1)-z)=(x+1)(x+2)(x+3)$ So my suggestion would be, test $x=1,2,3,\dots$ in turn. For each $x$, you only have to try values of $z$ with $z\lt2(x+1)$ and with $2(x+1)-z$ a factor of $(x+1)(x+2)(x+3)$. Then you just have to see whether $y$ comes out integral. You should be able to test lots of values of $x$ pretty quickly and get some idea of what the solution space looks like.

EDIT: Perhaps a better idea is to look at the equation modulo $x+2$. It becomes $z\equiv-2\pmod{x+2}$. Combined with $0\lt z\lt2(x+1)$, this proves $z=x$. So now we have $(x+2)y+2x+3=(x+1)(x+3)$ which is $y=x$, QED.

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    @fretty: you are right, I had a type, thank you2012-02-25