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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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up vote 5 down vote accepted

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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Amazing...I would have spent days seeing that! – fretty Feb 25 '12 at 11:30
It is a solution...the LHS is $((3)*1 + 2 + 3)(2(2) - 1) =8*3 = 24$ and the RHS is $2*3*4 = 24$. – fretty Feb 25 '12 at 11:44
Gerry's equation is just a rearrangement of the original equation...he probably just noticed that if you isolate the right things then you get a nice factorisation. Then the rest of the argument is pure genius. – fretty Feb 25 '12 at 11:50
@fretty: you are right, I had a type, thank you – miracle173 Feb 25 '12 at 12:23

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