Lets say I start with an equation $$3(1+x+x^2)(1+y+y^2)(1+z+z^2)=4x^2y^2z^2-1$$ and I get a solution in positive integers $(4,4^3,4^6)$. To go about showing that there are no solutions possible in odd positive integers $x_0,y_0,z_0$ I perform a substitution with $x_0$ being the least odd positive solution. So I get that: $$3(1+x_0+x_0^2)(1+x_0+c_1+(x_0+c_1)^2)(1+x_0+c_2+(x_0+c_2)^2)=4x_0^2\cdot(x_0+c_1)^2)(x_0+c_2)^2-1 $$ Doing this gets one variable and I can assume that $c_1,c_2$ constants are even and positive since$ x_0,y_0,z_0$ are odd. If I actually multiplied out and simplified it would show that there exist only one positive solution using Descartes rule of signs, and I know that one positive solution exists with $x_0=4$ but if I am assuming something about $x_0$ being odd does that mean I'm bouncing between cases? In other words when I rewrite everything in terms of $x_0$ then my constants are somehow tied to what I am assuming about $x_0$? So I cannot say that I have found the only positive root hence none other exist because the constants depend on what $x_0$ is?
Asked
Active
Viewed 35 times
2
-
In case someone wonders about Descartes rule of signs I will deal with that later so this is sort of hypothetical – thestar May 24 '20 at 22:12
-
Maybe there is a map between powers of 2 and some set of positive integers – thestar May 24 '20 at 23:35
-
1Another question with this precise equation was asked a few days ago and also got answered. Its poster asked a few more questions of similar nature and also mentioned Descartes rule of signs in one of them. The coincidence is somewhat strange; what is the actual origin of this problem? – Peter Košinár May 27 '20 at 22:14