Descartes rule of sign multivariable If I have a polynomial say $$7(1+x+x^{2})(1+y+y^{2})-8x^2y^2=0$$ is there a way to determine how many possible positive, negative, $\underline {integer}$ solutions exist? 
 A: The equation
$$7(1+x+x^{2})(1+y+y^{2})=8x^2y^2$$
does not have any integer solutions.
This is because 
$$\text{LHS}=7\bigg(1+\underbrace{x(x+1)}_{\text{even}}\bigg)\bigg(1+\underbrace{y(y+1)}_{\text{even}}\bigg)=\text{odd}$$
while RHS is even.
A: 568 has set of factors $\{1,2,4,8,71,142,284,568\}$
It is not possibe for $1+y+y^2 < 0$
Therefore we don't have to look at the negatives of the factor pairs.
There are as many as 2 roots for $1+y+y^2 = c$ for each of the facors, and 3 roots for $1+x+x^2 + x^3 = d$ where $cd = 568.$  That gives six potetial $(x,y)$ pairs for each $(c,d)$ pair.
Or, $48$ potential $(x,y)$ pairs.
A: The question is very deep. In fact, tenth Hilbert's problem states:

Is there an algorithm which determines, of an arbitrary diophantine equation, 
  whether it has a solution in the natural numbers?

The answer is negative, see the linked article for details.
For a specific 

Question: For which $d, k \in \mathbb{N}$ is there an algorithm to determine
  whether a polynomial diophantine equation 
  $$
  P(x_1, \dots, x_k)  = 0, \ \ \ P \in \mathbb{Z}[x_1, \dots, x_k]
$$
  of total degree $d$ in $k$ variables has a solution in integers? --
  And for which $d$, $k$ is there no such algorithm,
  respectively, is it unknown whether such algorithm exists or not?

see this MathOverflow thread.
