Polynomial equation in many variable. No algorithm at all? I read an interesting paper by S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani on NP-problems (https://people.eecs.berkeley.edu/~vazirani/algorithms/chap8.pdf) and found there the following statement:
"At least an NP-complete problem can be solved by some algorithm—the trouble is that this algorithm will be exponential. But it turns out there are perfectly decent computational problems for which no algorithms exist at all! One famous problem of this sort is an arithmetical version of SAT. Given a polynomial equation in many variables, perhaps $x^3yz + 2y^4z^2 − 7xy^5z = 6$, are there integer values of $x$, $y$, $z$ that satisfy it? There is no algorithm that solves this problem. No algorithm at all, polynomial, exponential, doubly exponential, or worse! Such problems are called unsolvable."
I wonder what do you guys think about it? 
I dare to suggest that "No algorithm at all" is a slight exaggeration.  The appropriate algorithm surely exists, otherwise I wouldn’t be able to find out integer values of $x,y,z$ for the following polynomial equation which is similar to the one given as an example:
x3yz + 2y4z2 - 7xy5z = 8

-3, -1, -4 = 8
-3, -1, 1 = 8
-3, 1, -1 = 8
-3, 1, 4 = 8
-2, -1, -1 = 8
-2, -1, 4 = 8
-2, 1, -4 = 8
-2, 1, 1 = 8
-1, -1, -1 = 8
-1, -1, 4 = 8
-1, 1, -4 = 8
-1, 1, 1 = 8
0, -1, -2 = 8
0, -1, 2 = 8
0, 1, -2 = 8
0, 1, 2 = 8
1, -1, -4 = 8
1, -1, 1 = 8
1, 1, -1 = 8
1, 1, 4 = 8
2, -1, -4 = 8
2, -1, 1 = 8
2, 1, -1 = 8
2, 1, 4 = 8
3, -1, -1 = 8
3, -1, 4 = 8
3, 1, -4 = 8
3, 1, 1 = 8
All 28 sets of results for the equation are found by 3-dimentional iteration within the range (-30, 30). 
Wouldn't it be considered a proof that algorithm does exist? It's greedy and brute-force, and it's not even the best of its kind, but it does exist.  
Here is a link to the polynomial equations generator that I've written for the purpose:
http://www.etymologia.net/polynomial/polynomial_equation_generator.php 
Please, guys, tell me what do you think about the problem itself. Your ideas would be very much appreciated. 
PS.
I found an interesting polynomial equation of high degree with interesting results. 
4x3y5z4 + 2y6z6 - 2x3y6z3 = 0
Assign any (but the same) value to x,y,z variables. Flip the sign of y and z. You've just solved the polynomial equation of high degree. 
Any set of numbers makes correct solution for the above equation, where y=0-x and z=0-x.  
So, is it true after all that NO algorithm at all, polynomial, exponential, doubly exponential, or worse for polynomial equations?
 A: The proposed algorithm would have to work on any polynomial, not just the ones you chose. You can't do better than trying all possible combinations of integers (of which there are infinite) on an arbitrary polynomial.
In analogy to the halting problem: we can tell algorithmically if some programs halt, but not all of them, such as in this paper.
A: What you are looking for are undecidable problems in this sense; and yes, the given problem is one of them. To make it clearer, it could be rephrased as

Given a polynomial equation $f$ (in multiple variables) over the integers. Then there is no algorithm to decide if $f$ has solutions.

That does not mean that you can't compute solutions, ever. To compare it to SAT, look at the following SAT problem:
$$x_1 + x_2 + x_3.$$
Of course this is satisfiable, that is easy to see. But still, you can't give an easy/efficient algorithm that will answer every instance of SAT.
It's just the same with the polynomials: You can't give an algorithm that works for all of them - even though you can solve some cases.
A: You can certainly iterate over all integers and try them all.  
However, that doesn't give you an algorithm, but a semi-algorithm.  
A semi-algorithm is similar to an algorithm except that it is only guaranteed to halt if it finds a solution. Otherwise, it may keep on searching forever.  
A good example of a semi-algorithm is the Risch algorithm. It is used for indefinite integration, but it is not guaranteed to halt. In practice, computers use heuristics for figuring out when to "give up" a search, such as simply cutting off the computation when it takes too long, or otherwise avoiding solutions that can grow arbitrarily complicated.
So yes, this type of problem has no algorithm in the computational sense unless you allow for false negatives, because it cannot tell you whether or not a solution exists.  
(However, as others have already said, do note that any particular polynomial will have an an algorithm for detecting whether or not it has a solution. Specifically, either the algorithm "always return yes" and the algorithm "always return no" will return the correct result! The trouble comes when you try to generalize this to an infinite set of polynomials.)
