# Lower bound on number of solutions to diophantine equations with all but one linear constraint

Hi,

I want to ask a simple question in diophantine systems. I have tried to search under different headings, but was unable to find a suitable answer to my question.

I have a set of $m$ linear equations, in $n$ variables: $m \leq n$. Say this multivariate system is expressed as:

$$\sum_{i=0}^{n}c_{1i}x_{i} = 1$$

$$\sum_{i=0}^{n}c_{2i}x_{i} = 1$$

$$\vdots$$

$$\sum_{i=0}^{n}c_{mi}x_{i} = 1$$

where $x_{i}\in\mathbb{Z}$ are integers.

If the above system of solutions is solvable, the number of free variables left (in the solutions) would be $n-m$. Now I want to add a $(m+1)^{\text{th}}$ constraint, but it would be a non-linear (quadratic) constraint.

$$\sum_{i_{1}+ \cdots + i_{n} = 2}^{n}C_{i_{1}}\cdots C_{i_{n}}x_{1}^{i_{1}}\cdots x_{n}^{i_{n}} = 1$$

Can I tell anything about atleast how many free variables will be left out after solving the new system of $(m+1)$ equations. (Note: $m+1 \leq n$).

I can suppose determine how many integers satisfy the first set of equations. Can I now determine atleast how many integers would satisfy the new set of equations. Some sort of a lowerbound which depends on the coefficients of the equations would be better. (Zero is obviously a lowerbound when the system is consistent, but I want something that would depende on $\{C_{i}\}$.

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