# Prove that if polynomial with integral coefficients has a solution in integers, then the congruence is solvable for any value of modulus $m$. [duplicate]

Prove that if $$F(x_1, x_2, \ldots, x_n) = 0$$, where $$F$$ is a polynomial with integral coefficients, has a solution in integers, then the congruence $$F(x_1, x_2, \ldots, x_n) \equiv 0 \pmod{m}$$ is solvable for any value of modulus $$m$$.

I am not sure how to prove this.

My attempt: Let $$F(x_1, x_2, \ldots, x_n) = \sum_{i_1, i_2, \ldots, i_n} a_{i_1, i_2, \ldots, i_n} x_1^{i_1} x_2^{i_2} \ldots x_n^{i_n}$$ be the polynomial with integer coefficients. Suppose there exists a solution $$(a_1, a_2, \ldots, a_n)$$ such that $$F(a_1, a_2, \ldots, a_n) = 0$$.

Consider the congruences:

$$x_1 \equiv a_1 \pmod{m}$$

$$x_2 \equiv a_2 \pmod{m}$$

$$\vdots$$

$$x_n \equiv a_n \pmod{m}$$

By the definition of congruence, we have $$x_i = a_i + m \cdot t_i$$ for some integers $$t_i$$. Substitute these expressions into $$F(x_1, x_2, \ldots, x_n)$$:

$$F(a_1 + m \cdot t_1, a_2 + m \cdot t_2, \ldots, a_n + m \cdot t_n) = 0$$ Is this correct so far? Not sure what to do next.

• Question is already closed, no need to be mean to nominate it for deletion. We all start from somewhere and there might be some people whom it may help. Mar 7 at 20:44

Can't you just say $$F(x_1,\ldots, x_n) = 0$$ $$\implies$$ $$F(x_1, \ldots, x_n) \equiv_m 0$$ $$\forall m \in \mathbb{N}$$, as $$0 \equiv_m 0$$ for all such $$m$$?
• You CAN show that if $F(x_1,\ldots, x_n) = 0$ then $F(y_1, \ldots, y_n) \equiv_m 0$ for all $y_1,\ldots, y_n$ such that $y_i \equiv_m x_i$.