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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.

Bagaringa
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  • 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. – Bagaringa Mar 07 '24 at 20:44

1 Answers1

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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 $m$?

Mike
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  • It makes sense, thanks. maybe the problem wasn't as hard as i initially thought. Btw I am wondering is it possible to show something useful with my attempt or is it already doomed to fail? – Bagaringa Feb 27 '24 at 21:37
  • 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 $y_i \equiv_m x_i$. – Mike Feb 27 '24 at 21:47
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Feb 27 '24 at 21:57