Consider an infinite set of primes $\{p_{1}, p_{2}, \ldots, p_{s}, p_{s + 1}, \ldots\}$ and the system of congruences $x \equiv a_{i} \mod p_{i}$ for $i = 1, 2, \ldots, s$. Then by the Chinese Remainder Theorem there is a solution to this system of linear congruences. But what if I add the condition that $x \not\equiv 0 \mod p_{s + k}$ for all $k \geq 1$. Does such an $x$ still exist?
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It is not necessarily true that such an $x$ exists. Take for example $s=1$ and $p_1=5$, $a_1=2$, and let $p_2,p_3,\dots$ be all of the other primes. The only integers $x$ that are not divisible by any of the primes $p_2,p_3,\dots$ are those of the form $\pm 5^k$, and none of those are congruent to $2$ (mod $5$). (I see Yuval Filmus had the same idea.) |
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Suppose $p_1 = 3$, $a_1 = 2$, and $p_2,\ldots$ are the rest of the primes. What do you get? |
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