Timeline for Applications of the Chinese remainder theorem
Current License: CC BY-SA 4.0
15 events
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| May 9, 2023 at 17:12 | history | edited | José Hdz. Stgo. | CC BY-SA 4.0 |
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| Dec 12, 2013 at 6:57 | comment | added | KConrad | The connection of this with CRT is that the following two conditions on a polynomial $f(x)$ with integral coefficients are equivalent: (i) there isn't a prime $p$ such that every $f(a)$ is divisible by $p$, and (ii) there isn't a finite set of primes $p_1,\dots, p_k$ such that every $f(a)$ is divisible by some $p_i$ (allowed to vary with $a$). Likewise, we have the equivalence of (i) there isn't a prime $p$ such that every $f(a)$ is divisible by $p^2$, and (ii) there isn't a finite set of primes $p_1,\dots, p_k$ such that every $f(a)$ is divisible by some $p_i^2$ (allowed to vary with $a$). | |
| Dec 12, 2013 at 6:51 | comment | added | KConrad | would be simpler than infinitude of prime values, but both conjectures are still open in general. Though Granville did prove the conjecture about infinitude of squarefree values under the $ABC$ conjecture, while nobody has derived Bunyakowsky's conjecture from the $ABC$ conjecture or other conjectures not obviously related to Bunyakowsky's conjecture. | |
| Dec 12, 2013 at 6:49 | comment | added | KConrad | I rewrote my second entry. What I think I may have had in mind when I initially wrote it is the following. Let $f(x)$ be irreducible with integral coefficients. Bunyakowsky conjectured that if there is no prime $p$ dividing $f(a)$ for all $a$ in $\mathbf Z$ then $f(a)$ is prime infinitely often (allow negative primes). This is unsolved if $\deg f > 1$. It is also conjectured that if there is no prime $p$ such that $p^2|f(a)$ for every integer $a$ then $f(a)$ is squarefree infinitely often, and this is wide open if $\deg f > 3$. You'd think proving infinitude of squarefree values [contd.] | |
| Dec 12, 2013 at 6:42 | history | edited | KConrad | CC BY-SA 3.0 |
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| Dec 12, 2013 at 3:49 | comment | added | KConrad | @AntonKlyachko: Ой, конечно я ошибся. Спасибо. Мне нужно пересмотреть то, что я имел в виду... | |
| Dec 11, 2013 at 17:03 | comment | added | Anton Klyachko | Excuse me, KConrad (and @Zack), I do not understand the problem about squares. Suppose that we have integers $a$ and $b$ such that $f(a)\ne0$ modulo $p^2$ and $f(b)\ne0$ modulo $q^2$, then CRT provides us with an integer $c$ equal to $a$ modulo $p^2$ and equal to $b$ modulo $q^2$. Thus, $f(c)=f(a)\ne0$ modulo $p^2$ and $f(c)=f(b)\ne0$ modulo $q^2$, which is a contradiction. Or I miss something? | |
| Jul 12, 2012 at 2:51 | comment | added | KConrad | @Zack: Thanks for pointing that out. I fixed the statement of 2. And as far as I know the question on squares is still unproved. | |
| Jul 12, 2012 at 2:50 | history | edited | KConrad | CC BY-SA 3.0 |
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| Jul 12, 2012 at 1:36 | comment | added | Zack Wolske | Since this is a few years old, do you know if the question on squares is still open? | |
| Jul 12, 2012 at 1:35 | comment | added | Zack Wolske | The statement starting at "That is," in 2. is not what you meant to write, based on what you say in the first sentence. It says "If $6$ divides $f(a)$ for all $a$, then either $2$ divides every $f(a)$, or $3$ divides every $f(a)$", which is certainly true and does not use CRT. I think that's why Idoneal was suggesting that it was a modified version of congruence. You could instead say "That is, if $f(a) \equiv 0, 2, 3$, or $4 \mod{6}$ for each $a$, then either..." | |
| Jul 30, 2010 at 0:11 | history | edited | Gerry Myerson | CC BY-SA 2.5 |
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| Jan 17, 2010 at 6:17 | comment | added | KConrad | Huh? The notation looks correct to me. It's the congruence notation from modular arithmetic. To say a number is 0 mod m means it is a multiple of m, and that's the situation I'm describing in 2. In what way does the usage look nonstandard? (I said at the start that a runs over integers, and that remains true anywhere later on as well.) | |
| Jan 17, 2010 at 5:49 | comment | added | Idoneal | I think you are using the symbol $\equiv$ in a nonstandard way in 2 above. | |
| Jan 16, 2010 at 19:35 | history | answered | KConrad | CC BY-SA 2.5 |