Timeline for Linear combinations of reducible polynomials
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2016 at 17:16 | comment | added | Pat Devlin | Can we fix this by modifying the question to be about polynomials over the rationals? | |
| Nov 18, 2016 at 10:33 | comment | added | Ilya Bogdanov | And surely such constants exist: they can be constructed via the Chinese remainders theorem. | |
| Nov 18, 2016 at 10:31 | comment | added | Ilya Bogdanov | This does not cure. If you know two composite coprime constants $a$ and $b$ such that $ua+vb$ are composite (say, divisible by $p_{uv}q_{uv}$) for all small $u$ and $v$, then you may take $f=ax^d$ and $g=b(x-N)^d$, where $N$ is the product of all $p_{uv}q_{uv}$.Then $uf+vg$ will still be divisible by $p_{uv}q_{uv}$. | |
| Nov 18, 2016 at 6:52 | comment | added | Stanley Yao Xiao | @Gerhard Paseman I put in the condition $d \geq 2$ | |
| Nov 17, 2016 at 23:25 | comment | added | Gerhard Paseman | I'm not sure C exists for d=0. Do you know of a bound that permits ua+vb to be prime when a and b are composite and coprime integers? Gerhard "Thinking Of Really Simple Cases" Paseman, 2016.11.17. | |
| Nov 17, 2016 at 23:01 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
deleted 9 characters in body
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| Nov 17, 2016 at 23:00 | comment | added | Stanley Yao Xiao | Ah yes, $C$ should be a function of the degree, I will fix that. And yes, co-prime is the appropriate condition | |
| Nov 17, 2016 at 22:27 | comment | added | rlo | Presumably you mean coprime, not non-proportional. Regardless, surely $C$ must be allowed to grow with the degree. For example, choose $f$, $g$, and $uf+vg$ for small $u$ and $v$ to vanish at distinct integers. | |
| Nov 17, 2016 at 22:22 | comment | added | Gerry Myerson | I'm not sure what you mean by "non-proportional", but if $f(x)=x(x-1)$ and $g(x)=x(x-2)$.... | |
| Nov 17, 2016 at 22:15 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |