Timeline for representability of consecutive integers by a binary quadratic form
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 6, 2010 at 23:28 | comment | added | Wadim Zudilin | @Kevin: Fedja got a better time zone. :) | |
| May 6, 2010 at 18:34 | comment | added | Kevin Buzzard | @Robin: you're right x2. | |
| May 6, 2010 at 16:06 | comment | added | Robin Chapman | Kevin, an easier trick is to note that the $p_n$ needn't be prime; all they need is to be pairwise coprime. Also $(2x+1)(3y+1)$ does work. It does not represent $0$ but represents all other integers: write $N=2^r a$ where $a$ is odd. Then the $3y+1$ is either $2^r$ or $-2^r$. | |
| May 6, 2010 at 15:37 | comment | added | Kevin Buzzard | @Wadim: :-) | |
| May 6, 2010 at 15:28 | comment | added | Kevin Buzzard | Explicitly: if there are only finitely many primes that end in 3 or 7 then multiply them all together, raise to the power 4, add 2, and then consider a divisor. Of course I'm sure you're well aware of this trick, as you indicate :-) | |
| May 6, 2010 at 15:28 | comment | added | Kevin Buzzard | @Robin: isn't the general rule that it's possible to prove using "Euclid's trick" that you can avoid any proper subgroup of (Z/NZ)^* infinitely often? If H is a proper subgroup of (Z/NZ)^* then I claim there are infinitely many primes not in H mod N and the proof is that if there were finitely many then times them all together, raise to the power phi(N), and then add some small number t to get you outside H, and this number is not in H and so has a prime factor not in H. So there's an elementary proof that there are inf many primes which end in either 3 or 7, which is good enough. | |
| May 6, 2010 at 15:27 | comment | added | Wadim Zudilin | Allright, I'll find a proof in dreams. | |
| May 6, 2010 at 15:16 | comment | added | Kevin Buzzard | @Wadim: "if the question is rectricted to positive definite forms, then a C exists. ". Prove it! | |
| May 6, 2010 at 15:14 | comment | added | Wadim Zudilin | @Kevin: And if the question is rectricted to positive definite forms, then a $C$ exists. That what I mean in my starting remark on whether the question is specific to a class of quadratic forms--sign(in)definite and degenerate. | |
| May 6, 2010 at 15:06 | vote | accept | Ewan Delanoy | ||
| May 6, 2010 at 14:31 | comment | added | Robin Chapman | Sorry Kevin, you're right. As an alternative, you could let $C_n=p_n^3$ where $p_n$ is the $n$-th prime congruent to $2$ mod $5$ (hmm, this depends on Dirichlet's theorem but I can see how to get round that!) | |
| May 6, 2010 at 14:10 | comment | added | Kevin Buzzard | I was worried about sign issues! Any number is odd*(+-power of 2) and you change the sign to make the power of 2 equal to 1 mod 3 :-/ | |
| May 6, 2010 at 14:06 | comment | added | Robin Chapman | Kevin, wouldn't $(2x+1)(3y+1)$ be easier? Let $C_n=p_n$ where $p_n$ is the $n$-th prime congruent to 2 mod 3, and proceed in roughly the same manner. | |
| May 6, 2010 at 13:57 | comment | added | Kevin Buzzard | @Wadim: we are interpreting the question in different ways! My interpretation is that one is supposed to find C such that in any block of C consecutive integers, one is not represented. Just because -3<=n<=-1 are not represented does not mean that C=3 is OK, because 8,9,10 can all be represented. | |
| May 6, 2010 at 13:57 | comment | added | Robin Chapman | Wadim, you need to show that either every block of $C$ consecutive integers has an element not in the range of $f$, or that some particular block has all its elements represented (as Kevin does). | |
| May 6, 2010 at 13:20 | comment | added | Wadim Zudilin | @Kevin: Taking $f(x,y)=x^2+y^2$ one gets no negative integer represented by the form. So, $C$ again does not exists by taking the range $-C\le n\le-1$. | |
| May 6, 2010 at 12:37 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |