Timeline for When two non-equivalent binary forms primitively represent the same infinite subset of the integers
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| Jul 6, 2016 at 6:36 | comment | added | Daniel Loughran | @Anton: Well you can come up with silly examples like $F(x,y) = x^{2n} + y^{2n}$ and $G(x,y) = - x^{2n} - y^{2n}$: these clearly do not simultaneously represent infinitely many integers. But of course here there is an obstruction at the real place. The interesting case is where there are no local obstructions, but as Felipe pointed out the problem here is essentially a special case of the Bombieri-Lang conjecture. | |
| Jul 6, 2016 at 2:52 | comment | added | Anton | @FelipeVoloch thanks, for bringing this up. Rrecently Stanley Yao Xiao also pointed out to me that the problem I posed should be quite difficult, since it basically reduces to Bombieri-Lang Conjecture. | |
| Jul 6, 2016 at 2:44 | comment | added | Anton | @DanielLoughran I do not have an example of F and G which, under specified conditions, cannot primitively represent an infinite subset of the integers. I need to think about it, but I bet that such examples can be made. It was proved by Mahler in 1930's that irreducible binary forms of degree $\geq 3$ primitively represent only 0% of the integers. So I guess that if two thin sets that F and G represent intersect infinitely many times, then F and G should be connected to each other through some relation, like in the example above. | |
| Jul 3, 2016 at 7:08 | comment | added | Felipe Voloch | If $d>4$ you could try applying the Bombieri-Lang conjecture to the projective surface defined by $F(x,y)=G(z,w)$ and see what you get. | |
| Jul 3, 2016 at 2:24 | history | edited | Anton | CC BY-SA 3.0 |
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| Jun 30, 2016 at 22:42 | comment | added | Will Jagy | Not the restrictions you wanted, but take a look at mathoverflow.net/questions/152889/… which, if ever proved, give counterexamples to a conjecture of Hsia that positive ternary quadratic forms are determined by their primitively represented numbers. I did check for primitivity, not sure i mentioned that in my original question. | |
| Jun 30, 2016 at 12:34 | comment | added | Anton | @DanielLoughran I do know an example when this is not true. Stanley Yao Xiao gave an example of two binary forms of degree 4 with coefficient vectors F=(2401,98,3,4,5) and G=(1,2,3,196,12005). They are not equivalent and don't have any GL2(Q)-automorphisms. However, they are connected by means of the relation F(x,7y)=G(7x,y), and because of that there are infinitely many integers which can be primitively presented by both forms. | |
| Jun 30, 2016 at 12:14 | comment | added | Daniel Loughran | Thanks for clarifying. Do you know examples of $F$ and $G$ where this property does not hold? | |
| Jun 30, 2016 at 11:28 | comment | added | Anton | @DanielLoughran the condition that $P_F$ and $P_G$ consist of primes is too restrictive. Instead, let those two sets contain all numbers primitively represented by F and G, respectively. I'm asking under which conditions $P_F \cap P_G$ can be infinite. | |
| Jun 30, 2016 at 8:48 | comment | added | Daniel Loughran | Could you please clarify your question? Let $P_F$ (resp. $P_G$) be the set of primes primitively represented by $F$ (resp. $G$). Are you asking whether it is possible for $P_F = P_G$, or whether it is possible for $P_F \cap P_G$ to be infinite? | |
| Jun 30, 2016 at 3:32 | history | edited | Anton | CC BY-SA 3.0 |
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| Jun 30, 2016 at 2:15 | history | edited | Anton | CC BY-SA 3.0 |
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| Jun 30, 2016 at 1:51 | history | edited | Anton | CC BY-SA 3.0 |
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| Jun 30, 2016 at 1:25 | history | edited | Anton | CC BY-SA 3.0 |
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| Jun 30, 2016 at 1:11 | history | edited | Anton | CC BY-SA 3.0 |
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| Jun 30, 2016 at 1:03 | history | asked | Anton | CC BY-SA 3.0 |