Timeline for Proving conditions on $(r+s)^2 \mid (4r^4+1)$, related to Pell oblongs
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 22, 2014 at 1:44 | vote | accept | Kieren MacMillan | ||
| Feb 5, 2014 at 0:11 | answer | added | Noam D. Elkies | timeline score: 8 | |
| Feb 4, 2014 at 21:56 | history | edited | user9072 |
edited tags
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| Oct 20, 2013 at 2:48 | comment | added | Kieren MacMillan | @Lucia: Thanks! You gave me a great idea on how to attack the problem: [try to] find the precise conditions such that $2r^2 + 1 \pm 2r$ are both squareful. | |
| Oct 18, 2013 at 14:41 | comment | added | Lucia | @Kieren MacMillan: I'm not terribly sure why that should be true, and in any case don't have a good intuition for this problem. | |
| Oct 18, 2013 at 14:37 | comment | added | Kieren MacMillan | So, @Lucia, any thoughts on how one might prove the [apparent] condition $s/r < 1/2$ when $r > 3$? Thanks. | |
| Oct 18, 2013 at 14:15 | comment | added | Kieren MacMillan | @Lucia: Excellent point — thanks for the correction! | |
| Oct 18, 2013 at 14:10 | comment | added | Lucia | @KierenMacMillan: I don't see why you can't have $(r+s)=ab$ and then $a^2|(2r^2+1+2r)$ and $b^2|(2r^2+1-2r)$. For any prime square your statement is fine. | |
| Oct 18, 2013 at 11:48 | comment | added | Gerry Myerson | Looks that way. Also, letting $r+s=y$, we have $y^2=2r^2\pm2r+1$, which leads to a couple of Pellians, which will give you many (but apparently not all) solutions. | |
| Oct 18, 2013 at 11:43 | comment | added | Kieren MacMillan | @GerryMyerson: Thanks! I recalled that just after I posted… Since those two factors are odd, they are evidently relatively prime. Hence any square dividing $4r^4+1$ divides exactly one of $2r^2+1 \pm 2r$ (i.e., the square is not "split" across the two factors). Now if $s/r > 1/2$, then $(r+s)^2 > (3r/2)^2 = (9/4)r^2$, and hence this [alleged] square factor $2r^2 < (r+s)^2 \mid (2r^2+1\pm 2r)$. Unless $(r+s)^2 = (2r^2+1+2r)$, wouldn't that be an immediate contradiction? In other words, is that a valid proof of the condition $s/r < 1/2$ for $r > 3$? | |
| Oct 18, 2013 at 11:30 | comment | added | Gerry Myerson | It may be worth noting that $4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$. | |
| Oct 18, 2013 at 10:44 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
clarified last paragraph, added s/r bound question
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| Oct 18, 2013 at 10:28 | history | asked | Kieren MacMillan | CC BY-SA 3.0 |