Timeline for $x^4+y^4$ powerful for relatively prime $x,y$
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2023 at 3:03 | comment | added | Benjamin Dickman | maybe worth posting a pointer to people.maths.bris.ac.uk/~jb12407/ANTS-XV/papers/… in case others encounter this question and wonder whether it's still open [even tho that is easily answered by googling the conjectured number] | |
| Jul 1, 2022 at 13:08 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Apr 16, 2022 at 1:28 | comment | added | CHUAKS | Are there coprime $x,y$ with $x^3+y^3$ $3$-powerful? $1735^2+2463^3=2^4.7^4.73^3$, $397^3+683^3=2^3.3^4.5.7^6$ has average power 3.69522.. | |
| Apr 9, 2022 at 14:15 | answer | added | taf | timeline score: 1 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 11, 2015 at 23:45 | comment | added | Noam D. Elkies | Yes I checked, but not that way: you don't want to wait for gp to count to $10^{16}$, let alone factor every number of at most $16$ digits! Much better to try all coprime $(x,y)$ of opposite parity with $x<y$ and $x^4 + y^4 < 10^{16}$; that's only 40 million or so factorizations, which take a few hours to try (and as expected find nothing). Still it's hopeless to reach $3 \cdot 10^{36}$ this way... Now that it's a couple of months since I posted this question, I should post a partial answer evaluating different strategies, the best of which might make the computation barely feasible. | |
| Mar 11, 2015 at 15:50 | comment | added | Moritz Firsching | @NoamD.Elkies so you checked already up to $10^{16}$? For small values one might use pari as follows: for(m=2,20000000,n=8*m+1;if(ispowerful(n),for(x=1,(n/2)^.25,for(y=1,(n-x^4)^.25,if(gcd(x,y)==1,if(n==x^4+y^4,print1(n,":",factor(n),":",x,"^4 + ",y,"^4",","))))))) | |
| Jan 9, 2015 at 15:34 | answer | added | John R Ramsden | timeline score: 0 | |
| Jan 3, 2015 at 0:45 | comment | added | Noam D. Elkies | [Of course I meant $x^4 + 1 = 17 z^2$, not $x^2 + 1 = 17 z^2$.] | |
| Jan 2, 2015 at 17:54 | comment | added | Noam D. Elkies | Um, the curves with $m=17$, $17^3$, $17^5$ are isomorphic (just write $z=17z_1$ or $17^2 z_2$ in $x^2+1=17z^2$), and thus give rise to exactly the same candidates... | |
| Jan 2, 2015 at 17:14 | comment | added | joro | I suspect $x^4+1=17^5 z^2$ has rational points, but clearing the gcd after homogenizing won't give answer, is this true? | |
| Jan 2, 2015 at 16:59 | comment | added | joro | Well, in my approach with m=17^3 i got solution smaller than yours, but after clearing the gcd 17 remained to first power. This doesn't answer your question. | |
| Jan 2, 2015 at 16:56 | comment | added | Noam D. Elkies | I used squarefree $m$, not powerful $m$. This gives coprimality for free but you then need to impose the condition that the numerator of $z$ has a factor of $m$, which is why (in this approach) the solutions tend to be large. | |
| Jan 2, 2015 at 16:54 | comment | added | joro | This doesn't surprise me. Do you need to clear the gcd or you get coprime solutions for free? | |
| Jan 2, 2015 at 16:15 | comment | added | Noam D. Elkies | As I'll report, I did try the first few $m$ past $17$ for which the curve has rational points, and in each case the smallest example produced of powerful=$x^4+y^4$ was even larger. | |
| Jan 2, 2015 at 15:49 | comment | added | joro | I am currently testing this without luck so far. Find birational map from $x^4+1=m z^2$ to a Weierstrass model without known point. Maple found such map over quadratic extension of Q. Find the generators over the extension, try to map small multiples to Q, clear the gcd, check. My pain is sage is slow for finding the generators. Of course homogenize first. | |
| Jan 2, 2015 at 15:45 | comment | added | Noam D. Elkies | So what search strategy are you proposing? (There is a way to go at it with elliptic curves that might be barely feasible, which I'll note when I next edit the question to remove typos like "for for" etc.) | |
| Jan 2, 2015 at 15:41 | comment | added | joro | No :-). I mean to to use the affine quartic model $x^4+1=m z^2$ with $m$ powerful. | |
| Jan 2, 2015 at 15:39 | comment | added | Noam D. Elkies | You mean, try all powerful $m$ up to about $3 \cdot 10^{36}$? Unfortunately there's too many of them (must be at least $10^{16}$ even when we require that each prime factor be $1 \bmod 8$). | |
| Jan 2, 2015 at 13:36 | comment | added | joro | Is it known that working with powerful $m$ doesn't help? (you need not care about $z$ in this case). | |
| Jan 2, 2015 at 12:28 | comment | added | joro | Google still doesn't recognize your first solution, though it has indexed your question according to search for the title. | |
| Jan 1, 2015 at 5:13 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
3*10^36, not 3*10^35...
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| Jan 1, 2015 at 5:03 | history | asked | Noam D. Elkies | CC BY-SA 3.0 |