[math-fun] Diophantine equation a^4+b^4=d*c^2 and the specialness of 17, 68, 82...
Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive integer solutions. Also a^4-b^4=2*c^2 has no positive integer solutions. These results are shown in T.Nagell: Intro to number theory, Wiley 1951, pp.227-230. However, a^4+b^4=2*c^2 trivially has a ton of solutions a=b, c=a^2. If we exclude the trivial by demanding b>a>0 though, I don't know of any solutions. Let us generalize. Ask about a^4 + b^4 = d*c^2 with a,b,c,d integers and b>a>0. My computer looked for solutions. Search I: a^4+b^4=d*c^2 with 0<a<b<65536 and 0<d<65. Result: a ton of solutions for d=17, but zero solutions for any other d. a=1, b=2, d=17, c=1 a=2, b=4, d=17, c=4 a=2, b=13, d=17, c=41 a=3, b=6, d=17, c=9 a=4, b=8, d=17, c=16 ... Well, that's interesting: 17 is special. Search II: a^4+b^4=d*c^2 with 0<a<b<=1024 and 0<d<=1024. Result: The following d are "special": 17, 68, 82, 97, 113, 153, 193, 257, 272, 274, 328, 337, 388, 425, 433, 452, 514, 577, 593, 612, 626, 641, 673, 706, 738, 772, 833, 873, 881, 914, 1017. [This sequence is not in OEIS.] What is going on here? I do not really know, but an initial speculation is this. When d=17, there is the easy solution a=1, b=2, d=17, c=1. But here is the thing: if there is one solution, then it can usually be used to generate an infinite set of solutions. I got that from Tito Piezas's online book http://sites.google.com/site/tpiezas/018 where he attributes it to A.Desboves (probably about 1880?). So there are "special" d with an infinity of solutions, and there are "usual" d with zero solutions (and conceivably there are a few "crazy" d with a finite nonzero number of solutions because the below Desboves method for generating more solutions happens merely to regenerate old ones, which would be an amazing miracle). How to generate more solutions from one: Piezas says Desboves said (as edited by me): if p^4+q^4 = d*r^2, then x^4+y^4 = d*z^2 where x = p*(4*m^2-3*(m+n)^2), y = q*(4*n^2-3*(m+n)^2), z = r*(4*(m+n)^4-3*(m-n)^4) where m=p^4 and n=q^4. So for example starting from p=1, q=2, d=17, r=1 we get x=863, y=314, z=182209 and sure enough 863^4 + 314^4 = 17*182209^2 is another solution. Questions: Q1. Can you find a parameterized family of d=17 solutions? Q2. Can you understand the set of special d? Q3. Can you prove there are no "crazy" d (or find them all)?
I added the sequence from "Search II" as A209078. Comments welcomed. Neil On Sat, Dec 10, 2011 at 11:56 AM, Warren Smith <warren.wds@gmail.com> wrote:
Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive integer solutions. Also a^4-b^4=2*c^2 has no positive integer solutions. These results are shown in T.Nagell: Intro to number theory, Wiley 1951, pp.227-230. However, a^4+b^4=2*c^2 trivially has a ton of solutions a=b, c=a^2. If we exclude the trivial by demanding b>a>0 though, I don't know of any solutions.
Let us generalize. Ask about a^4 + b^4 = d*c^2 with a,b,c,d integers and b>a>0.
My computer looked for solutions. Search I: a^4+b^4=d*c^2 with 0<a<b<65536 and 0<d<65. Result: a ton of solutions for d=17, but zero solutions for any other d. a=1, b=2, d=17, c=1 a=2, b=4, d=17, c=4 a=2, b=13, d=17, c=41 a=3, b=6, d=17, c=9 a=4, b=8, d=17, c=16 ...
Well, that's interesting: 17 is special.
Search II: a^4+b^4=d*c^2 with 0<a<b<=1024 and 0<d<=1024. Result: The following d are "special": 17, 68, 82, 97, 113, 153, 193, 257, 272, 274, 328, 337, 388, 425, 433, 452, 514, 577, 593, 612, 626, 641, 673, 706, 738, 772, 833, 873, 881, 914, 1017. [This sequence is not in OEIS.]
What is going on here? I do not really know, but an initial speculation is this. When d=17, there is the easy solution a=1, b=2, d=17, c=1.
But here is the thing: if there is one solution, then it can usually be used to generate an infinite set of solutions. I got that from Tito Piezas's online book http://sites.google.com/site/tpiezas/018 where he attributes it to A.Desboves (probably about 1880?). So there are "special" d with an infinity of solutions, and there are "usual" d with zero solutions (and conceivably there are a few "crazy" d with a finite nonzero number of solutions because the below Desboves method for generating more solutions happens merely to regenerate old ones, which would be an amazing miracle).
How to generate more solutions from one: Piezas says Desboves said (as edited by me): if p^4+q^4 = d*r^2, then x^4+y^4 = d*z^2 where x = p*(4*m^2-3*(m+n)^2), y = q*(4*n^2-3*(m+n)^2), z = r*(4*(m+n)^4-3*(m-n)^4) where m=p^4 and n=q^4.
So for example starting from p=1, q=2, d=17, r=1 we get x=863, y=314, z=182209 and sure enough 863^4 + 314^4 = 17*182209^2 is another solution.
Questions: Q1. Can you find a parameterized family of d=17 solutions? Q2. Can you understand the set of special d? Q3. Can you prove there are no "crazy" d (or find them all)?
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
I think the square-free terms, namely: 17, 82, 97, 113, 193, 257, 274, 337, 433, 514, 577, 593, 626, 641, 673, 706, 881, 914, ... make a more interesting sequence, and it is already in OEIS as A184982. On Sun, Mar 4, 2012 at 6:16 PM, Neil Sloane <njasloane@gmail.com> wrote:
I added the sequence from "Search II" as A209078. Comments welcomed. Neil
On Sat, Dec 10, 2011 at 11:56 AM, Warren Smith <warren.wds@gmail.com> wrote:
Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive integer solutions. Also a^4-b^4=2*c^2 has no positive integer solutions. These results are shown in T.Nagell: Intro to number theory, Wiley 1951, pp.227-230. However, a^4+b^4=2*c^2 trivially has a ton of solutions a=b, c=a^2. If we exclude the trivial by demanding b>a>0 though, I don't know of any solutions.
Let us generalize. Ask about a^4 + b^4 = d*c^2 with a,b,c,d integers and b>a>0.
My computer looked for solutions. Search I: a^4+b^4=d*c^2 with 0<a<b<65536 and 0<d<65. Result: a ton of solutions for d=17, but zero solutions for any other d. a=1, b=2, d=17, c=1 a=2, b=4, d=17, c=4 a=2, b=13, d=17, c=41 a=3, b=6, d=17, c=9 a=4, b=8, d=17, c=16 ...
Well, that's interesting: 17 is special.
Search II: a^4+b^4=d*c^2 with 0<a<b<=1024 and 0<d<=1024. Result: The following d are "special": 17, 68, 82, 97, 113, 153, 193, 257, 272, 274, 328, 337, 388, 425, 433, 452, 514, 577, 593, 612, 626, 641, 673, 706, 738, 772, 833, 873, 881, 914, 1017. [This sequence is not in OEIS.]
What is going on here? I do not really know, but an initial speculation is this. When d=17, there is the easy solution a=1, b=2, d=17, c=1.
But here is the thing: if there is one solution, then it can usually be used to generate an infinite set of solutions. I got that from Tito Piezas's online book http://sites.google.com/site/tpiezas/018 where he attributes it to A.Desboves (probably about 1880?). So there are "special" d with an infinity of solutions, and there are "usual" d with zero solutions (and conceivably there are a few "crazy" d with a finite nonzero number of solutions because the below Desboves method for generating more solutions happens merely to regenerate old ones, which would be an amazing miracle).
How to generate more solutions from one: Piezas says Desboves said (as edited by me): if p^4+q^4 = d*r^2, then x^4+y^4 = d*z^2 where x = p*(4*m^2-3*(m+n)^2), y = q*(4*n^2-3*(m+n)^2), z = r*(4*(m+n)^4-3*(m-n)^4) where m=p^4 and n=q^4.
So for example starting from p=1, q=2, d=17, r=1 we get x=863, y=314, z=182209 and sure enough 863^4 + 314^4 = 17*182209^2 is another solution.
Questions: Q1. Can you find a parameterized family of d=17 solutions? Q2. Can you understand the set of special d? Q3. Can you prove there are no "crazy" d (or find them all)?
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
="James Buddenhagen" <jbuddenh@gmail.com> I think the square-free terms, namely: 17, 82, 97, 113, 193, 257, 274, 337, 433, 514, 577, 593, 626, 641, 673, 706, 881, 914, ... make a more interesting sequence, and it is already in OEIS as A184982.
I haven't been following this thread closely but this comment seems worth yet another reminder: Not submitting a sequence because it seems "less interesting" isn't really in the best interests of the OEIS and its users. The bar should be set just high enough to filter out truly silly sequences. But borderline cases should be given the benefit of the doubt and submitted. We simply have no way of knowing today what sequences will turn up as valuable serendipitous hits in future searches, particularly from queries generated by superseeker and other automation over the next few centuries. Each sequence that's included in the OEIS increases the chances for it contributing to mathematical knowledge in this way. The potential benefit of that is much greater than the small overhead of carrying another entry. Heh. There's still plenty of A-numbers available, so go for it!
participants (4)
-
James Buddenhagen -
Marc LeBrun -
Neil Sloane -
Warren Smith