Concordant Form

Citrix · 2006-03-18, 22:14

http://mathworld.wolfram.com/ConcordantForm.html

According to this page 60 is a concordant number but I can't find any solutions, could some one please help me find a solution.I personally think this is an error.

Thanks,Citrix

grandpascorpion · 2006-03-19, 01:33

http://134.129.111.8/cgi-bin/wa.exe?...thry&O=A&P=329

Citrix · 2006-03-19, 01:56

It does not provide a solution for 60? I am not sure what to do with the link?

Citrix

jchein1 · 2006-03-19, 01:58

Hi Citrix,

The first two solutions is given by

(x,y) = (187,84)
(x,y) = (748,336).

cheers,

Joesph E.Z. Chein

Citrix · 2006-03-19, 02:07

Thanks, my program was too slow to find one? Can your program find a solution for 103? (This is an open problem)

Citrix

grandpascorpion · 2006-03-19, 05:26

Sorry, I thought you needed more info on this worked.

103 isn't solvable (if the Birch-Swinnerton-Dyer conjecture is valid)
http://www.mathpages.com/home/286tab1.htm

mfgoode · 2006-03-19, 11:49

Quote:

Originally Posted by jchein1

Hi Citrix,

The first two solutions is given by

(x,y) = (187,84)
(x,y) = (748,336).

cheers,

Joesph E.Z. Chein

:surprised Fantastic Joe, Keep up the good work.
Maybe you should have added N=60 for those following this thread.
Mally

jchein1 · 2006-03-19, 15:33

Hi Citrix and mfgoode,

If one put a = x^2 – y^2 and b = 2xy into the concordance form 103, you would have

x^4 + 410*x^2*y^2 + y^4 = d^2 (1).

The necessary condition for that equation to be solvable is that

5 | x or 5 | y.

It appears that there may be no solutions for x, y <= 100000 (my program still cooking but almost done). The general solution for equation (1) (either has infinitely solutions or no solutions) depends upon the algebraic number theory and is connected with the units in certain algebraic number fields.

cheers,

Joseph E.Z. Chein

Citrix · 2006-03-19, 20:45

I have reduced the problem to this simple equation

if (103*r*r+k*r)/(k*r+k*k)=(a/b)^2
Then only is there a solution to the 103 case.

Clearly r and k must not have a common factor ie. they must be coprime, this eliminates alot of steps.

There are other rules you can develop and eliminate some steps.

Ultimately it comes to this
(103*r*r+k*r)=a^2
and (k*r+k*k)=b^2

Which is equally difficult to solve. So no real progress here.

2006-03-18, 22:14	#1
Citrix Jun 2003 2·7·113 Posts	Concordant Form http://mathworld.wolfram.com/ConcordantForm.html According to this page 60 is a concordant number but I can't find any solutions, could some one please help me find a solution.I personally think this is an error. Thanks,Citrix Last fiddled with by Citrix on 2006-03-18 at 22:15

2006-03-19, 01:33	#2
grandpascorpion Jan 2005 Transdniestr 503 Posts	Googled it http://134.129.111.8/cgi-bin/wa.exe?...thry&O=A&P=329

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2006-03-19, 01:56	#3
Citrix Jun 2003 1582₁₀ Posts	It does not provide a solution for 60? I am not sure what to do with the link? Citrix

2006-03-19, 01:58	#4
jchein1 May 2005 3C₁₆ Posts	Hi Citrix, The first two solutions is given by (x,y) = (187,84) (x,y) = (748,336). cheers, Joesph E.Z. Chein

2006-03-19, 02:07	#5
Citrix Jun 2003 2×7×113 Posts	Thanks, my program was too slow to find one? Can your program find a solution for 103? (This is an open problem) Citrix

2006-03-19, 05:26	#6
grandpascorpion Jan 2005 Transdniestr 503 Posts	Sorry, I thought you needed more info on this worked. 103 isn't solvable (if the Birch-Swinnerton-Dyer conjecture is valid) http://www.mathpages.com/home/286tab1.htm

2006-03-19, 15:33	#8
jchein1 May 2005 2²·3·5 Posts	Hi Citrix and mfgoode, If one put a = x^2 – y^2 and b = 2xy into the concordance form 103, you would have x^4 + 410x^2y^2 + y^4 = d^2 (1). The necessary condition for that equation to be solvable is that 5 \| x or 5 \| y. It appears that there may be no solutions for x, y <= 100000 (my program still cooking but almost done). The general solution for equation (1) (either has infinitely solutions or no solutions) depends upon the algebraic number theory and is connected with the units in certain algebraic number fields. cheers, Joseph E.Z. Chein

2006-03-19, 20:45	#9
Citrix Jun 2003 3056₈ Posts	I have reduced the problem to this simple equation if (103rr+kr)/(kr+kk)=(a/b)^2 Then only is there a solution to the 103 case. Clearly r and k must not have a common factor ie. they must be coprime, this eliminates alot of steps. There are other rules you can develop and eliminate some steps. Ultimately it comes to this (103rr+kr)=a^2 and (kr+kk)=b^2 Which is equally difficult to solve. So no real progress here.