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[Artur]

P.S.
Exactly "Don't existed such integer x,y that
1+2xy+x^2 y+y^2=0 "
ARTUR




Artur pisze:
> Dear Max and Rest of Seqfans
>
> Do you able to proove  my conjecture that form  1+2xy+x^2 y+y^2  don't 
> have integer solutions or find one of these solution as countersample?
>
> BEST WISHES
> ARTUR
>
>
> Max Alekseyev pisze:
>> On Nov 16, 2007 11:06 AM, Artur <grafix at csl.pl> wrote:
>>
>>  
>>> I'm apologize but but I was do mistake in formula
>>> Find number m such that 5n^4(4+n^2) is square
>>> Mathemtica sample Code:
>>> a={};Do[If[IntegerQ[Sqrt[5n^4 (4+n^2)]],AppendTo[a,n]],{n,1,1364}];a
>>>
>>> Out: {1,4,11,29,76,199,521,1364}
>>>     
>>
>> If I got you correctly, you're looking for positive integer solutions to
>> 5n^4(4+n^2) = m^2
>> It implies that n^2 divides m. Let m=n^2*k. Then the equation is 
>> equivalent to
>> 5(4+n^2) = k^2
>> or
>> k^2 - 5n^2 = 20.
>> This is a generalized Pell equation and you can plug its coefficients
>> to Dario Alpern's on-line tool:
>> http://www.alpertron.com.ar/QUAD.HTM
>> to get solution in the form of recurrent sequences.
>>
>> With respect to n, the solution consists of the following 3 sequences,
>> satisfying the same recurrent relation n(j+1)=18*n(j)-n(j-1) with
>> different initial conditions:
>>
>> n(0) = 1, n(1) = 29: { 1, 29, 521, 9349, 167761, 3010349, ... }
>>
>> n(0) = 4, n(1) = 76: { 4, 76, 1364, 24476, 439204, 7881196, ... }
>>
>> n(0) = 11, n(1) = 199:  { 11, 199, 3571, 64079, 1149851, 20633239, ... }
>>
>> Your sequence is the union of these three.
>>
>> Max
>>
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>>
>>
>>   
>
> __________ NOD32 Informacje 2699 (20071203) __________
>
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> http://www.nod32.com lub http://www.nod32.pl
>
>