[seqfan] Proof or counter sample needed

Artur grafix at csl.pl
Mon Oct 27 00:19:31 CET 2008


Hypergeometric2F1[1/5, 4/5, 1/2, 3/4]=GoldenRatio=(1+Sqrt[5])/2
Artur

Artur pisze:
> P.S.
> 1 + 10 x^3 - 12 x^5 + 5 x^6 is square if and only when x belonging to 
> A081016
> Artur
>
> Artur pisze:
>> P.S.
>> Easest to proof should be conjecture that:
>> Quintic polynomial
>>
>> 4 k - k^2 + 5 k^2 x + (20 k - 20 k^2) x^3 + (16 - 32 k + 16 k^2) x^5
>> have one rational root if and only when k belonging to finite set 
>> {2,4,243}
>>
>> If so than:
>>
>> 1) existed such rational x that 1 + 10 x^3 - 12 x^5 + 5 x^6  is 
>> square <x belonging to set{0,3/11,1}>
>> and
>> 2) (2 (-1 - 5 x^3 + 8 x^5 + Sqrt[1 + 10 x^3 - 12 x^5 + 5 x^6]))/(-1 +
>> 5 x - 20 x^3 + 16 x^5)
>> or (2 (-1 - 5 x^3 + 8 x^5 - Sqrt[1 + 10 x^3 - 12 x^5 + 5 x^6]))/(-1 +
>> 5 x - 20 x^3 + 16 x^5)
>> is integer
>>
>> Best wishes
>> Artur
>>
>>
>> Artur pisze:
>>> Dear Seqfans,
>>>
>>> Who is able to proof or find counter-sample following conjecture
>>> Quintic polynomial:
>>> (4 k - k^2 + 5 k^2 x + (20 k - 20 k^2) x^3 + (16 - 32 k + 16 k^2) x^5
>>> is factorizable if and only when k belonging to finite set {2,4,243}
>>> no more such k up to k=10^7
>>>
>>> Who have any idea let me know
>>>
>>> Best wishes
>>> Artur
>>>
>>>
>>> _______________________________________________
>>>
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>>
>
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