That makes sense - I was trying to solve using the cubic polynomials, rather
than the sextic one given. Then p = 2122943609450249057621 and the rest is
straightforward.
-Alasdair
2011/9/15 Biche de C?rynie <bichedecerynie at gmail.com>
> Maxima :o)
>
> No magick here. Actually, I read the page given by "Matematicas". There is
> an interesting relation :
>
> *M = (1800p3 - 1310p2 - 50p - 1) (1800p3 - 1220p2 - 130p - 1)
> M-1 = 2 * 5 * p * (90p2 - 61p - 6) (3600p3 - 2620p2 - 120p - 3)*
>
> With this the problem is quite easy. This gives (almost) p :
> inrt(floor(n/1800^2), 6);
> But you may find p by "direct" factorization, since it's quite small.
>
>
>
> 2011/9/14 Alasdair McAndrew <amca01 at gmail.com>
>
>> Nice - what software did you use for the factorization?
>>
>> -Alasdair
>>
>>
>> 2011/9/14 Biche de C?rynie <bichedecerynie at gmail.com>
>>
>>> Prime factors for your numbers :
>>>
>>> 2
>>> 5
>>> 2122943609450249057621
>>> 405620061201508643988375124643817406843652803
>>> 34444340671702467488860342883734479061576134487870291451728421045657
>>>
>>> And
>>>
>>> 17222170335851233744430171441867239530788067265164581820366701099039
>>> 17222170335851233744430577061928441039432055599953297058218812657049
>>>
>>> Jean-Claude Arbaut
>>>
>>>
>>>
>>> On Wed, Sep 14, 2011 at 2:21 AM, Alasdair McAndrew <amca01 at gmail.com>wrote:
>>>
>>>> The number is simply too big to be factored by Maxima's algorithm. You
>>>> might like to try other software with better support for computational
>>>> number theory, such as Pari/GP. I suspect that the numbers in question
>>>> (from here <http://www.immortaltheory.com/NumberTheory/RuthAaron.htm>;)
>>>> were produced by a polynomial, so all you need to do is to solve the
>>>> appropriate polynomial f(x)=n, where n is the smaller of the two numbers.
>>>>
>>>> -Alasdair
>>>>
>>>> On Wed, Sep 14, 2011 at 7:25 AM, Matematicas <metodosu at gmail.com>wrote:
>>>>
>>>>> Numbers are called Ruth-Aaron pair, such as {714, 715}. Are
>>>>> consecutive numbers {n, n +1} such that the sum of the prime factors
>>>>> of each are equal.
>>>>>
>>>>> Let's see
>>>>>
>>>>> 714 = 2x3x7x17
>>>>> 715 = 5x11x13
>>>>>
>>>>> Sum of divisors of 714 = 29 = Sum of divisors of 715
>>>>>
>>>>>
>>>>>
>>>>> FACTORS FOR EACH OF THE FOLLOWING TWO NUMBERS. COULD NOT DERIVE
>>>>>
>>>>> A number of 135 digits es1 Aoron
>>>>>
>>>>> *
>>>>> 296603151077074197308684124715622560525294501766069587161483751777320391554230009758029564174782423278870115094641644418265512090475910
>>>>> *
>>>>>
>>>>>
>>>>> The following of the former is
>>>>>
>>>>> *
>>>>> 296603151077074197308684124715622560525294501766069587161483751777320391554230009758029564174782423278870115094641644418265512090475911
>>>>> *
>>>>>
>>>>> The problem is that the command ifactor(n) don't work, generate an
>>>>> error, can help me??
>>>>>
>>>>> 1<#13268eec65d80c87_13267fb41fb33c74_13267b72b91849ef_132654fce756ae77_13264ae44588bf30_sdfootnote1anc>
>>>>> http://www.immortaltheory.com/NumberTheory/RuthAaron.htm
>>>>>
>>>>>
>>>>>
>>>>> _______________________________________________
>>>>> Maxima mailing list
>>>>> Maxima at math.utexas.edu
>>>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>>>>
>>>>>
>>>>
>>>>
>>>> --
>>>> Blog: http://amca01.wordpress.com
>>>> Web: http://bit.ly/Alasdair
>>>> Facebook: http://www.facebook.com/alasdair.mcandrew
>>>>
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>>>>
>>>>
>>>
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>>>
>>>
>>
>>
>> --
>> Blog: http://amca01.wordpress.com
>> Web: http://bit.ly/Alasdair
>> Facebook: http://www.facebook.com/alasdair.mcandrew
>>
>
>
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>
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