Re: diofantea

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#39

From: Dr. Wolfgang Hintze
Date: Mon, 07 Nov 2011 21:42

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"superpollo" <superpollo@tznvy.pbz> schrieb im Newsbeitrag 
news:4eb84dae$0$1385$4fafbaef@reader1.news.tin.it...
> Ignacio Larrosa Ca�estro ha scritto:
>> Ignacio Larrosa Ca�estro wrote:
>>> superpollo wrote:
>>>> superpollo ha scritto:
>>>>> risolvere per x,y interi positivi:
>>>>>
>>>>> x^4/y = (y+4)*(y^2-1)*(y^2-4)*(y^2-9)
>>>>
>>>> hola! quieria extender el problema a e.c.m.
>>>
>>>
>>> Multiplicando por y, con lo que no deberemos tener en cuenta las
>>> soluciones con y = 0, queda
>>>
>>> x^4 = (y+4)(y+3)(y+2)(y+1)y(y-1)(y-2)(y-3)
>>>
>>> La �nica posibilidad de que el producto de 8 enteros consecutivos 
>>> sea
>>> cero,
>>
>> quise decir "una cuarta potencia", no "cero".
>
> como se puede demonstrar?

Poniendo k = y + 3 queda

x^4 = k(k+1)...(k+7)

o

k! x^4 = (7+k)!

de donde

x^4 = C(7+k,k) 7! = C(7+k,k) 2^4.3^2.5.7

==> con x = 2 t

t^4 = C(7+k,k) 3^2.5.7

los factores excepto C(,) no son quartos.
Bueno ... � y qu� eso nos da !

Estoy atascado ... tal vez vosotros podeis continuar.

Saludos,
Wolfgang

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From: "Dr. Wolfgang Hintze" <weh@snafu.de>
Newsgroups: it.scienza.matematica,es.ciencia.matematicas
Subject: Re: diofantea
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