In article <14ovf5tai6d3ucp0c11ck7bb20vq8mh...@4ax.com>,
AP <marc.picher...@wanadoo.fr.invalid> wrote: > if a, p, q, are i n Z > If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with > S_0=3
> it is easy to prove that S_n is in Z
> but, also, if n is prime, n divide S_n+a > if u,v, w are in Z it is obvious cf Fermat > but in the general case it is more difficult (I have read a proof)
> this result ( n prime divide S_n+a) is only an exercice or something > with a general interest?
Is this really about cubics, or does a corresponding result hold for polynomials of arbitrary degree?
-- Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
<ge...@maths.mq.edi.ai.i2u4email> wrote: >In article <14ovf5tai6d3ucp0c11ck7bb20vq8mh...@4ax.com>, > AP <marc.picher...@wanadoo.fr.invalid> wrote:
>> if a, p, q, are i n Z >> If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with >> S_0=3
>> it is easy to prove that S_n is in Z
>> but, also, if n is prime, n divide S_n+a >> if u,v, w are in Z it is obvious cf Fermat >> but in the general case it is more difficult (I have read a proof)
>> this result ( n prime divide S_n+a) is only an exercice or something >> with a general interest?
>Is this really about cubics, or does a corresponding result hold >for polynomials of arbitrary degree?
I don't know : but the proof I have seen for cubics use Cardan
On Nov 15, 3:05 am, AP <marc.picher...@wanadoo.fr.invalid> wrote:
> if a, p, q, are i n Z > If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with > S_0=3
> it is easy to prove that S_n is in Z
> but, also, if n is prime, n divide S_n+a > if u,v, w are in Z it is obvious cf Fermat > but in the general case it is more difficult (I have read a proof)
> this result ( n prime divide S_n+a) is only an exercice or something > with a general interest?
> Thanks
If x_1, ..., x_n are the roots of any monic polynomial of degree n with integer coefficients, then (x_1)^r + ... + (x_n)^r is easily seen to be an integers by the symmetric functions theorem, that is any symmetric polynomial in n variables with integer coefficients is a polynomial in the elementary symmetric functions of x_1, ..., x_n with integer coefficients. Since these elementary symmetric functions in the roots are just the coefficients of the polynomial of which they are the roots. This theorem was certainly known to Gauss - he provided a beautiful proof - but I suspect it was known much earlier.
However, I fail to understand the rest of your post. It seems to me that that if x = y = z = 1, then all of your S_r are 1^n + 1^n + 1^n = 3 and your conclusion fails. What am I missing?
> If x_1, ..., x_n are the roots of any monic polynomial of degree n > with integer coefficients, then (x_1)^r + ... + (x_n)^r is easily > seen to be an integers by the symmetric functions theorem, that is any > symmetric polynomial in n variables with integer coefficients is a > polynomial in the elementary symmetric functions of x_1, ..., x_n with > integer coefficients. Since these elementary symmetric functions in > the roots are just the coefficients of the polynomial of which they > are the roots. This theorem was certainly known to Gauss - he > provided a beautiful proof - but I suspect it was known much earlier.
I think the method for expressing S_n in terms of the coefficients is usually ascribed to Newton.
-- Timothy Murphy e-mail: gayleard /at/ eircom.net tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Snake" <ach...@hotmail.com> wrote: >On Nov 15, 3:05 am, AP <marc.picher...@wanadoo.fr.invalid> wrote: >> if a, p, q, are i n Z >> If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with >> S_0=3
>> it is easy to prove that S_n is in Z
>> but, also, if n is prime, n divide S_n+a >> if u,v, w are in Z it is obvious cf Fermat >> but in the general case it is more difficult (I have read a proof)
>> this result ( n prime divide S_n+a) is only an exercice or something >> with a general interest?
>> Thanks
>If x_1, ..., x_n are the roots of any monic polynomial of degree n >with integer coefficients, then (x_1)^r + ... + (x_n)^r is easily >seen to be an integers by the symmetric functions theorem, that is any >symmetric polynomial in n variables with integer coefficients is a >polynomial in the elementary symmetric functions of x_1, ..., x_n with >integer coefficients. Since these elementary symmetric functions in >the roots are just the coefficients of the polynomial of which they >are the roots. This theorem was certainly known to Gauss - he >provided a beautiful proof - but I suspect it was known much earlier.
>However, I fail to understand the rest of your post. It seems to me >that that if >x = y = z = 1, then all of your S_r are 1^n + 1^n + 1^n = 3 and your >conclusion fails. What am I missing?
> On Mon, 16 Nov 2009 20:35:06 -0800 (PST), "Achava Nakhash, the Loving
> Snake" <ach...@hotmail.com> wrote: > >On Nov 15, 3:05 am, AP <marc.picher...@wanadoo.fr.invalid> wrote: > >> if a, p, q, are i n Z > >> If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with > >> S_0=3
> >> it is easy to prove that S_n is in Z
> >> but, also, if n is prime, n divide S_n+a > >> if u,v, w are in Z it is obvious cf Fermat > >> but in the general case it is more difficult (I have read a proof)
> >> this result ( n prime divide S_n+a) is only an exercice or something > >> with a general interest?
> >> Thanks
> >If x_1, ..., x_n are the roots of any monic polynomial of degree n > >with integer coefficients, then (x_1)^r + ... + (x_n)^r is easily > >seen to be an integers by the symmetric functions theorem, that is any > >symmetric polynomial in n variables with integer coefficients is a > >polynomial in the elementary symmetric functions of x_1, ..., x_n with > >integer coefficients. Since these elementary symmetric functions in > >the roots are just the coefficients of the polynomial of which they > >are the roots. This theorem was certainly known to Gauss - he > >provided a beautiful proof - but I suspect it was known much earlier.
> >However, I fail to understand the rest of your post. It seems to me > >that that if > >x = y = z = 1, then all of your S_r are 1^n + 1^n + 1^n = 3 and your > >conclusion fails. What am I missing?
> in this case a=-3 : (X-1]^3=X^3-3X^2....
> >Regards, > >Achava
Thank you for replying. I was misreading your problem. I would write this as (using p for prime instead of n because I am like that) p divides (S_p) + a, or alternatively S_p = -a (mod p), where a = -(x + y + z). This is indeed true for monic polynomials of any degree, say degree n, as long as you call the coefficient of x^(n-1) by the name a, and note that here also, with roots x_1, ..., x_n, so that a = -(x_1 + ... + x_n), so that we are trying to show that
where the symbol H is to chosen to stand for Horrific Mess, but H is nevertheless easily seen to be a symmtric polynomial in the x_i with integer coefficients, and hence is an integer, again by the symmetric polynomial theorem. Hence
Where the first equals sign is definitions, the second is Fermat's little theorem, and the third follows from the fact that H is an integer.
The more modern approach would exploit a little Galois theory to show imnmediately that a symmetric function in the roots of any polynomial over the rationals must be a rational number, and the bare beginnings of the theory of integral ring extensions, or in this case you could say the beginnings of algebraic number theory which says that any algebraic integer, which any polynomial in the roots of a monic polynomial over the integers must be, which is also rational, must be integral.
Thank you Timothy Murphy, for your historical comment on S_n being attributed to Newton. I didn't know that, but the Newton identities, which relate the S_n and the elementry symmetric functions, are always presented using this notation, which is certainly suggestive. I was giving a no later than date for the symmetric functions theorem, which might well predate Newton for all I know. Does anyone out there know?
> if a, p, q, are i n Z > If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with > S_0=3
> it is easy to prove that S_n is in Z
> but, also, if n is prime, n divide S_n+a > if u,v, w are in Z it is obvious cf Fermat > but in the general case it is more difficult (I have read a proof)
> this result ( n prime divide S_n+a) is only an exercice or something > with a general interest?
It's a well-known very old result - going back to at least 1839 according to Dickson, cf. his History... v.1, p.69 where he wrote:
Th. Schonemann [1] proved by use of symmetric functions of the roots that if z^n + b1 z^(n-1)+... = 0 is the equation for the pth powers of the roots of x^n + a1 x^(n-1)+... = 0,, where the a's are integers and p is a prime, then bi = ai^p (mod p). If the latter equation is (x-1)^n = 0, the former is z^n - (n^p + pQ) z^(n-1)+... = 0, and yet is evidently (z-l)^n = 0. Hence n^p = n (mod p).
>where the symbol H is to chosen to stand for Horrific Mess, but H is >nevertheless easily seen to be a symmtric polynomial in the x_i with >integer coefficients, and hence is an integer, again by the symmetric >polynomial theorem. Hence
>> if a, p, q, are i n Z >> If S_n=u^n+v^n+w^n (u, v, w roots, in C, of X^3+aX^2+pX+q), with >> S_0=3
>> it is easy to prove that S_n is in Z
>> but, also, if n is prime, n divide S_n+a >> if u,v, w are in Z it is obvious cf Fermat >> but in the general case it is more difficult (I have read a proof)
>> this result ( n prime divide S_n+a) is only an exercice or something >> with a general interest?
>It's a well-known very old result - going back to at least 1839 according >to Dickson, cf. his History... v.1, p.69 where he wrote:
> Th. Schonemann [1] proved by use of symmetric functions of the roots that if > z^n + b1 z^(n-1)+... = 0 is the equation for the pth powers of the roots of > x^n + a1 x^(n-1)+... = 0,, where the a's are integers and p is a prime, then > bi = ai^p (mod p). If the latter equation is (x-1)^n = 0, the former is > z^n - (n^p + pQ) z^(n-1)+... = 0, and yet is evidently (z-l)^n = 0. Hence > n^p = n (mod p).
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