polynomial with rational coefficients and integer values
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From: christian.bau on 12 Jul 2010 16:42 On Jul 12, 8:22 pm, Axel Vogt <&nore...(a)axelvogt.de> wrote: > Let be P(x) = Sum(a(k)*x^k, k=0 .. N) a polynomial function, > where the a(k) are rational numbers. > > I am looking for a 'handy' criterion when P maps integers > to integers - say 'true for enough integers or large primes > then yes' or so (sorry for being vague). > > Certainly something known for that (probably it is classical), > but I am not aware of it. > > My example (why I stumbled into that) is P(x) = A(x)/120 = > = 2+107/10*x-13/3*x^2+43/24*x^3-1/6*x^4+1/120*x^5 where > A(x) = 240+1284*x-520*x^2+215*x^3-20*x^4+x^5 > > Using the little Fermat for p=3 and p=5 as well as Euler's > generalization using the totienten function shows it here > (120 = 8*3*5, A(2*k) = 0 mod 8 and n^4 = 1 mod 8, n odd). > > The only special thing I could see for that A(x): it has > only 1 real root and seems to irreducible. > > Any educating or enlightening hint for me? If the degree of the polynomial is n, and p (x) is an integer for n+1 consecutive integers, then p (x) is an integer for all integers x.
From: William C Waterhouse on 12 Jul 2010 16:59 The binomial coefficients x choose m must give integer values for integer x, since that coefficient has a combinatorial meaning. It's not hard to show by induction that sums of integer multiples of these coefficients are the only polynomials sending integers to integers. You can then take any polynomial, starting at the highest term, and see whether you can cancel that term by subtracting an integer multiple of the binomial of that same degree. If not, stop. If so, repeat with the highest remaining term, and so on. You get either a disproof or an explicit expression as a sum of integer multiples of binomials. William C. Waterhouse Penn State On 07/12/2010 03:58 PM, Axel Vogt wrote: > Arturo Magidin wrote: >> On Jul 12, 2:22 pm, Axel Vogt <&nore...(a)axelvogt.de> wrote: >>> Let be P(x) = Sum(a(k)*x^k, k=0 .. N) a polynomial function, >>> where the a(k) are rational numbers. >>> >>> I am looking for a 'handy' criterion when P maps integers >>> to integers - say 'true for enough integers or large primes >>> then yes' or so (sorry for being vague). >>> >>> Certainly something known for that (probably it is classical), >>> but I am not aware of it. >>> >>> My example (why I stumbled into that) is P(x) = A(x)/120 = >>> = 2+107/10*x-13/3*x^2+43/24*x^3-1/6*x^4+1/120*x^5 where >>> A(x) = 240+1284*x-520*x^2+215*x^3-20*x^4+x^5 >>> >>> Using the little Fermat for p=3 and p=5 as well as Euler's >>> generalization using the totienten function shows it here >>> (120 = 8*3*5, A(2*k) = 0 mod 8 and n^4 = 1 mod 8, n odd). >>> >>> The only special thing I could see for that A(x): it has >>> only 1 real root and seems to irreducible. >>> >>> Any educating or enlightening hint for me? >> >> I'm not sure if this is what you want, but... >> >> Let q(P) = r/s be the content of the polynomial; that is, r/s is a >> rational, gcd(r,s)=1, r and s integers, s>0, and P(x) = (r/s)Q(x), >> where Q(x) is a primitive polynomial with integer coefficients. >> (Primitive here means that the gcd of all the coefficients is 1). If >> P(n) is an integer for every integer n, then that means that for every >> integer n, s|Q(n) (since s must divide r*Q(n), and r is relatively >> prime to s). Thus, it follows that for every integer n, Q(n) = 0 (mod >> s). That is, the reduction modulo s of Q(x) must be the zero function >> in Z/sZ. >> >> Conversely, suppose that the reduction modulo s of Q(x) is the zero >> function in Z/sZ. Then for every integer n, Q(n) = 0 (mod s), hence s >> divides Q(n), hence s divides r*Q(n), hence (r/s)Q(n) is an integer. >> >> Thus, P(x) sends integers to integers if and only if Q(x) induces the >> zero function on Z/sZ. >> >> In particular, you would only need to check one integer per congruence >> class modulo s, giving that "sufficiently large" bound of integers. >> >> Hope this helps. >> >> -- >> Arturo Magidin > > Thank you for the fast reply. > > Hm ... (using Maple) the content of my P is 1/120 and the primitive > part is the A, so it is your Q. May be it is to late and hot, but > it seems that I just worked through what you sketched as recipe.
From: Bill Dubuque on 13 Jul 2010 12:28 Axel Vogt <&noreply(a)axelvogt.de> writes: > > Let be P(x) = Sum(a(k)*x^k, k=0 .. N) a polynomial function, > where the a(k) are rational numbers. > > I am looking for a 'handy' criterion when P maps integers > to integers - say 'true for enough integers or large primes > then yes' or so (sorry for being vague). > > Certainly something known for that (probably it is classical), > but I am not aware of it. [...] One classical criterion is simply to check that it has an integral difference table, i.e. use the finite difference calculus. Googling kewyords should yield more, e.g. http://deltaepsilons.wordpress.com/2009/07/21/integer-valued-polynomials/ --Bill Dubuque
From: Herman Jurjus on 13 Jul 2010 15:22 On 7/13/2010 6:28 PM, Bill Dubuque wrote: > Axel Vogt<&noreply(a)axelvogt.de> writes: >> >> Let be P(x) = Sum(a(k)*x^k, k=0 .. N) a polynomial function, >> where the a(k) are rational numbers. >> >> I am looking for a 'handy' criterion when P maps integers >> to integers - say 'true for enough integers or large primes >> then yes' or so (sorry for being vague). >> >> Certainly something known for that (probably it is classical), >> but I am not aware of it. [...] > > One classical criterion is simply to check that it has an > integral difference table, i.e. use the finite difference > calculus. Googling kewyords should yield more, e.g. > http://deltaepsilons.wordpress.com/2009/07/21/integer-valued-polynomials/ > > --Bill Dubuque A commonly known Z-basis of these integer-mapping-polynomials is the set of the polynomials 'X over n', for n = 0, 1, 2, ... (i.e. 1/n! (X*(X-1)*...*(X-n)) (That -is- a well known result, so you probably already know it. But perhaps others don't.) -- Cheers, Herman Jurjus
From: Axel Vogt on 13 Jul 2010 15:45
christian.bau wrote: > On Jul 12, 8:22 pm, Axel Vogt <&nore...(a)axelvogt.de> wrote: >> Let be P(x) = Sum(a(k)*x^k, k=0 .. N) a polynomial function, >> where the a(k) are rational numbers. >> >> I am looking for a 'handy' criterion when P maps integers >> to integers - say 'true for enough integers or large primes >> then yes' or so (sorry for being vague). .... > > If the degree of the polynomial is n, and p (x) is an integer for n+1 > consecutive integers, then p (x) is an integer for all integers x. Does it follow from the answers given or is there some direct proof as well? It seems to be a quite handy criterion. PS: Thanks to all (I hope it is ok that lame & general way)! |