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Is it possible to construct a polynomial of degree N, with all of them as integer coefficient have a root as the given value. The root value provided is not necessarily a rational number.

For example, if the root is 28.552622898861801 we can have a polynomial of degree 10 whose one root will be the given value.

10000 x^10 - 280000 x^9 - 150000 x^8 - 220000 x^7 - 40000 x^6 - 790000 x^5 - 160000 x^4 - 320000 x^3 - 270000 x^2 - 250000 x - 251271 = 0

Wolfram alpha link for this equation

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If your rational number is $p/q$ you can take the polynomial $(px-q)^N$. Or did I misunderstand the question. – Michael Bächtold Oct 20 '11 at 7:30
I meant $(qx-p)^N$. – Michael Bächtold Oct 20 '11 at 7:32
oops. updated the description. – Arif Oct 20 '11 at 7:32
See Not all real numbers are algebraic. – Michael Bächtold Oct 20 '11 at 7:34
Your welcome. For the future you might want to try asking at instead of this site which is more for research level questions. – Michael Bächtold Oct 20 '11 at 8:39
up vote 2 down vote accepted

The problem can be solved by running some Integer Relation algorithm (e.g., PSLQ) on the numbers $1, r, r^2, \dots, r^N$ where $r$ is a given root.


For example, here is computation in PARI/GP which gives a better result than the polynomial shown in question:

? r = 28.552622898861801; algdep(r,10)

%1 = 3*x^10 + 38*x^9 - 3695*x^8 + 4582*x^7 + 3016*x^6 + 1435*x^5 + 4552*x^4 - 1219*x^3 - 9920*x^2 - 2402*x + 3087

? subst(%1,x,r)

%2 = -2.7334689816478450022 E-24

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