Assume that is a rational function with rational coefficients. Suppose moreover that takes integer values at the integers. Then in fact is a *polynomial* with rational coefficients.

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# Rational function over Q

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2 thoughts on “Rational function over Q”

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Assume that is a rational function with rational coefficients. Suppose moreover that takes integer values at the integers. Then in fact is a *polynomial* with rational coefficients.

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The main ingredient to prove this is the following:

Let a polynomial with integer coefficients and consider the set . Then is an infinite set.

The proof of this fact is easy. An sketch of a proof is the following: suppose is finite and enumerate its elements. Consider for sufficiently large such that is greater than the independent coefficient of and also greater the valuation of each prime of the independent coefficient. This will give the desired contradiction.

Now, suppose is a rational function which is not a polynomial, then there exist such that . Moreover, we can suppose that are coprime and thus by Euclid algorithm, there are polynomials such that . By clearing denominators we obtain polynomials such that for a certain integer and is an integral multiple of . Since for every we must have that every prime divisor of divides . But we saw that this set must be infinite. This is a contradiction and we are done.

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Let be a rational function and suppose that it takes integer values and has degree greater than . Then . Indeed, , so must have infinitely many roots, and . Taking now we can assume has degree at most that of . Writing with rational polynomials and of degree less than , we can assume by multiplying by a common denominator that is a polynomial in integers. Then takes integral values at the integers, and we conclude that it is zero by the above.

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