Polynomials with integer coefficients Hello, I want to prove that for P, Q two unitary polynomials with rational coefficients, if PQ's coefficients are integers then both P and Q are of integer coefficients. I'm trying to look for a starting point. Any advice? Thanks! 
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As example: (2*4)^5 = 2^3 * 4^6 = 32768 
Clear denominators of $P$ and $Q$. That is, take the smallest positive integers $a$ and $b$ such that $aP$ and $bQ$ have integral coefficients. Then try to show $ab$ can't have any prime factors (which will imply $ab = 1$, and so $a = b = 1$ which means $P$ and $Q$ have integral coefficients). 
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@cjem: Thanks. I will try to build on that using the smallest common multiples of the denominators. Do you think P and Q being unitary will be of use? 
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However, you could replace the assumption that $P$ and $Q$ are unitary by the weaker assumption that $PQ$ is primitive (i.e. no prime divides all of its coefficients) and the result would be true again. 
But the second assumption would imply P and Q are unitary only if P and Q actually had integral coefficients (which we are trying to prove here)? 
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All I meant was that the stronger statement "if $P$ and $Q$ are rational polynomials such that $PQ$ is integral and primitive, then $P$ and $Q$ are integral" is true  you don't actually need to assume that $P$ and $Q$ are unitary, it suffices to just assume that $PQ$ is primitive. (Note $P$ and $Q$ being unitary implies that $PQ$ is primitive, but the reverse implication isn't true, as shown by the above example.) 
Ah, I see now. Thanks. Does P being unitary imply aP is primitive? 
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I wrote aP in coefficients, and I supposed that it is not primitive, meaning there is a prime number p such that p divides the greatest common factor of aP's coefficients. That factor divides a too (since a is a aP coefficient due to P being unitary). Any help in finding a contradiction? 
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