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A polynomial is denoted by $f(x)$. The coefficients of the polynomial are positive integers. $$f(1) =17$$ $$f(20)=421350$$ Could you tell if such a polynomial is possible? If ye, find the degree of the polynomial and also it's coefficients.

My inference:Using $f(20)=421350$ we can determine, that the degree of the polynomial cannot exceed $4$. Also since every coefficient is positive, therefore each individual coefficient $< 17$. The constant coefficient independent of $x$ is $10$. Since $421350$-constant term should be divisible by $20$.

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Note that for any polynomial $P$ with integer coefficients, and any integers $a$ and $b$, $P(a)-P(b)$ is divisible by $a-b$. This is because $a^k-b^k$ is divisible by $a-b$.

Thus if there is a polynomial $f(x)$ with the given values, then $421350-17$ must be divisible by $19$. But it isn't. So no such polynomial exists, even without the positivity restriction.

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    excellent answer!2012-11-15
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    SO the polynomial doesn't exist.2012-11-15
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    @AdwaitKumar: That's right. Asking for the coefficients to be positive was a mean little trick, it makes the problem look harder than it is.2012-11-15
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    I do not see the connection. What I wrote is that **if** $f(20)=421350$ and $f(1)=17$, where $f$ is a polynomial with integer coefficients, **then** $20-1$ must divide $421350-17$. It doesn't, which means there is no such polynomial $f(x)$.2012-11-15
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Brute force. Take your constant coefficient 10, and look at $$\frac{f(20)-10}{20}=21 067$$ This gives the coefficient of $x$ as 7. Total of coefficients so far is 17, which is given by $f(1)$, but you need higher terms to get $f(20)$ right.

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    Can the person who downvoted this please explain why. This is a perfectly reasonable way of using the positivity and integrality of the coefficients - extending the method in the original post, which I suggested because it might illuminate why the question was asked in the way it was (eg to avoid having to notice the approach André Nicholas put in his answer, by making the positivity do some work).2012-11-15
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    Strongly agree. I upvoted, of course, but that only partly negates the effect of the incorrect downvote.2012-11-15