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I've been thinking and trying to solve this problem for quite sometime ( like a month or so ), but haven't achieved any success so far, so I finally decided to post it here.

Here is my problem:

If $f(x)$ is a polynomial with integer coefficients and $f( 2)= 3$ and $f(7) = -5$ then prove that $f(x)$ has no integer roots.

All I can think is that if we want to prove

  1. that if $f( x)$ has no integer roots, then by the integer root theorem its coefficient of highest power will not be equal to 1, but how can I use this fact ( that I don't know)?

  2. How to make use of given data that $f( 2)= 3$ and $f(7) = -5$?

  3. Assuming $f(x)$ to be a polynomial of degree $n$ and replacing $x$ with $2$ and $7$ and trying to make use of given data creates only mess.

    Now, if someone could tell me how to approach these types of problems other than giving a few hints on how to solve this particular problem , I would greatly appreciate his/her help.

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up vote 5 down vote accepted

Assume by contradiction that $f(x)$ has a root $x=a$. Then

$$2-a | f(2)-f(a)=3 \Rightarrow 2-a \in \{ -3,-1,1,3 \} \,.$$ $$7-a | f(7)-f(a)=-5 \Rightarrow 7-a \in \{ -5,-1,1,5 \} \,.$$


$$a \in \{ -1,1,3,5 \} \cap \{2,6,8,12 \}$$

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THANKS FOR THE ANSWER – shrey Dec 22 '12 at 6:09

Let's define a new polynomial by $g(x)=f(x+2)$. Then we are told $g(0)=3, g(5)=-5$ and $g$ will have integer roots if and only if $f$ does. We can see that the constant term of $g$ is $3$. Because the coefficients are integers, when we evaluate $g(5)$, we get terms that are multiples of $5$ plus the constant term $3$, so $g(5)$ must equal $3 \pmod 5$ Therefore there is no polynomial that meets the requirement. As the antecedent is false, the implication is true.

This is an example of the statement that for all polynomials $p(x)$ with integer coefficients, $a,b \in \mathbb Z \implies (b-a) | p(b)-p(a)$

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Your answer shows that, in fact, there is no polynomial with integer coefficients, $f(2) = 3$ and $f(7) = -5$. – David Speyer Dec 21 '12 at 19:01

The excellent answer of @N. S. makes it seem that the solution is really a matter of even and odd. Look at the supposed polynomial $f$ modulo $2$ and you see that $f(0)\equiv1\pmod2$ and $f(1)\equiv1\pmod2$, so that the polynomial always takes odd values at integers, and consequently has no integer roots. For a more elementary treatment, I’ll use the letter $e$ with subscripts to represent unspecified even numbers. Then since $f(2)=3$, $f(e_0)=f(2+e_1)=e_2+3$, odd, while $f(7)=-5$ shows that $f(1+e_3)=f(7+e_4)=-5+e_5$, also odd.

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Thanks for the enlightenment. – shrey Dec 22 '12 at 6:08

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