Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 4 down vote accepted

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

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


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

share|improve this answer
THANKS FOR THE ANSWER –  shrey Dec 22 '12 at 6:09
add comment

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, $a,b \in \mathbb Z \implies (b-a) | p(b)-p(a)$

share|improve this answer
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
add comment

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.

share|improve this answer
Thanks for the enlightenment. –  shrey Dec 22 '12 at 6:08
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.