The following is a homework question:

Let $P(x)$ be a polynomial with integer coefficients and $P(x_1)=P(x_2)=P(x_3)=P(x_4)=P(x_5)=P(x_6)=P(x_7)=7$ where $x_i$ are distinct integers. Determine if $P(x)$ has integer zeros.

I've never done questions like this before. I started with this:

If $\deg(P) = 7$,


where $\alpha$ is some integer.

However, the question doesn't state that the polynomial must be of the seventh degree. Even then, I don't see how I can determine if $P(x)$ has integer zeros without knowing all the $x_i$.

Can someone please help me? Thanks.

Edit: Is this a valid solution? $$P(x)=Q(x)(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$$

If $P(n)=0$, $$Q(n)(n-x_1)(n-x_2)(n-x_3)(n-x_4)(n-x_5)(n-x_6)(n-x_7)=-7$$

where $Q(x)$ is a polynomial of integer coefficients (therefore $Q(n)$ is an integer)

And since all the terms on the LHS are integers and the $x_i$ are distinct integers, it follows that some of the factors on the LHS $\ne$ {$\pm 1, \pm 7$}. And $7$ is a prime number, therefore $n$ cannot be a zero.

  • 1
    $\begingroup$ I think you need to add that $P$ is monic, i.e. that the lead coefficient is $1$. Otherwise if $Q(x)=(x-x_1)\dots(x-x_7)+7=R(x)+7$, the poly $((nx-1)R(x)+1)Q(x)$ has $\frac1n$ as a root, and evaluates as $7$ at the $x_i$. $\endgroup$ Apr 17, 2013 at 2:23
  • $\begingroup$ @TomOldfield Why is $1/n$ a root? If I read correctly, it gives $Q(1/n)\neq 0$. $\endgroup$
    – Julien
    Apr 17, 2013 at 2:38
  • 2
    $\begingroup$ Does $P(x) = 7$ count as a polynomial? $\endgroup$
    – user17762
    Apr 17, 2013 at 2:39
  • 1
    $\begingroup$ $7$ is a prime number. $\endgroup$ Apr 17, 2013 at 2:41
  • 1
    $\begingroup$ What you should prove is that if $x$ is a root of $P$, then it can't be an integer. So take such an $x$. Evaluate. And consider achille hui's comment again. $\endgroup$
    – Julien
    Apr 17, 2013 at 2:50

1 Answer 1



$P(x)= Q(x) (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$

Idea : You can't get $-7$ by multiplying $7$ distinct integers..

  • $\begingroup$ We'll save the argument for the day when $7$ will be replaced by some badly composed integer, +1. $\endgroup$
    – Julien
    Apr 17, 2013 at 2:56
  • $\begingroup$ I think what I included after editing my original post is similar. @julien Can you please check if it's correct? Thanks. $\endgroup$
    – Angie
    Apr 17, 2013 at 3:03
  • 1
    $\begingroup$ @Angie Nicely done. Just add $-7$ to the list of values $\{\pm 1,\pm 7\}$. $\endgroup$
    – Julien
    Apr 17, 2013 at 3:05
  • 1
    $\begingroup$ @Angie, you need to mention $Q(x)$ is a polynomial of integer coefficients. $\endgroup$ Apr 17, 2013 at 3:08
  • 1
    $\begingroup$ I think you're using the "Rational Root theorem". The term $Q(x)(x-x_1)\cdots(x-x_7)$ has a constant term which adds to 7, so the possible roots may be other than 1, -1, 7 and -7... $\endgroup$ Apr 17, 2013 at 3:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .