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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 $\deg(P) = 7$,

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

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.

Angie
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    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 $x_i$. – Tom Oldfield Apr 17 '13 at 02:23
  • @TomOldfield Why is $1/n$ a root? If I read correctly, it gives $Q(1/n)\neq 0$. – Julien Apr 17 '13 at 02:38
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    Does $P(x) = 7$ count as a polynomial? –  Apr 17 '13 at 02:39
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    $7$ is a prime number. – achille hui Apr 17 '13 at 02:41
  • @user17762 Well even if it does, it wouldn't make a difference, right? $P(x)=7$ doesn't have any integer zeros. – Angie Apr 17 '13 at 02:42
  • @Angie $P(x) = 7$ has no zeros (let alone integer zero). Hence, $P(x)$ need not have integer zeros. –  Apr 17 '13 at 02:44
  • @TomOldfield Maybe. I'm not sure. I wrote it in the exact way the problem was stated. – Angie Apr 17 '13 at 02:44
  • @user17762 I think the question is asking if there is any $P(x)$ that does have integer zeros. – Angie Apr 17 '13 at 02:48
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    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. – Julien Apr 17 '13 at 02:50

1 Answers1

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Hint:

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

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

Halil Duru
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