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A question from pathfinder for Olympiad mathematics:

Prove that if a polynomial of degree $7$ over $\mathbb{Z}$ is equal to $+1$ or $-1$ for $7$ different integers then it is irreducible over $\mathbb{Z}$.

My Attempt:

$a,b,c,d,e,f,g$ be the seven integers.

Let $f(a) = 1,f(b) = -1$

$f(a) - f(b)= 2$

But $a-b\mid f(a)-f(b)$

Therefore $a-b\mid 2$

Therefore $a-b = 1,-1,2,-2$

I am stuck here... Please help me...

Blue
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Ouestion Asker
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    Assume it factors into a product of polynomials of positive degrees with integer coefficients. Note that the values of the factors at those points should also be $\pm1$. Look at the factor of smallest degree. Its degree is at most $3$, but at least one of the two values $+1$ or $-1$ is taken at least at $4$ of those seven points. – plop Jun 08 '21 at 15:34
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    Seems appropriate for an answer, not just a comment. – Robert Shore Jun 08 '21 at 15:36
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    Who is pathfinder ? – Jean Marie Jun 08 '21 at 18:54

2 Answers2

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Let $f$ be the polynomial of degree 7, and let $x_1,x_2,\ldots, x_7$ be the integers in increasing order satisfying $f(x_k) \in \pm 1$ for each $k=1,\ldots,7$. Suppose $f=f_1f_2$, where $f_1$ and $f_2$ also have all integral coefficients, and both have positive degree strictly less than 7. Then $f_1(x_k) \in \pm 1$ for $x_k$ for all $k\in\{1,2, \ldots, 7\}$ as well [because $f_1(x_k)$ and $f_2(x_k)$ are each integral and $f_1(x_k)f_2(x_k) = f(x_k) \in \pm 1$]. Then as $f_1$ has degree strictly less than 7, it follows that the $f(x_k)$ cannot all be the same for all $k=1,\ldots, 7$. Thus, it follows [from the fact that $(x_7-x_k)|f(x_7)-f(x_k)$ and the fact that the $x_7-x_k$s are all positive and distinct integers for any $k \not = 7$] that $$f_1(x_1)=f_1(x_2)=f_1(x_3)=f_1(x_4)=f_1(x_7)$$ [as $x_7-x_k$ cannot divide $\pm 2$ for $k \le 4$]. Likewise, it follows that $$f_1(x_1) = f_1(x_4)=f_1(x_5)=f(x_6)=f_1(x_7).$$ So from these we conclude that all of the $f_1(x_k)$s must be the same after all, which implies that $f_1$ must be either degree 0 or 7, which is a contradiction. Thus $f$ is irreducible.

Mike
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  • You say $f(x_k)=0.$ I think you mean $f(x_k)=\pm 1.$ – Thomas Andrews Jun 15 '21 at 22:32
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    The rest of your argument might work, but it is easier to show that, since either $\deg f_1<4$ or $\deg f_2<4,$ and for each $i,$ either $f_i(x)+1$ or $f_i(x)-1$ has four roots, contradicting that condition on the degree. – Thomas Andrews Jun 15 '21 at 22:37
  • Yes fixed that. But anyways yes noting that the degree of $f_1$ is at most 3 and finishing that way would have been simpler. – Mike Jun 16 '21 at 00:03
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Suppose $p(x),q(x),r(x)\in\Bbb Z[x]$ with $p(x)=q(x)r(x)$ where $0\le \deg(q)\le 3<\deg (r)\le\deg (p)=7$.

Suppose $n_1,...,n_7$ are $7$ distinct integers and $|p(n_j)|=1$ for $1\le j\le 7.$

Then $q(n_j),r(n_j)\in \Bbb Z$ for each $j$ and $|q(n_j)r(n_j)|=1.$ So $q(n_j)=\pm 1.$

So either $\{n_j: q(n_j)=1\}$ or $\{n_j: q(n_j)=-1\}$ has at least $4$ members, but $deg(q)\le 3.$ So $q$ must be the constant $1$ or $-1,$ and $r=\pm p.$

DanielWainfleet
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