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Problem:

Suppose $f(x)$ is a polynomial with integer coefficients. If $f(a)=17$ for some integer $a$, show that $f(x)$ has at most $3$ integer roots.

I tried to solve this problem but ended up getting the result as a maximum of $4$ integer roots. Here's the method I used:

Assuming that the three integer roots are $\alpha$,$\beta$,$\gamma$
$f(\alpha) = f(\beta) = f(\gamma) = 0$

As $f(m) - f(n) = k * (m-n)$, where $m,n,k$ are integers

$f(17) - f(\alpha) = k * (17-\alpha) 17 - 0 = k * (17-\alpha)$

$(17-\alpha)$ can now be $-17 , -1 , 1 \text{ or } 17$
Thus $\alpha$ can be $34 , 18 , 16 \text{ or } 0$

Similarly $\beta$ and $\gamma$ can take any of the remaining values, leaving one last value for a fourth root.

Could anyone suggest any other methods or point out where exactly I went wrong?

nonuser
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3 Answers3

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Let $g(x)=f(x+a)$. Then $g$ is a polynomial with integer coefficients and $g(0)=17$ and with as many integral roots as $f$ has. By the rational root theorem, all integer roots of $g$ must be divisors of the constant term, i.e., they must be among $\{17,1,-1,-17\}$. So far, there are still four roots possible. It seems this is what you found with your method as well.

However, suppose $17$ is one of the integer roots. Then we can divide out the linear factor $(x-17)$, i.e., we have $g(x)=(x-17)h(x)$ for some polynomial $h$ with integer coefficients. Then $h(0)=-1$, which now implies that all integral roots of $h$ (i.e., all remaining integral roots of $g$) are among $\{1,-1\}$, and so $-17$ cannot be a root of $g$. By the same method, we find that $g(-17)=0$ implies $g(+17)\ne 0$. Hence at most three of the numbers $-17,-1,1,17$ can be roots of $g$.

Hagen von Eitzen
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    More generally, if $g$ is a polynomial with integer coefficients, the product of its integer roots must divide the constant term. – Robert Israel Feb 13 '20 at 19:44
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Say exists $f$ with at least 4 integer roots $b,c,d,e$. Then $$f(x) = g(x)(x-b)(x-c)(x-d)(x-e)$$ where $g(x)$ is another polynomial with integer coefficents. So plugging $a$ in to this formula we get $$17= |f(a)| = |g(a)||a-b||a-c||a-d||a-e|$$ which means that at least three linear factors are $1$, WLOG say $$ 1=|a-c|=|a-d|=|a-e|\implies $$ $$a-c,a-d,a-e\in\{1,-1\}$$ But then two of them are, by pigeonhole principle, equal, say $$a-c=a-d\implies c=d$$

A contradiction.

nonuser
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0

Well, if $f(x)$ has at least $4$ four distinct integer roots, $b,c,d,e$ then $f(x)= p(x)(x-b)(x-c)(x-d)(x-e)$.

If $f(a) = 17$ then $p(a)(a-b)(a-c)(a-d)(a-e) = 17$ which is prime.

But $b,c,d,e$ are distinct so $(a-b),(a-c),(a-d),(a-e)$ are distinct. But $17$ is prime and can not have $4$ distinct factors. We can have $1,-1,\pm 17$ as three distinct factors, but ... no more.

If you want to be more thorough:

Let $\{p(x), (a-b), (a-c), (a-d), (a-e)\}= \{m_1, m_2, m_3, m_4,m_5\}$ so

$m_1m_2m_3m_4m_5 = 17$. So $17$ is prime so it divides one of the $m_i$. Let's label that value $m_1$. (It could be that $m_1 = p(a)$ or it could be that $m_1$ is one of the $(a-$ something $)$ terms. We don't know.)

So $\frac {m_1}{17}$ is an integer and $\frac{m_1}{17}*m_2m_3m_4m_5=1$. The only factors of $1$ are $1$ and $-1$ so $m_2, m_3, m_4, m_5$ are all either $1$ or $-1$.

So either all four of $m_2, m_3, m_4, m_5$ are the same and at least three of $(a-$ something $)$ are the same.

Or three of $m_2,m_3,m_4,m_5$ are the same and at least two of $(a-$ something $)$ are the same.

Or two of $m_2,m_3, m_4, m_5$ are $1$ and the other two are $-1$. And at least one of these pairs are both $(a-$ something $)$ terms. So those two are the same.

So no matter how we cut it $2$ of $(a- $something$)$ terms are the same and at least two of $b,c,d,e$ are the same.

fleablood
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