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Given a polynomial $f(x)$ with integral coefficients and $f(x) = 37$ has 5 distinct integral roots, find the number of integral roots of $f(x) = 41$?

My Approach: Say $f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) + 37$, where $r_i$ are the distinct integers.

Now for $f(x) = 41$ we have $(x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) = 4$, so the factors can be $\pm 1, \pm2$ or $\pm 4$. Given $r_i$ are distinct at most two of them will give $\pm1$, then there can be both of $\pm 2$ or one of $\pm 4$. This is where I get lost, since even if I use all of $\pm1, \pm2$, I will be still be left with one $x-r_i$ factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence?

Thanks in advance.

Morty
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  • What kind of function is $f$? A polynomial? If so, does it have integer coefficients? – Alex Becker Apr 17 '14 at 00:34
  • Yes, sorry for that omission. – Morty Apr 17 '14 at 00:35
  • Does the polynomial have $5$ roots, each of which are integral, or more roots, of which five are integral? – MT_ Apr 17 '14 at 00:37
  • The question states that there are 5 distinct integers. Nothing is been said about whether they are only 5 integral solutions. + The coefficients of $f(x)$ are integers. – Morty Apr 17 '14 at 00:40

2 Answers2

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You are essentially finished. There cannot be an integer $a$ such that $(a-r_1)(a-r_2)(a-r_3)(a-r_4)(a-r_5)g(a)=4$. For the $a-r_i$ are distinct integers, and no product of $5$ distinct integers can divide $4$.

André Nicolas
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  • So it had nothing to do with 37 and 41 being prime. Just their difference mattered. In general suppose I would have another number say 12 (instead of 4), I could have got maximum 5 integral roots (1, -1, 2, -2, 3). But can I say that the answer is 5 in this case? – Morty Apr 17 '14 at 00:52
  • Or $-8$, since $g(x)$ is at our disposal. Again, there will not be a solution, but for numbers ricer in divisors than $8$, it will be possible. – André Nicolas Apr 17 '14 at 00:54
  • Thanks for the clarification. – Morty Apr 17 '14 at 00:56
  • You changed the $8$ to $12$ while I was typing! Yes, the max number of integral roots is $5$ for the $12$ case. The roots can also be $1,-1,2,-2,-3$ if we construct $g$ so that $g(x)=-1$ at suitable places. – André Nicolas Apr 17 '14 at 00:58
  • Thanks a lot for the clarification. – Morty Apr 17 '14 at 01:01
  • You are welcome. – André Nicolas Apr 17 '14 at 01:09
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Hint $\ $ Finish by applying the following

Key Idea $\ $ The possible factorizations of a polynomial $\in\Bbb Z[x]$ are constrained by the factorizations of the integer values that the polynomial takes. For a simple example, if some integer value has few factorizations (e.g. a unit $\,\pm1 $ or prime $p$) then the polynomial must also have few factors, asssuming that that the factors are distinct at the evaluation point. More precisely

If $\, f(x) = f_1(x)\cdots f_k(x)\,$ and $\,f_i\in\Bbb Z[x]\,$ satisfy $\color{#0a0}{f_i(n) \ne f_j(n)}\,$ for $\,i\ne j,$ all $\,n\in \Bbb Z$

$\quad \color{}{f(n) =\pm1}\,\Rightarrow\, k\le 2\ $ else $1$ would have $\rm\,3\,\ \color{#0a0}{distinct}$ factors $\,f_1(n),f_2(n),f_3(n)$

$\quad f(n) = \pm p\,\Rightarrow\, k\le \color{#c0f}4\ $ since a prime $p$ has at most $\,\color{#c0f}4\,$ distinct factors $\,\pm1,\pm p$

Remark $\ $ One can push the key idea to the hilt to obtain a simple algorithm for polynomial factorization using factorization of its integer values and Lagrange interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. See this answer for references.

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Bill Dubuque
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