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Question: Suppose $P(x)$ is a polynomial with $P(2)=2017$ and $P(5)=2002$. If it is given that $P(x)=0$ has exactly one integer root, find that root.

My approach: I tried solving the problem by considering that $P(x)\in\mathbb{Z}[x]$ and that $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0.$$ Then $P(5)\equiv a_0 \pmod 5\implies 2002\equiv a_0 \pmod 5\implies a_0\equiv 2\pmod 5.$ Again, $P(2)\equiv a_0 \pmod 2 \implies 2017\equiv a_0 \pmod 2\implies a_0\equiv 1 \pmod 2.$ Thus, we have $$\begin{cases}a_0\equiv 1\pmod 2,\\a_0\equiv 2\pmod 5.\end{cases}$$ By solving the same we can conclude that $a_0\equiv 7 \pmod {10}.$

Also if $a$ is the only integer root of $P(x)$, then there exists $Q(x)\in \mathbb{Z}[x]$ with $\deg (Q(x))=n-1$ such that $$P(x)=(x-a)Q(x).$$ Thus, we have $$(2-a)Q(2)=2017,\text{ and }(5-a)Q(5)=2002=2\times 7\times 11\times 13.$$

After this I couldn't make any significant approach. I also understand that I shouldn't have made the assumption of $P(x)\in\mathbb{Z}[x]$, but that's the best I could think of.

Can anyone help me progress (initially a small hint would be nice)?

Sanket Biswas
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    Hint: we have $(x-2),|,(P(x)-2017)$ so, if $x_0$ is the root, $(x_0-2),|,2017$. And we get a similar result from knowing $P(5)=2002$. – lulu Aug 21 '20 at 20:40
  • Thanks @lulu. This approach directly helps in concluding the answer to be 3. – Sanket Biswas Aug 21 '20 at 21:01
  • Yes...just list the divisors of each, add the right constant and look for a common element. – lulu Aug 21 '20 at 21:09
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    Just to settle the integrality question, note that $P(x)=\frac {14139 }{10}x - \frac {2027}{10}x^2$ satisfies your two conditions, and $P(0)=0$. Indeed, this example was obtained by interpolating those three values . We need $P(x)\in \mathbb Z[x]$. – lulu Aug 21 '20 at 21:53

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The second approach is more fruitful. $2017$ is prime whence $2- a=1, -1, -2017, 2017$. If it is 2017 then $a=-2015$ and $a-5= -2020$, does not divide $2002$ , a contradiction. If it is $-2017$ the $a=2019, 5-a= - 2014$ , a contradiction. If $2-a=-1$, $a=3, 5-a=2$ is Ok. Finally if $2-a=1, a=1, 5-a=4,$ a contradiction. Answer $a=3$.

markvs
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