For polynomial $P$ with whole-number coefficients, it is given that $P(1)=1$. Prove that this polynomial doesn't have three distinct whole-number roots.
Any help would be appreciated.
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Are they integer roots, complex roots or what? You need to specify. – jijijojo Jan 06 '20 at 16:04
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Question. What can you conclude from $a $ is a root of $P$? – jijijojo Jan 06 '20 at 16:05
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I'm not sure what you mean by that mr.pigeon. – AdamPedrovec Jan 06 '20 at 16:08
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If $a$ is a root of $P$, then $x-a$ divides $P(x)$. Try using this. – jijijojo Jan 06 '20 at 16:09
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2If $n$ is a root then you can write $P(x) = (x-n) Q(x)$ where $Q$ also has integer coefficients. Rinse & repeat twice. Now evaluate the result at $x=1$ and use properties of integers. – copper.hat Jan 06 '20 at 16:13
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Key to understand is why it was asked for whole numbers, and not for more popular natural numbers? – Bhaskar Vashishth Jan 06 '20 at 16:19
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Assume there are three distinct roots, $a,b,c$.
$$P(x) = (x-a)(x-b)(x-c)Q(x)$$
$$P(1) = (1-a)(1-b)(1-c)Q(1) = 1$$
This implies $(1-a), (1-b), (1-c) \in \{-1,1\}$, which implies at most two distinct roots.
SlipEternal
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1Why does $Q$ have to be an integer then? That is what a proper proof of this have to demonstrate. – Winther Jan 06 '20 at 16:23
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That's what happens when you have a polynomial with integer coefficients is divided by a linear expression of $x-r$ where $r$ is a root. You wind up with a polynomial with integer coefficients. This is already discussed in the comments above. – SlipEternal Jan 06 '20 at 16:24
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Let $$p(x) = a_0x^n + a_1x^{n-1} +....+a_k$$
At $x=1$ , we have :
$$p(1) = a_0 + a_1+....+a_k = 1$$
Since $a_0 , a_1 , ... ,a_k $ are whole numbers , it follows that exactly one of them is $1$ while the others are $0$ .
If $a_k = 1$ , then $p(x) = 1$ , which has no roots.
If any other term is $1$ , then $p(x) = x^n$ , which has only one root.
The Demonix _ Hermit
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I think by whole number OP means an integer. The term 'whole number' is not very clear. – bjorn93 Jan 06 '20 at 16:17