12

$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients. If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.

I tried by assuming a fifth degree polynomial but got stuck after that.

The question was asked by my friend.

hardmath
  • 37,015
Ramanujam Ganit Prashikshan Ke
  • 509
  • 6
    "The question was asked by my friend." Sorry but why are you mentioning this? Is this suposed to make any difference (for example, say, to alleviate the lack of personal input)? – Did Jan 01 '19 at 11:49

3 Answers3

30

If $u$ and $v$ are integer roots of $P$, then $P(x)=(x-u)(x-v)Q(x)$, where $Q$ is a polynomial with integer coefficients. From $P(2)=3$ we get $(u-2)(v-2)\mid 3$, and then WLOG either $u-2=1$ or $u-2=-1$, implying $u\in\{1,3\}$. Now $P(10)=11$ gives $(u-10)(v-10)\mid 11$, showing that $u-10$ is a divisor of $11$. However, neither $1-10=-9$, not $3-10=-7$ is a divisor of $11$, a contradiction.

W-t-P
  • 4,629
  • 2
    No idea why your answer has gotten so many fewer votes than mine! Your approach of considering both factors at once is quite slick. – Eric Wofsey Dec 30 '18 at 16:54
  • 2
    @EricWofsey: Don't worry, that's fine. Your answer provides some theoretical background and looks "more scientific". Besides, my empirical observation is that the number of upvotes is proportional to the present reputation of the person answering :-) – W-t-P Dec 30 '18 at 17:19
24

The assumption that $P$ has degree $5$ is irrelevant and unhelpful.

If $r$ is a root of $P$, we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$, $P(a)=(a-r)Q(a)$ must be divisible by $a-r$. Taking $a=2$ and $a=10$, we see easily that the only possible value of $r$ is $-1$.

Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$.

Eric Wofsey
  • 330,363
2

Key Idea $\ $ (Kronecker) $ $ How polynomials can factor is constrained by how their values factor, $ $ e.g. as below, in some cases if $\,P\,$ takes a prime value then it has at most one integer root.

Hint $ $ If $\,P\,$ has more roots than $\,P(2)\,$ has prime factors then factoring $P$ & evaluating at $x\!=\!2$ $\,\Rightarrow\,P(1)\!=\!0\,$ or $P(3)\!=\!0.\,$ But $P(1)\!\neq\! 0\,$ else $\,10\!-\!1\mid P(10)\!-\!P(1) = 11.\,$ $P(3)\!\neq\! 0\,$ similarly.

Theorem $ $ Suppose $P(x)$ is a polynomial with integer coefficients and $a$ is an integer with $\,P(a)\neq 0\,$ and there exists an integer $b$ such that neither of $\,b\!-\!a\pm 1$ divides $P(b).$

$$\begin{align} {\rm Then}\ \ &P(a)\,\ \text{has $\,\ k\,\ $ prime factors (counting multiplicity)}\\ \Longrightarrow\ \ &P(x)\, \text{ has $\le\! k\,$ integer roots (counting multiplicty)} \end{align}\qquad $$

Proof $ $ If not then $P$ has at least $\,k+1\,$ roots $\,r_i\,$ so iterating the Factor Theorem yields $$\,P(x) = (x-r_0)\cdots (x-r_k)\,q(x)\qquad$$

for a polynomial $\,q(x)\,$ with integer coefficients. Evaluating above at $\,x = a\,$ yields

$$\,P(a) = (a-r_0)\cdots (a-r_k)\,q(a)\qquad$$

If all $\,a-r_i\neq \pm1\,$ then they all have a prime factor yielding at least $k+1$ prime factors on the RHS, contra LHS $\,P(a)\,$ has $\,k\,$ prime factors (prime factorizations are unique). So some $\,a-r_j = \pm1\,$ so $\,r_j = a\pm 1.\,$ Evaluating at $\, x = b\,$ yields $\,b-r_j = b-a\pm1\,$ divides $\, P(b),\,$ contra hypothesis.

Bill Dubuque
  • 272,048