Showing a polynomial having at least one integer root under certain conditions has precisely one integer root $P(x) = 0$ 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.
 A: 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. 
A: 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 $(2+1)^2=9$.
A: 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. 
