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$P(x) = 0$ 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.

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 $(2+1)^2=9$.

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 $uin{1,3}$. 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.

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 $,10!-!1mid P(10)!-!P(1) = 11.,$ $P(3)!neq! 0,$ 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 $,b!-!apm 1$ 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 $,a-r_ineq pm1,$ 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 $,r_j = apm 1.,$ Evaluating at $, x = b,$ yields $,b-r_j = b-apm1,$ divides $, P(b),,$ contra hypothesis.