Argthtjtr

$P(x) = 0$ 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.