For $a$, $b$, $c$ distinct integers, and $P$ a polynomial with integer coefficients, $P(a)=b$, $P(b)=c$, $P(c)=a$ cannot be satisfied simultaneously

Question

I have pasted the problem and its solution below. My question is about (1), (2), and (3). What does $P_1(x)$, $P_2(x)$, and $P_3(x)$ represent? Near the end they say it's a positive integer. So couldn't we just put $k$ there instead, where $k$ represents some positive integer?

Thanks

---

Answer 1

Here $P_1$, $P_2$, and $P_3$ each represent a polynomial (with integer coefficients), and equations (1,2,3) are equations of polynomials in $x$ (not just single numbers). For instance, $P_1(x)$ is the polynomial you obtain by doing polynomial long division to divide $P(x)-b$ by $x-a$ (there will be no remainder since $P(a)=b$).

In particular, if you plug in any integer value for $x$, these polynomials will give integer outputs. That's why $P_1(c)$ is an integer.

Answer 2

From the given condition the polynomial $P(x)$ can be factorized as follow

• $P(a)-b=0 \implies P(x)-b=P_1(x)\cdot (x-a)$

where $P_1(x)$ is a polynomial of degree $n-1$ (assuming $P(x)$ is of degree $n$).

Since $P_1(c)\neq 0$ is an integer we have $|P_1(c)|\ge 1$.

Note that $P_1(c)\neq 0$ since

• $P(x)-b=P_1(x)\cdot (x-a)\implies P(c)-b=a-b=P_1(c)\cdot (c-a)\neq 0$