If $P$ is a polynomial with $P(3)=10$ and $P(1)=1$, then why can't all the coefficients of $P$ be integers?

Question

If $P$ is a polynomial with $P(3)=10$ and $P(1)=1$, then why can't all the coefficients of $P$ be integers?

---

This question was deleted for not enough details half a year ago, therefore I'm providing them. In this question specifically, I'm asking for your help to solve this problem in 8th grader way, because I'm sure that this question might help other students to understand polynomials much better, without any higher, university-level knowledge (some provided answers there are very elegant and understandable even for a 6th grader).

To mention more, right now it would be quite hypocritical to say that 'I tried < insert any theorem > but got stuck, hence I'm asking for your help'. So I'm not saying it now, instead of that just simply asking for you to undelete this question for the reasons I mentioned above.

Thank you!

Answer 1

If $p(x)$ is a polynomial with integer coeficients then for all integers $a,b$ you have $$a-b\mid p(a)-p(b)$$

In particular you have $$3-1\mid p(3)-p(1) = 9$$

A contradiction.

Answer 2

If the coefficients of $P$ are all integers, then $P(odd)=odd$ (that is, the value of $P$ at an odd integer is odd) if and only if an odd number of coefficients are odd. So if $P(1)=1$, there must be an odd number of odd coefficients, but if $P(3)=10$ there must be an even number of odd coefficients. This is a contradiction, so the coefficients cannot all be integers.

Answer 3

Suppose all coefficients of $P$ are integers. Then $Q(x) := P(x+1)-1$ is also a polynomial with integer coefficients (just expand and simplify). We have $$Q(0) = P(1)-1 = 0,$$ which means the zero-th degree coefficient of $Q$ is $0$, so $Q(x)/x$ is also a polynomial with integer coefficients. In particular this implies that $Q(2)/2$ must be an integer. But $$\frac{Q(2)}{2} = \frac{P(3)-1}{2} = \frac{10-1}{2} = \frac{9}{2}.$$ Contradiction.