parmenides51 wrote: (a) Let $a_0, a_1,a_2$ be real numbers and consider the polynomial $P(x) = a_0 + a_1x + a_2x^2$ . Assume that $P(-1), P(0)$ and $P(1)$ are integers. Prove that $P(n)$ is an integer for all integers $n$. (b) Let $a_0,a_1, a_2, a_3$ be real numbers and consider the polynomial $Q(x) = a0 + a_1x + a_2x^2 + a_3x^3 $. Assume that there exists an integer $i$ such that $Q(i),Q(i+1),Q(i+2)$ and $Q(i+3)$ are integers. Prove that $Q(n)$ is an integer for all integers $n$. for $a$ , trivial i guess you can see that ${2^n}a_1$ and ${2^n}a_2$ are integers ,and also $ a_0 , a_1 +a_2 \in \mathbb Z$ hence the conclusion , does it make any sense?

a) By shifting one can replace $-1,0,1$ by $0,1,2$ which makes things easier to write down. We have $P=a{x\choose 0}+b{x\choose 1}+c{x\choose 2}$ where $a,b,c$ are suitable constants. Then $a=P(0)\in {\bf Z},a+b=P(1)\in {\bf Z}$ hence $b\in {\bf Z}$ and $a+2b+c=P(2)\in {\bf Z}$ hence $c\in {\bf Z}$. It follows that $a,b,c\in {\bf Z}$ and $P( {\bf Z})\subset {\bf Z}$. b) can be done in the same way.

If we have (a) we have (b). Take a look The theorem is also trivial for polynomial of degree $0$ or $1$. Now define $G(x)=Q(x+1)-Q(x)$. It's a polynomial of degree at most $2$. Because $G(i),G(i+1),G(i+2)\in Z$ we have $G(x)\in Z$ for all $x\in Z$. Hence by induction $\forall n\in Z \ Q(n)\in Z$