Let $a,b,c$ rational numbers.Define$f:\mathbb{Z}\mapsto \mathbb{Z}$ by $f(s)=as^{2}+bs+c$.Find the condition for $a,b,c$ such that $f$is well defined.

Question

Let $\mathbb{Z}$ be the set of all integers and $a,b,c$ rational numbers.Define$f:\mathbb{Z}\mapsto \mathbb{Z}$by $f(s)=as^{2}+bs+c$.Find necessary and sufficient conditions on $a,b,c,$ so that $f$ defines a mapping on $\mathbb{Z}$.

When I define $a,b,c$ are integers then the map is well defined. When I define $a=b=\frac{1}{2}$ then also $f(s)\in \mathbb{Z}$ . How to find the other possibilities? Please help me.

Answer 1

Substituting $s=0$ we see that $c$ has to be an integer. So, we may ignore $c$ henceforth. Taking $s=1$, we see that $$a+b\in\mathbb{Z},$$ and taking $s=2$, $$4a+2b\in\mathbb{Z}.$$ This should be enough to convince yourself that you have found all the solutions.

Answer 2

Trivially, $c$ must be an integer. Now, if $f(s)$ is integral for all $s$, then $g(s) = f(s+1)-f(s)$ must also be an integer for all $s$. But $g(s) = (a(s+1)^2+b(s+1)+c)-(as^2+bs+c)$ $= 2as+a+b$ must also be integral for all $s$. But this implies that $a+b$ must be an integer (set $s=0$) and then $2a$ must also be an integer (repeat the same process). So $a$ is either an integer or a half-integer, and $b$ must be $a$ plus an integer.

But then any polynomial $f(s)=as^2+bs+c$ with $2a, b-a, c\in\mathbb{Z}$ will be integral - this is easily proven by case analysis or by representing it in terms of binomials.