A necessary and sufficient on the co-efficients of a quadratic to give an integer Le $f(n) := an^2 + bn + c$ for all integers $n$, where $a$, $b$, $c$ are rational. What are the necessary and sufficient conditions on $a$, $b$, and $c$ such that $f(n)$ be an integer for all $n$? 

We are told that $an^2+bn+c$ is an integer forevery integer value of $n$. Set $n=0$. This tells us that $c$ must be an integer.
Since $c$ is an integer, $an^2+bn+c$ is an integer for all $n$ if and only if $an^2+bn$ is an integer for all $n$. 
Suppose that $an^2+bn$ is an integer for all $n$. Then in particular it is an integer for $n=1$ and for $n=-1$. Thus
$a+b$ is an integer, say $k$, and $a-b$ is an integer, say $l$. 
From $a+b=k$, $a-b=l$ we get $a=\frac{k+l}{2}$ and $b=\frac{k-l}{2}$. There are two possibilities (i) $k$ and $l$ have the same parity, that is they are both even or both odd and (ii) $k$ and $l$ have opposite parities.
In Case (i), the numbers $a$ and $b$ are integers.
In Case (ii), the numbers $a$ and $b$ are each half of an odd integer.
Finally, we check that (i) if $a$ and $b$ are integers, then $an^2 +bn$ is an integer for all integers $n$. This is obvious.
We also check that (ii) if $a$ and $b$ are each half of an odd integer, then $an^2+bn$ is an integer for all $n$.
So $a=\frac{c}{2}$ and $b=\frac{d}{2}$ where $c$ and $d$ are odd integers. 
We have then $an^2+bn=\frac{1}{2}n(cn+d)$. If $n$ is even, then certainly $\frac{1}{2}n(cn+d)$ is an integer. And if $n$ is odd, then $cn+d$ is even, so again $\frac{1}{2}n(cn+d)$ is an integer.
Conclusion: we have $an^2+bn+c$ is an integer for all integers $n$ if and only if $c$ is an integer and $a$ and $b$ are each integers, or each half of an odd integer.
