I have two related questions on the representability of integers 
by quadratic forms in two variables :


(1) Let $f: {\mathbb Z} \times {\mathbb Z} \to {\mathbb Z} $ be such a quadratic
form, i.e. we have $f(x,y)=ax^2+bxy+cy^2+dx+ey+g$ for some integer constants
$a,b,c,d,e,g$. Suppose that $f$ is not surjective, i.e. some integer is not represented
by $f$. Is it true that there is an integer constant $C$ such that in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?


(2) If the answer to (1) is yes, is there a uniform bound ? In other words,
is there a uniform constant $C$ such that for any non-surjective $f$, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?


**Update** : Good answers to my original questions appeared quickly. It seems the only
interesting subquestion left is the one asked by fedja : 
(2') is there a universal $C$ such that for any $f$ with positive definite quadratic part, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?


One may also ask, 
(3) is there a universal $C$ such that for any non-surjective and irreducible $f$, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?



representability of consecutive integers by a binary quadratic form
Such a $C$ does not exist, if I got it right.


My example is the function $f(x,y)=(2x+1)(5y+1)$. An integer is represented by $f$ iff it can be written as the product of an odd integer and a number which is 1 mod 5. So for example it is not hard to check that 2 cannot be written in this way (the odd number had better be $\pm1$, and neither of $\pm2$ are 1 mod 5). However a tedious construction using the Chinese Remainder Theorem gives arbitrarily large sequences of consecutive integers which are representable by $f$. 


Let me explain a bit more about the CRT argument. Clearly any positive integer $N$ can be written as the product $M2^r$ of an odd integer and a power of two. If furthermore this odd integer $M$ is divisible by three distinct primes, one of which is 2 mod 5, one of which is 3 mod 5 and one of which is 4 mod 5, then moving one of these factors $p$ from $M$ to the power of two gives $N=(M/p)(p2^r)$, and if $2^r$ isn't 1 mod 5 then there will be some $p$ such that $p2^r$ will be 1 mod 5, so we have our decomposition of $N$ which is hence representable by $f$.


So it suffices to find arbitrarily large strings of integers each of which has a prime factor which is 2 mod 5, a prime factor which is 3 mod 5 and a prime factor which is 4 mod 5 (in fact by using sign changes one can even get away with primes which are 2 mod 5). But now this is easy: enumerate the primes which are 2 mod 5, 3 mod 5 and 4 mod 5, let C\_n be the product of the n'th 2-mod-5, the n'th 3-mod-5 and the n'th 4-mod-5, and now choose a big positive $N$ with $N=0$ mod C\_1, $N+1=0$ mod $C\_2$,... $N+10^6=0$ mod $C\_{10^6+1}$ (such an $N$ exists by CRT) and there you have it.


