Write $ x^2 + Ax + B = 2y^2 + 2y + C \Leftrightarrow$ $ 4x^2 + 4Ax + B = (2x + A)^2 + (B - A^2) = 2(2y + 1)^2 - 2 + 4C \Leftrightarrow$ $ (2x + A)^2 - 2(2y + 1)^2 = 2(2C - 1) + A^2 - B$. Lemma: No integer of the form $ a^2 - 2b^2$ is congruent to $ 3 \bmod 9$. Proof: Suppose that $ a^2 - 2b^2 \equiv 3 \bmod 9$. Clearly $ a, b$ are not both divisible by $ 3$. If one of $ a, b$ was divisible by $ 3$, then the other would also have to be, so neither is. Then $ a^2 \equiv 2b^2 \bmod 3 \Leftrightarrow \left( \frac {a}{b} \right)^2 \equiv 2 \bmod 3$, but $ 2$ is not a quadratic residue $ \bmod 3$; contradiction. Since $ C$ allows us to increase or decrease the value of the RHS by $ 4$ and $ 4$ is invertible $ \bmod 9$, it follows that for any $ A, B$ we can find $ C$ so that $ 2(2C - 1) + A^2 - B \equiv 3 \bmod 9$.

another solution: note that $ 2x^2 + 2x$ is divisible by $ 4$ for all $ x$. let: $ N_1=$ the set $ M_1$ mod 4. therefore: $ N_1 = \{B,A + B + 1,2A + B,3A + B + 1\} \mod 4$ if $ A$ is odd,then all elements of $ N_1$ have the same parity,therefore $ N_1$ is not complete residue system mod 4. and if $ A$ is even then obviously $ B\equiv2A + B \mod 4$ therefore $ N_1$ is not complete residue system mod 4. since $ N_1$ is not complete residue system mod 4 and $ 4|2x^2 + 2x$ obviously there exist $ C$ such that $ M_1$ and $ M_2$ do not intersect.

orl wrote: Let $ \mathbb{Z}$ denote the set of all integers. Prove that for any integers $ A$ and $ B,$ one can find an integer $ C$ for which $ M_1 = \{x^2 + Ax + B : x \in \mathbb{Z}\}$ and $ M_2 = {2x^2 + 2x + C : x \in \mathbb{Z}}$ do not intersect. I think this works:

The QuattoMaster 6000 wrote: orl wrote: Let $ \mathbb{Z}$ denote the set of all integers. Prove that for any integers $ A$ and $ B,$ one can find an integer $ C$ for which $ M_1 = \{x^2 + Ax + B : x \in \mathbb{Z}\}$ and $ M_2 = {2x^2 + 2x + C : x \in \mathbb{Z}}$ do not intersect. I think this works:

I think you must consider $ x^2+Ax+B=2y^2+2y+C$

Why is every solution keeping b? Can't we just assume b is 0 since we could adjust c to make up for it?

The problem states that for any $ A,B$..., i.e., given fixed $ A,B$...

Right, but even if b is fixed we could choose an appropriate c to make up for it. For example, if when b=0 we would choose c=3, then when b=4 we would choose c=7, right?

ANd why should this specific $ C$ work¿

dgreenb801 is saying that $ \{ x^2 + Ax + B \}$ and $ \{ 2x^2 + 2x + C \}$ intersect if and only if $ \{ x^2 + Ax \}$ and $ \{ 2x^2 + 2x + (C - B) \}$ intersect, which is true but does not significantly shorten the solution (as far as I can see).

If $A$ is odd, pick $C = B - 1$. If $x^2 + Ax + B = 2y^2 + 2y + C$ has a solution, then $4y^2 + 4y + 2C = 2x^2 + 2Ax + 2B$, so $(2y+1)^2 - 2x(x+A) = 2B-2C+1 = 3$. $(2y+1)^2 \equiv 1 \pmod{4}$. If $x$ is even, $4 | 2x$, and if $x$ is odd, $2 | x+A$, so $4 | 2x(x+A)$. In either case, we get that $3 = (2y+1)^2 - 2x(x+A) \equiv 1 \pmod{4}$, a contradiction. If $A$ is odd, let $A = 2a$, and let $C = B + a - 1$. Rearrange $x^2 + Ax + B = 2y^2 + 2y + C$ to get $2(2y+1)^2 - ((2x+A)^2 - A^2) = 4(B-C) + 2$, which simplifies to $(2y+1)^2 - 2(x+a)^2 = 2a + 1 + 2(B-C) = 2 - 2a + 2a + 1 = 3$. $(2y+1)^2 \equiv 1 \pmod{8}$, so $-2(x+a)^2 \equiv 2 \pmod{8}$, so $(x+a)^2 \equiv 3 \pmod{4}$, which is a contradiction.

Hey, this is rather trivial... all you need to notice is that $x(x+A)+B$ can't form a CRS(complete residue system) modulo $4$ (just check for different values of $x \pmod{4}$ to get $B,1+A+B, 2A+B, 1-A+B$ and now check for different values of $A \pmod{4}$) and $2x(x+1) \equiv 0 \pmod{4} => 2x^2+2x+C \equiv C \pmod{4}$, hence there's always a way to choose the required $C$