Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9$ as we only care $\pmod 9.$
So $(x+a)(x+b)(x+c)\equiv 1,2,0\mod 9.$
Note that $$\frac{\prod_{x=0}^8 (x+a)}{(a+2)(a+5)(a+8)}\cdot  \frac{\prod_{x=0}^8(x+b)}{(b+2)(b+5)(b+8)}\cdot \frac{\prod_{x=0}^8(x+c)}{(c+2)(c+5)(c+8)}= [1\cdot 2\cdot 4\cdot 5\cdot 7\cdot 8]^3\equiv -1^3\equiv -1 $$
I am stuck. Solutions are appreciated.

The $0 \leq a,b,c<9$ step is fine, after all if $a,b,c$ satisfy the equation then so do $a \%9,b\%9$ and $c\%9$. (Where $\%9$ denotes the remainder upon division by $9$).

I don't think, however, that the step that realizes that $x^3+1$ can only be $0,1,2$ and reflects this on the RHS of the modulo, is a good step (even if it is correct), because this step loses the individuality of the term $x^3+1$. Note that the equations given by $$
(x+a)(x+b)(x+c) \equiv x^3+1 \pmod{9} \\
(x+a)(x+b)(x+c) \equiv 0,1,2 \pmod{9}
$$
are different equations, and it's likely that the second is significantly weaker than the first one, weak enough to actually admit solutions. I don't think the fact that the $x^3+1$ term depends on $x$ can be compromised, which is what that particular step was doing.

If we went with $(x+a)(x+b)(x+c) \equiv x^3+1 \pmod{9}$ for all $x$ (i.e. $x=0,1,...,8$) , then this is likelier to be helpful, particularly when you substitute values of $x$ and check for inconsistencies. Note that the above holds for all $x$ if and only if it holds for $x=0,1,2,...,8$.
The way we use the term $x^3+1$ , for example, is to find when it's zero.  For what values of $x$ is $x^3+1 \equiv 0 \pmod{9}$? You can check that $x=2,5,8$ are the solutions to this equation.
Therefore, $(2+a)(2+b)(2+c) \equiv 0 \pmod{9}$. In particular, either :

*

*At least one of the terms is a multiple of $9$, or :

*At least two of the terms are multiples of $3$.

This translates to : either one of $a,b,c = 7$ OR at least two of $a,b,c$ are among the values $1,4,7$.
Repeating the same arguments with $2$ replaced by $5,8$ , which are the other numbers that yield $x^3+1= 0$ would give us that (putting the $x=2$ conclusion in here as well) all of these conclusions are true :

*

*Either one of $a,b,c = 1$ ​OR at least two of $a,b,c$ are among the values $1,4,7$.

*Either one of $a,b,c = 4$ ​OR at least two of $a,b,c$ are among the values $1,4,7$.

*Either one of $a,b,c = 7$ ​OR at least two of $a,b,c$ are among the values $1,4,7$.

What can we conclude about $a,b,c$ from here? I claim that we can conclude that at least two of $a,b,c$ are among the values $1,4,7$. I'll hide the answer below if you can't deduce this from the given statements.

 If this were NOT the case, then the first part (coming before the OR) of each of the above statements must be true, but in that case we know that at least one of $a,b,c$ is $1$ and $4$ and $7$, so $a,b,c$ must be $1,4,7$ in some order, a contradiction because then the second part is satisfied! Thus, it is definitively true that at least two of $a,b,c$ are $1,4,7$ in some order.

We then substitute $x=0$ in the original equation to get that $abc \equiv 1 \pmod{9}$. Now, the symmetry of $a,b,c$ means that it's enough to show that each of the cases

*

*$a = 1,b=4$

*$a=1,b=7$

*$a=4,b=7$

*$a=1,b=1$

*$a=4,b=4$

*$a=7,b=7$
are impossible. But there is , in fact, something happening here which is very interesting.

You see, IF $a=1,b=4$, then $abc \equiv 1 \pmod{9}$ forces $c=7$, since this is the only value of $c$ which can satisfy this equation! Similarly, if $a=1,b=7$, then $c=4$ is forced, and similarly the third case.
For the fourth case, we are led to similar conclusions : indeed, each of $1,4,7$ satisfy $x^3 \equiv 1 \pmod{9}$, therefore IF any of the last three cases held, we must have $a=b=c = 1,4,7$.
Therefore, we conclude that if we consider the equations for just $x=0,2,5,8$, $a=1,b=4,c=7$ (or a symmetric change-up) and $a=b=c=1,4,7$ are the only possibilities for solutions to the initial equation . We've just used four values of $x$.
So if we can prove that the equation
$$
(x+1)(x+4)(x+7) \equiv x^3+1 \pmod{9}
$$
doesn't hold for some other $x$ which isn't $0,2,5,8$, then we are done with the first case. What better $x$ than $x=1$ to choose, where the RHS is $2$ and the LHS is $2 \times 5 \times 8 = 80$, and of course $80-2$ is not a multiple of $9$.
For the other cases, you can check that $(x+a)^3 = x^3+1 \pmod{9}$ is contradicted by $x=1$, for each of $a=1,4,7$ as well. In particular, the equation cannot hold for any value of $a,b,c$, for just the values $x=0,1,2,5,8$ alone. This proves the result.
