# Solutions to $x^4 + ax^3 + bx^2 + cx \forall x$ mod $6$ where $0 < a, b, c < M$ [closed]

I want to be able to generate triplets $(a, b, c)$ for the polynomial $f(x) = x^4 + ax^3 + bx^2 + cx$ mod $6$ where $0 < a, b, c < M$. Trying values of $x$ up to $6$ I get these equations:

$a + b + c + 1 \equiv 0$ mod $6$

$4 + 2a + 4b + 2c \equiv 0$ mod $6$

$3 + 3a + 3b + 3c \equiv 0$ mod $6$

$4 + 4a + 4b + 4c \equiv 0$ mod $6$

$1 + 5a + b + 5c \equiv 0$ mod $6$

I can see that the obvious solutions satisfy $b + 1 \equiv 0$ mod $6$ and $a + c \equiv 0$ mod $6$, but what about the not-so-obvious ones?

Any help is appreciated. My background is not in this. Picked up a book on Number Theory two days ago. So far it's an interesting read, but I'm far from being able to do this quickly on my own.

Edit: I'm looking for solutions $\forall x$. Inductively I think it's enough to find $(a, b, c)$ such that $a + b + c + 1 \equiv 0$ mod $6$ and $f(x+1) - f(x) \equiv 0$ mod $6$ or $(4n^3 + 6n^2 + 4n + 1) + a(3n^2 + 3n + 1) + b(2n + 1) + c \equiv 0$ mod $6$. Trying values of $x < M$ gives these:

$3 + a + 3b + c \equiv 0$ mod $6$

$5 + a + 5b + c \equiv 0$ mod $6$

$1 + a + b + c \equiv 0$ mod $6$

$3 + a + 3b + c \equiv 0$ mod $6$

$5 + a + 5b + c \equiv 0$ mod $6$

Not sure where to take this from here...

-

## closed as not a real question by Will Jagy, robjohn♦Nov 25 '12 at 18:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

It's a little unclear what you mean by "solution triplets". Do you mean that there is some x? Some nonzero x? – Nathan Grigg Nov 22 '12 at 3:05
I think he is looking for triplets $(a,b,c)$ such that $x^4+ax^3+bx^2+cx\equiv 0$ for all $x$. – Jeff Tolliver Nov 22 '12 at 3:37

If $f_{a,b,c}(x)=x^4+ax^3+bx^2+cx$, then the number of triples $(a,b,c)$ with $a,b,c \in \{0,1,\ldots,5\}$ such that $f_{a,b,c}(x) \equiv y \pmod 6$ is given in the table below.

$$\begin{array}{r|rrrrrr} & y=0 & 1 & 2 & 3 & 4 & 5 \\ \hline x=0 & 216 & 0 & 0 & 0 & 0 & 0 \\ 1 & 36 & 36 & 36 & 36 & 36 & 36 \\ 2 & 72 & 0 & 72 & 0 & 72 & 0 \\ 3 & 108 & 0 & 0 & 108 & 0 & 0 \\ 4 & 72 & 0 & 72 & 0 & 72 & 0 \\ 5 & 36 & 36 & 36 & 36 & 36 & 36 \\ \hline \end{array}$$

I computed this using GAP, where the pattern is made clear. We find $$6^2 \gcd(6,x)$$ triples $(a,b,c)$ satisfy $f_{a,b,c}(x) \equiv y \pmod 6$ provided $\gcd(6,x)$ divides $y$, and $0$ otherwise.

Why? Essentially, the choices for $a$ and $b$ don't matter. Once they are chosen, there's an equal number of choices for $c$ such that $f_{a,b,c}(x)$ belongs in one of the $6/\gcd(6,x)$ residue classes. So there are $6/(6/\gcd(6,x))=\gcd(6,x)$ ways to choose $c$ such that $f_{a,b,c}(x)$ ends in any one of the possible residue classes.

-
Thank you. I was actually looking for a way to find $(a, b, c)$ where $y = 0 \forall{x}$. – McTrafik Nov 22 '12 at 16:50

From the way you proceeded, I'm assuming you're looking to find the restrictions on $a,b,c$ so that $f(x)=0$ mod 6 for all integers $x$, and that after that you wish to count the number of such triples where $0<a,b,c<M$. The latter part may be difficult.

But the approach so far does give the necessary and sufficient equations. now note that the terms $x=a+c$ and $y=b+1$ each appear only together in the five equations. So there can be no hope of further relations on them. That is, your equations in terms of $x,y$ are

[1] $x+y=0$

[2] $2x+4y=0$

[3] $3x+3y=0$

[4] $4x+4y=0$

[5] $5x+y=0$

Note that equations [3], [4] are consequences of [1], so the system is equivalent to the subsystem consisting of [1] [2] and [5].

If we subtract [1] from [5] we obtain $4x=0$, which itself doesn't imply $x=0$ mod 6, because 4 is not invertible. But 4 is invertible mod 3 so from $4x=0$ you can say that $x=0$ mod 3. Similarly we can subtract [5] from 5*[1] and get $4y=0$, so that $y=0$ mod 3.

Looked at mod 2, equation [2] says nothing, while [1] and [5] only say that $x,y$ have the same parity.

EDIT: in fact you cannot get zero mod 6, since putting $x=y=3$ makes all six equations true mod 6.

I think from the above all you can say is that $x,y$ have the same parity, and are zero mod 3. It's getting late; I think this means $x,y$ are either $0,0$ or $3,3$ mod 6.

In any case it is clear from the above that there can be no other relations than those involving the sums $a+c$ and $b+1$ which you refer to.

-