# 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...

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

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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.

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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.

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