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Can every quadratic with integer coefficients be written as a sum of two polynomials with integer roots? (Any constant $k \in \mathbb{Z}$, including $0$, is also allowed as a term for simplicity's sake.)

(In other words, is any given $P(x) = A + Bx + Cx^2$ expressible as

$$P(x) = \color{red}{k(x-r_1)(x-r_2)\cdots(x-r_n)} + \color{blue}{\ell(x-s_1)(x-s_2)\cdots(x-s_m)}$$

where all variables other than $x$ are integers?) As an example of such a decomposition, if $C = 1$ then $P(x) = (A - Ax) + (Ax + Bx + x^2) = \color{red}{-A(x-1)} + \color{blue}{(x)(x+A+B)}$. The "two polynomials" restriction is essential; expressions like $P(x) = \color{red}{(A)} + \color{green}{(Bx)} + \color{blue}{(Cx^2)}$ don't count.

I've been contemplating this statement for a while and could use some help. I'm having trouble whether trying to prove it or find a (verifiable) counterexample. (Note that the components can have arbitrarily high degrees $n,m$ but cancel out to give $P(x)$.) Variations on completing the square didn't help.

If the answer is affirmative, I would also be interested in the following generalizations:

  • In addition to quadratics, can higher-order polynomials be decomposed into two polynomials?

  • (Refinement of the above if it is true) If two polynomials do not suffice for $P(x)$ of arbitrary degree, is there a finite number $N$ that does?

Thanks in advance for any ideas or help.

Note: I have used the colors I can most easily distinguish in the question, but if they cause other people difficulty please feel free to change them or remove them.

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This is resisting everything I've thrown at it. I believe it's possible to decompose any quadratic into two quadratics in this way, and Mathematica has verified that this can be done up to $\bmod \gcd(C,n)$ for $n\leq 30$, but I am far from a proof. –  Alex Becker Jul 6 '12 at 21:23
Not sure if it matters, but the decomposition isn't unique. For example, $P(x) = x^2 + 6 x + 7$ can be written as $ \color{red}{(x+2)(x+3)} + \color{blue}{(x+1)} $ or $ \color{red}{(x+1)(x+4)} + \color{blue}{(x+3)} $. –  csd Jul 16 '12 at 21:02
@csd: Yes, I am aware of this. The existence is what I'd like to know about. –  Vandermonde Jul 21 '12 at 0:40

1 Answer 1

If $a$ is a multiple of $b$, so that $a =mb$, you can decompose in this way:
or, if you prefer:
But if $a$ can be expressed as $a=mb-n^2c$, where $m$ and $n$ are integers, then we can decompose the polynomial in this way: $$a+bx+cx^2=c(x-n)(x+n)+b(x+m)$$
Let's see an example that cannot be included in the previous cases ($a=mb$ and $a=mb-n^2c$).

Let $P(x)=2+57x+31x^2$, and let's decompose $P(x)$ in this way: $$2+57x+31x^2=k(x-r_1)(x-r_2)+\ell x+D \quad (1)$$ where $D=\ell s_1$, i.e. $D$ is a multiple of $\ell$.
From equation $(1)$ we can conclude: $$k=31$$ $$\ell = 57+31(r_1+r_2) \quad(2)$$ $$D=2-3(r_1 r_2)\quad(3)$$
But $D$ is a multiple of $\ell$ ($D=\ell s_1$), therefore we can conclude from that fact and from equations $(2)$ and $(3)$ the following: $$2-57s_1=31[r_1r_2+s_1(r_1+r_2)] \quad(4)$$ From equation $(4)$ we conclude that $2-57s_1$ is a multiple of $31$. Thus if $n=r_1 r_2+s_1(r_1+r_2)$ we reached the following equation: $$2=31n+57s_1 \quad(5)$$ A solution of equation $(5)$ is $n=-22$ and $s_1=12$ (Using Euclidean Algorithm).

Now we can solve equation $(4)$.
A solution is $r_1=-10$ and $r_2=49$ (Using factoring).
And replacing the values of $r_1$ and $r_2$ in $(2)$ we get: $$\ell = 1266$$ Therefore $P(x)=2+57x+31x^2$ can be decomposed in this way: $$2+57x+31x^2=31(x+10)(x-49)+1266(x+12)$$ A conclusion that we can draw from that example is:

If $a$ is not a multiple of $gcd(b,c)$, then the polynomial cannot be decomposed in this way: $$a+bx+cx^2=k(x-r_1)(x-r_2)+\ell(x-s_1)$$ since there won't be a solution to the Diophantine equation $(5)$.

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$2+57x+31x^2=(2-3x+x^2)+(60x+30x^2)$. –  Gerry Myerson Jul 25 '12 at 12:40
It always seems possible to do any particular example, but a proof that it's always possible.... –  Gerry Myerson Jul 25 '12 at 12:59
@Gerry. See my conclusion from the example I gave and try to decompose $4x^2+6x+1$ in that way. –  RicardoCruz Jul 25 '12 at 13:14
$4x^2+6x+1=(5x^2+10x+5)+(-x^2-4x-4)=5(x+1)^2-(x+2)^2$. –  Gerry Myerson Jul 26 '12 at 2:03
@Gerry. As you may have seen it was impossible to decompose $4x^2+6x+1$ in a second degree polynomial and a first degree polynomial, you needed two second degree polynomials. –  RicardoCruz Jul 26 '12 at 17:55

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