I apologize ahead of time that my question will be hazy, but I'm very much under an impression that it has a solution, or possibly even a class of solutions. Here, goes...

Long, long time ago, in middle school I recall we were given an equation like so:


where $m$, $n$, and $c$ were specific integer numbers. I am pretty sure each one of them was below 10, and $c$ was most definitely $1$

The task was to factor it out into a parenthesized expression, aka a product of two or more parenthesized terms.

I had no clue how to do that. But later, someone showed me a trick that made it possible to try. The trick was to add and then subtract $1$, and after that, it was more clear how to group and parenthesize it. I remember questioning this trick or why it was needed but somehow it helped.

For example, if $m=2$, $n=3$ and $c=-3$:

$$x^2+x^3-3$$ $$=x^2+x^3-3+1-1$$ $$=x^2+x^3+1-4$$ $$=(x^3+1)+(x^2-4)$$ $$=(x+1)(x^2-x+1)+(x-2)(x+2)$$ $$=(x+1)(x^2-x+1)+(x-2)(x+1+1)$$ $$=(x+1)(x^2-x+1)+(x-2)(x+1)+(x-2)$$ $$=(x+1)(x^2-1)+(x-2)$$

etc.... I am not sure in this case it's possible to form a product of parenthesized expressions.

My question is

Does such a such factorization exist for some numbers $m, n, c$?

And, as a side-note, what is this trick with adding and subtracting one and why did that help?

In my memory the equation looked something like $x^7+x^3+1$, but I am fairly sure that was not the case, the numbers were likely different.

  • $\begingroup$ Of course, take $c=-2$ $\endgroup$ – Aqua Dec 27 '17 at 17:34
  • $\begingroup$ it's also easy if $c=0$ :) $\endgroup$ – Vasya Dec 27 '17 at 17:38
  • $\begingroup$ If you allow us to use complex numbers then this is always possible according to the fundamental theorem of algebra. $\endgroup$ – Bumblebee Dec 27 '17 at 17:53
  • $\begingroup$ Also factors with complex numbers of your particular example can be found by solving a cubic equation. But there is no general way to factor such expressions. $\endgroup$ – Bumblebee Dec 27 '17 at 18:00

For naturals $m$ and $n$ we obtain that $x^{3m-1}+x^{3n-2}+1$ is divisible by $x^2+x+1$

because $$x^{3m-1}+x^{3n-2}+1=x^{3m-1}-x^2+x^{3n-2}-x+x^2+x+1=$$ $$=\left((x^3)^{m-1}-1\right)x^2+\left((x^3)^{n-1}-1\right)x+x^2+x+1.$$ Here $$\mathbb N=\{1,2,3,...\}$$

  • $\begingroup$ I am curious, how or why did you decide to use the $x^2+x+1$ for divisibility? $\endgroup$ – Dennis Dec 29 '17 at 16:40
  • 1
    $\begingroup$ Because I saw that $x^3-1=(x-1)(x^2+x+1).$ We can use also $x^2-x+1$ for divisibility. $\endgroup$ – Michael Rozenberg Dec 29 '17 at 16:44

If $c=-2$ then $$x^m-1+x^n-1 = (x-1)(x^{m-1}+...x+1)+(x-1)(x^{n-1}+...x+1) = (x-1)[...]$$


Here's an example with $c$ positive:


(In addition, if $c=0$, one simply has $x^m+x^n=x^n(x^{n-m}+1)$ if $n>m$.)


First of all, factorization may or may not be possible according to the values of $m$, $n$ and $c$. For example if you have $x^2+x+2$, it has no real roots so cannot be factorized (actually it can be but by using complex values, which is not you are looking for I guess). But for example if you have $c=-2$ as stated in John Watson's answer, factorization may be possible.

Now, adding and subtracting a term, or decomposing the terms might help you because in that way, sometimes you get expressions like squares or which can be factorized so that you can bracket them with remaining terms if possible. I will give an example with square roots as well because sometimes you can factorize an expression with the help of square roots. If we have $x^2+x+1$, then by adding $x$ and subtracting $x$, we get $$x^2+x+1 = x^2+x+1+x-x = x^2+2x+1-x = (x+1)^2-(\sqrt x)^2 = (x+\sqrt x+1)(x-\sqrt x+1)$$


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