Factorisation of Monic Polynomials

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You can think of this in a similar way to the integers


That is to say, when you find integers that divide a number, you always find them in pairs... unless?

This reminds me of this:
[video="youtube;LejoPGtliTs"]https://www.youtube.com/watch?v=LejoPGtliTs[/video]

(Note, spoilers for OP past a certain point)

You can think of this in a similar way to the integers

If an integer $x$ is divisible by an integer $a$ then $x = ab$ for some integer $b$ and so you've just found another integer that divides $x$

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

Unless a = b and you get perfect squares.

Not familiar with this, so can't really help, but I suspect a misunderstanding.

E.g. in $\mathbb{Z}[X], x^4-1$ can be written as the product of 3 factors (2 linear, one quadratic), but it is actually divisible by 6 (7 if you include itself) different monic polynomials. If the quadratic/linear factors are a,b,c, then the polynomials factors are a,b,c,ab,bc,ac,abc, and are all different.

Apologies if this is of no use, or misleading. PM me if you wish me to delete it.

Yes, that's what made me give the hint using integers

(although it's an identical argument anyway)

Yeah that was helpful. I think it I was told r(x) had 9 roots then it would be okay to put what I said, I think I just got myself a bit confused!




We went through all of the maths behind the locker room problem (http://mathforum.org/alejandre/frisbie/student.locker.html) and I think that's the same as the video.

I can prove that if the number of divisors is odd, then the number itself is a square number (using unique prime factorisation):


So should I base this question on the proof for that?

If so, do you have any more hints?

Yes, it's not really any different for polynomials.

Look at the first eight unique monic polynomials that divide r(x), you must have

$r(x) = a_1(x)a_2(x)$
$r(x) = a_3(x)a_4(x)$
$r(x) = a_5(x)a_6(x)$
$r(x) = a_7(x)a_8(x)$

Now there's also a ninth monic polynomials that is different from the first eight, so

$r(x) = a_9(x)a_{10}(x)$

But there are only nine unique polynomials so $a_{10} = a_i$ for some $i \in \{1,2,\ldots,9\}$

You can easily show that i cannot take values 1 to 8 (inclusive) because otherwise you'd have (WLOG)

$r(x) = a_1(x) a_2(x) = a_9(x) a_2(x)$ with $a_1, a_9$ different, which clearly doesn't work (this is essentially saying, in the integers, if you divide a non-zero integer by two different integers you can get the same value).

Right I get it now, thank you.
I would never have come up with that in the exam though.