# Factorisation of Monic Polynomials

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#1

I think that being divisible by 9 different monic polynomials means I can write:

where there are supposed to be 9 brackets on the RHS, and a=b+c+...+j

Then I have no idea how to continue.
0
6 years ago
#2
(Original post by rayquaza17)
...
Not familiar with this, so can't really help, but I suspect a misunderstanding.

E.g. in 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.
0
6 years ago
#3
You can think of this in a similar way to the integers

If an integer is divisible by an integer then for some integer and so you've just found another integer that divides

That is to say, when you find integers that divide a number, you always find them in pairs... unless?
1
6 years ago
#4
(Original post by Noble.)
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:

(Note, spoilers for OP past a certain point)
0
6 years ago
#5
(Original post by lerjj)
This reminds me of this:

(Note, spoilers for OP past a certain point)
Yes, that's what made me give the hint using integers

(although it's an identical argument anyway)
0
6 years ago
#6
(Original post by Noble.)
You can think of this in a similar way to the integers

If an integer is divisible by an integer then for some integer and so you've just found another integer that divides

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.
0
6 years ago
#7
(Original post by bpbp)
Unless a = b and you get perfect squares.
Yeah
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#8
(Original post by ghostwalker)
Not familiar with this, so can't really help, but I suspect a misunderstanding.

E.g. in 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.
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!

(Original post by Noble.)
Yes, that's what made me give the hint using integers

(although it's an identical argument anyway)
We went through all of the maths behind the locker room problem (http://mathforum.org/alejandre/frisb...nt.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):

Spoiler:
Show

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

If so, do you have any more hints?
0
6 years ago
#9
(Original post by rayquaza17)
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/frisb...nt.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):

Spoiler:
Show

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

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

But there are only nine unique polynomials so for some

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

with 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).
0
#10
(Original post by Noble.)
Yes, it's not really any different for polynomials.

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

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

But there are only nine unique polynomials so for some

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

with 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.
0
6 years ago
#11
(Original post by rayquaza17)
Right I get it now, thank you.
I would never have come up with that in the exam though.
Yes, to be fair you anticipate the method being much more complicated and involved, I only jumped to thinking about it that way because I was familiar with the common brainteaser giving the problem on the integers.
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