rayquaza17
Badges: 17
Rep:
?
#1
Report Thread starter 6 years ago
#1
Image

I think that being divisible by 9 different monic polynomials means I can write:
r(x)=x^a+r_1x^{a-1}+...=(x^b+...)(x^c+...)...(x^j+...)
where there are supposed to be 9 brackets on the RHS, and a=b+c+...+j

Then I have no idea how to continue.
Attached files
0
reply
ghostwalker
  • Study Helper
Badges: 17
#2
Report 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 \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.
0
reply
Noble.
Badges: 17
Rep:
?
#3
Report 6 years ago
#3
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?
1
reply
lerjj
Badges: 13
Rep:
?
#4
Report 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
reply
Noble.
Badges: 17
Rep:
?
#5
Report 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
reply
bpbp
Badges: 0
Rep:
?
#6
Report 6 years ago
#6
(Original post by Noble.)
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.
0
reply
Noble.
Badges: 17
Rep:
?
#7
Report 6 years ago
#7
(Original post by bpbp)
Unless a = b and you get perfect squares.
Yeah
0
reply
rayquaza17
Badges: 17
Rep:
?
#8
Report Thread starter 6 years ago
#8
(Original post by ghostwalker)
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.
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

Image

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

If so, do you have any more hints?
Attached files
0
reply
Noble.
Badges: 17
Rep:
?
#9
Report 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

Image

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).
0
reply
rayquaza17
Badges: 17
Rep:
?
#10
Report Thread starter 6 years ago
#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

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. :eek:
0
reply
Noble.
Badges: 17
Rep:
?
#11
Report 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. :eek:
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.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (13)
7.69%
Run extra compulsory lessons or workshops (28)
16.57%
Focus on making the normal lesson time with them as high quality as possible (27)
15.98%
Focus on making the normal learning resources as high quality/accessible as possible (24)
14.2%
Provide extra optional activities, lessons and/or workshops (45)
26.63%
Assess students, decide who needs extra support and focus on these students (32)
18.93%

Watched Threads

View All