TT's Maths Thread

[Ahmad]

You can use polynomial modular arithmetic. We're trying to find . Now and upon multiplying by gives . So that .

[TrueTears]

Thank you Ahmad!!!

[TrueTears]

Let and be the roots of . Prove that and are the roots of

[Ahmad]

We can call a multivariable polynomial symmetric if where is a permutation of 1..n. Symmetric polynomials can always be written as a polynomial in the roots of a polynomial. (If this confuses you just ignore it completely).

[Ahmad]

As an exercise solve the problem using matrices. (Hint: eigenvalues of a matrix A satisfy , cube of the eigenvalues satisfy )

[Ahmad]

As a third way, similar to the first but requires less algebraic ingenuity (especially for similar, but more difficult problems involving higher powers):

To work out it's not obvious what factorisation one should use.

However we know that r and s satisfy multiply this by x giving , now this must be satisfied by r and s, plugging in each r and s:




Adding these together yields

So we've reduced expressing in terms of r+s and rs to expressing . You can plug in each of r and s into to yield an expression for in terms of r+s, completing the problem.

As a fourth way, have a look at the Newton-Girard formulas which has an interesting proof using generating functions (which as I recall is shown on planetmath). I tend to stay clear of this path though, simply because I never bothered to remember the formulas, and don't like looking them up and for small powers there are reasonably fast other ways.  

[TrueTears]

Thanks for the help Ahmad!

[TrueTears]

1. Let where is an integer. Prove that cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least .

2. Let be a polynomial with integral coefficients. Suppose that there exist four distinct integers with . Prove that there is no integer with .

My working:

Using division algorithm

Where denotes the quotient and denotes the remainder.

Thus

We find out that:









Now what...?

 

[zzdfa]

hint for the 2nd one:
consider the polynomial P(x)-5. we know that this has zeros at a,b,c,d. so we can factorize it, P(x)-5=(x-a)(x-b)(x-c)(x-d)Q(x) where Q(x) is some polynomial. now show that P(x)-5 cannot equal 3.

and for the first one, write it as
(x^p + ..... + c) * (x^q+ ...... +d )      where the mononomials are arranged in descending powers

if p,q,c,d are all integers, what must c and d be?
what is the relationship between p and q? edit: actually i didn't really think that through, might not work out.

[kamil9876]

Rough idea for q1:

I think we must invoke some property of the integers:

say say our polynomial is:



Now we know that

so let's assume WLOG that

we know that for n>2, the x^1 term must have coefficent of 0, and this coefficent is gotten by
and .

Hence we have:






so in our first polynomial, our first two terms must be multiples of 3, perhaps this argument can be continued to show that all the terms are multiples of 3. But i am not sure I haven't tried.

[TrueTears]

hint for the 2nd one:
consider the polynomial P(x)-5. we know that this has zeros at a,b,c,d. so we can factorize it, P(x)-5=(x-a)(x-b)(x-c)(x-d)Q(x) where Q(x) is some polynomial. now show that P(x)-5 cannot equal 3.

and for the first one, write it as
(x^p + ..... + c) * (x^q+ ...... +d )      where the mononomials are arranged in descending powers

if p,q,c,d are all integers, what must c and d be?
what is the relationship between p and q? edit: actually i didn't really think that through, might not work out.

Thanks!

Right I think I got Q 2

2. Let be a polynomial with integral coefficients. Suppose that there exist four distinct integers with . Prove that there is no integer with .

Using the division algorithm we see that

Let

.

has 4 distinct integer roots, .

Using the fundamental theorem of Algebra, we get:

where is some other polynomial so it "bumps up" the degree of

Now using contradiction to prove that there is no integer with

Assume







Now since are distinct integer roots, at least of MUST equal to or else the product will go above .

Ofcourse there could be other cases such as and and and , but they all lead to the same result.

So if at least 3 of equals 1.

Then WLOG assume and

Thus

.

But the question says the are 4 distinct integers.

by contradiction, there exists no k such that

[TrueTears]

Rough idea for q1:

I think we must invoke some property of the integers:

say say our polynomial is:



Now we know that

so let's assume WLOG that

we know that for n>2, the x^1 term must have coefficent of 0, and this coefficent is gotten by
and .

Hence we have:






so in our first polynomial, our first two terms must be multiples of 3, perhaps this argument can be continued to show that all the terms are multiples of 3. But i am not sure I haven't tried.

Thanks kamil, that gives me some new ideas!


Some new questions:

1. Let be a polynomial with integer coefficients satisfying . Show that has no integer zeros.

2. Let be a polynomial with real coefficients. Show that there exists a nonzero polynomial with real coefficients such that has terms that are all of a degree divisible by .

[kamil9876]

Rough idea for q1:

I think we must invoke some property of the integers:

say say our polynomial is:



Now we know that

so let's assume WLOG that

we know that for n>2, the x^1 term must have coefficent of 0, and this coefficent is gotten by
and .

Hence we have:






so in our first polynomial, our first two terms must be multiples of 3, perhaps this argument can be continued to show that all the terms are multiples of 3. But i am not sure I haven't tried.

expanding on this idea, I have found that the proof requires testing a few cases individually so it gets messy, each case is alright to deal with, just keeping track that you have taken everything into account is annoying :/.

Ok so you can prove that the polynomials must have degree greater than 1, since otherwise we would get a linear term and we know this is impossible since

Ok so first case assume, in other words the first polynomial(with constant ) has lesser or equal degree.

now recall our proof that and are multiples of , we will know prove by strong induction that all the are multiples of .

the coefficient of the x^m term in the product (where m<p<n-1) is:


however by induction, is a multiple of for all , thus the equation above tells us that is a multiple of , hence a_m is a multiple of 3 completing the induction. This can be used for m=p to show that is a multiple of , a contradiction thus this proves the question for this case.

For the other case, p>q, it is more difficult to show that all the are multiples of and requires a slightly more complicated variation of the trick we used above.

[TrueTears]

Thanks kamil, however just looking back at your previous post, the question says "...the product of two polynomials, each of which has all its coefficients integers and degree at least 1." so that means all the powers and coefficients of the 2 polynomials are all greater or equal to 1, ie all positive.

So then how can ever equal to 0?

Since

[Over9000]

Thanks kamil, however just looking back at your previous post, the question says "...the product of two polynomials, each of which has all its coefficients integers and degree at least 1." so that means all the powers and coefficients of the 2 polynomials are all greater or equal to 1, ie all positive.

So then how can ever equal to 0?

Since

"\left(a_0, b_1, a_1, b_0\right) \ge 1"
I think they just mean the degree of the polynomials is greater than 1 not the coefficients as well.