Thread: Integer coefficients

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 1.

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

3. 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 .

For Q 1. This is my working so far:

So assume that can be factorised into 2 polynomials with integer coefficients.



Now .

Assume or and

Now the 2 factors can not be linear. Why?

When is linear then

But or but none of these equal when subbed into . Thus there can be no linear factors.

Now for all the coefficient of the term in must be .

Thus

Take the first case and





This means and are multiples of .

Doing some experimenting:

When , that means the coefficient of is also .

Thus

Thus

Since and are both multiples of we can factor out a leaving

Which means is also a multiple of .

Now a few things can be noticed:

1. The subscript of a and b both add to up the power of the term we need to the find the coefficient of.

2. Continuing this we find that are also multiples of .

This means if we can prove that all of the coefficients of are multiples of then we can use contradiction to show that can not exist as the coefficient of is which is not a multiple of .

Say we want to find the coefficient of term in where . (I'm restricting the domain only up to because we just require to prove the coefficients of are multiples of .)

Now consider the polynomial where (In other words, the lowest degree of is m and it must be strictly smaller than or else would be a linear factor)

Using the pattern we saw earlier in the experimenting to find an equation for the coefficient of yields:



Now we can use strong induction to prove is a multiple of .

First Case: When

Base Case

when



We have already proved earlier using this that is a multiple of .

Inductive Hypothesis

Assume all are multiples of .

Proof

Since are multiples of then the equation can be rewritten into:

(Factoring out a , while calling the integer sum inside the bracket )

But





Therefore by induction, is also a multiple of .

Second Case: When .

Why consider this case? Let's pick a random value of , say and .

Now say we wanted to find the coefficient of where .

Then the equation would be:

However does not exist as the highest power of is so there is no such things as the coefficient of for .

This causes a problem, so we need to change our general equation a bit.

Now what?

If p(x) has an integral root, then

p(x) = (x-r)q(x)

with r integral and q another polynomial with integer coefficients. In particular, q(0) and q(1) must be an integer.

Note 1999 is prime.
1999 = p(0) = -rq(0) ==> r = 1 or -1 or 1999 or -1999
1999 = p(1) = (1-r)q(1), but none of the above possibilities for r work.