# Thread: Integer coefficients

1. ## 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 .

2. For Q 1. This is my working so far:

So assume that $f(x)$ can be factorised into 2 polynomials with integer coefficients.

$f(x) = p(x)q(x) = (x^p + a_{p-1}x^{p-1} + a_{p-2}x^{p-2} +...+a_2x^2 + a_1x+a_0)(x^q + b_{q-1}x^{q-1} + b_{q-2}x^{q-2} +...+b_2x^2 + b_1x+b_0)$

Now $a_0b_0 = 3$.

Assume $a_0 = \pm 3 \implies b_0 = \pm 1$ or $a_0 = \pm 1$ and $b_0 = \pm 3$

Now the 2 factors can not be linear. Why?

When $p(x)$ is linear then $(x+a_0)q(x) = f(x)$

But $a_0 = \pm 3$ or $\pm 1$ but none of these equal $0$ when subbed into $f(x)$. Thus there can be no linear factors.

Now for all $n>2$ the coefficient of the $x$ term in $f(x)$ must be $0$.

Thus $a_1b_0 + a_0b_1 = 0$

Take the first case $a_0 = \pm 3$ and $b_0 = \pm 1$

$(\pm 1)a_1 + (\pm 3)b_1 = 0$

$\pm a_1 = \mp 3b_1$

This means $a_1$ and $a_0$ are multiples of $3$.

Doing some experimenting:

When $n = 4$, that means the coefficient of $x^2$ is also $0$.

Thus $a_2b_0 + a_1b_1 + a_0b_2 = 0$

Thus $a_2 = -(a_1b_1 + a_0b_2)$

Since $a_1$ and $a_0$ are both multiples of $3$ we can factor out a $3$ leaving $a_2 = -3(\mbox{some positive integer})$

Which means $a_2$ is also a multiple of $3$.

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 $a_3, a_4...$ are also multiples of $3$.

This means if we can prove that all of the coefficients of $p(x)$ are multiples of $3$ then we can use contradiction to show that $p(x)$ can not exist as the coefficient of $x^p$ is $1$ which is not a multiple of $3$.

Say we want to find the coefficient of $x^m$ term in $f(x)$ where $0 < m \le p$. (I'm restricting the domain only up to $p$ because we just require to prove the coefficients of $p(x)$ are multiples of $3$.)

Now consider the polynomial $p(x)$ where $m \le p < n-1$ (In other words, the lowest degree of $p$ is m and it must be strictly smaller than $n-1$ or else $q(x)$ would be a linear factor)

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

$a_0b_m + a_1b_{m-1} + a_2b_{m-2} + ... + a_{m-1}b_1 + a_mb_0 = 0$

Now we can use strong induction to prove $a_m$ is a multiple of $3$.

First Case: When $p \le q$

Base Case

when $m = 1$

$a_0b_1 + a_1b_0 = 0$

We have already proved earlier using this that $a_1$ is a multiple of $3$.

Inductive Hypothesis

Assume all $a_0, a_1, a_2 ... a_{m-1}$ are multiples of $3$.

Proof

Since $a_0, a_1, a_2 ... a_{m-1}$ are multiples of $3$ then the equation can be rewritten into:

$3(S) + a_mb_0 = 0$ (Factoring out a $3$, while calling the integer sum inside the bracket $S$)

But $b_0 = \pm 1$

$\therefore 3S \pm a_m = 0$

$3S = \mp a_m$

Therefore by induction, $a_m$ is also a multiple of $3$.

Second Case: When $p>q$.

Why consider this case? Let's pick a random value of $p$, say $p = 6$ and $q = 3$.

Now say we wanted to find the coefficient of $x^m$ where $m = 4$.

Then the equation would be: $a_0b_4 + a_1b_3 + a_2b_2 + a_3b_1 + a_4b_0 = 0$

However $b_4$ does not exist as the highest power of $q$ is $3$ so there is no such things as the coefficient of $x^4$ for $q(x)$.

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

Now what?

3. ## Question 2

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.