Question

Let `P(x)=x^n+5x^(n-1)+3` where `n > 1` is an integer. Prove that `P(x)` cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least `1`?

Answer 1

See explanation...

Explanation:

`P(x) = x^n+5x^(n-1)+3`

Suppose `P(x) = Q(x)R(x)` where:

`Q(x) = a_hx^h+a_(h-1)x^(h-1)+...+a_1x+a_0`

`R(x) = b_kx^k+b_(k-1)x^(k-1)+...+b_1x+b_0`

`h + k = n`

`a_h = b_k = 1`

`a_0 = 3`

`b_0 = 1`

Note that the signs on the constants must be positive, since otherwise `Q(x)` and `R(x)` would have real positive zeros, which `P(x)` does not.

The coefficient of `x^m` in `Q(x)R(x)` is:

`a_mb_0+a_(m-1)b_1+...+a_1b_(m-1)+a_0b_m`

Now let `m` be the least integer such that `a_m` is not a multiple of `3`.

Then modulo `3` the coefficient of `x^m` in `Q(x)R(x)` is:

`a_mb_0 = a_m`

since all of the other terms are multiples of `3`.

Note however that the smallest non-zero coefficient of `P(x)` beyond the constant is that of `x^(n-1)`.

Hence `m >= n-1`, so `Q(x)` is of degree at least `n-1` and `R(x)` is of degree at most `1`.

So either `R(x) = x+1`, which would make `-1` a zero of `P(x)` - which it is not, or `R(x) = 1`, meaning that `P(x)` has no non-trivial polynomial factors with integer coefficients.

Answer 2

See below.

Explanation:

We will show a general procedure but applied to a case study for `n = 9`

Now considering

`P_n(x) = x^n +5x^(n-1)+3 = Q_u(x)R_v(x)`

with

`P_n(x) = sum_(k=0)^n a_k x^k`
`Q_u(x) = sum_(k=0)^u b_k x^k`
`R_v(x) = sum_(k=0)^v c_k x^k`

then

`a_k = sum_(i=0)^k b_(k-i)c_i` (convolution of coefficients)

or exemplifying for `n = 9, u = 5, v = 4`

`{(), (a_0=b_0c_0=3), (a_1=b_1c_0+b_0c_1=0), (a_2=b_2c_0+b_1c_1+b_0c_2=0),(a_3=b_3c_0+b_2c_1+b_1c_2+b_0c_3=0), (cdots), (a_7=b_5c_2+b_4c_3+b_3c_4 = 0), (a_8=b_5c_3+b_4c_4=5),(a_9=b_5 c_5 = 1):}`

Now assuming `b_0 = 3, c_0 = 1` (the other possibility is `b_0 = 1, c_0 = 3`) also `b_5=c_4=1` and `{b_k, c_k} in ZZ`

As we can observe, from the first equation, as a consequence in the second equation

`b_1c_0 equiv b_0 c_1 equiv 0 mod 3 rArr b_1c_1 equiv 0 mod 3` having `b_(k-i)c_i equiv 0 mod 3`

This behavior propagates until

`b_5c_3+b_4c_4=5`

with `b_5b_4 equiv c_3 c_4 equiv 0 mod 3` from the previous equation

As a consequence `b_5c_3+b_4c_4=5` cannot be attained in integers.