After several operations of differentiation and multiplying by (x + 1) performed in an arbitrary order the polynomial x^8 + x^7 is changed to ax + b. Prove that the difference between the integers a and b is always divisible by 49.
After several operations of differentiation and multiplying by (x + 1) performed in an arbitrary order the polynomial x^8 + x^7 is changed to ax + b. Prove that the difference between the integers a and b is always divisible by 49.
Define D to be the differention operator, i.e.
$\displaystyle Df = \frac{df}{dx}$.
Let's say a polynomial p(x) of degree n is "acceptable" if
$\displaystyle D^{n-1} p = ax + b$
where $\displaystyle a - b$ is divisible by 49. It is easy to
verify that $\displaystyle x^8 + x^7$ is acceptable.
It is an immediate consequence that if p is acceptable, then
so is $\displaystyle Dp$. If we can show that $\displaystyle (x+1) p$
is acceptable whenever p is acceptable, we will be done.
So suppose p is an acceptable polynomial of degree n; i.e.
$\displaystyle D^{n-1}p = ax + b$ where $\displaystyle a-b$ is divisible
by 49. By the extended rule for differentiation of a product
(which has a binomial-theorem-like look to it),
$\displaystyle D^n (x+1)p = (x+1)D^n p + \binom{n}{1} D (x+1) D^{n-1} p$
$\displaystyle = (x+1) a + n (1) (ax + b)$
$\displaystyle = (n+1)ax + a+nb$
and $\displaystyle (n+1)a - (a+nb) = n(a - b)$, which is divisible
by 49; so $\displaystyle (x+1)p$ is acceptable. We're done.