Demonstrate that $\int_0^1{\frac{(x^2+x+1)^{4n+1}- x}{x^2+1}dx}$ is a rational number

Question

I thought about proving $x^2+1$ divides $(x^2+x+1)^{4n+1}- x$ , but I don't know how.

Answer 1

Yes you want to divide, the quotient will be a polynomial and its integral will give a rational number. The problem is that there may be a remainder. So let us calculate

$$(x^2+x+1)^{4n+1}-x \pmod{x^2+1}$$ first note that $x^2\equiv -1 \pmod{x^2+1}$ and so $x^{4n}\equiv 1 \pmod{x^2+1}$.

Now

$$(x^2+x+1)^{4n+1}-x \equiv x^{4n+1}-x \equiv x-x\equiv 0\pmod{x^2+1}$$

So there is no remainder and the integral is a rational number.

Answer 2

$$(x^2 + x + 1)^{4n + 1} - x = \\ [(x^2 + 1) + x]^{4n + 1} - x = \\ \sum _{k = 0} ^{4n} \binom {4n + 1} k (x^2 + 1)^k x^{4n + 1 - k} - x = \\ x^{4n + 1} + \sum _{k = 1} ^{4n + 1} \binom {4n + 1} k (x^2 + 1)^k x^{4n + 1 - k} - x = \\ x (x^{4n} - 1) + (x^2 + 1) \underbrace {\sum _{k = 1} ^{4n + 1} \binom {4n + 1} k (x^2 + 1)^{k - 1} x^{4n + 1 - k}} _{P(x)} = \\ x [(x^2)^{2n} - 1] + (x^2 + 1) P(x) = x\{[(x^2 + 1) - 1]^{2n} - 1\} + (x^2 + 1) P(x) = \\ x \left( \sum _{k = 0} ^{2n} \binom {2n} k (x^2 + 1)^k (-1)^{2n-k} - 1 \right) + (x^2 + 1) P(x) = \\ x \left( (-1)^{2n} + \sum _{k = 1} ^{2n} \binom {2n} k (x^2 + 1)^k (-1)^{2n-k} - 1 \right) + (x^2 + 1) P(x) = \\ x (x^2 + 1) \sum _{k = 1} ^{2n} \binom {2n} k (x^2 + 1)^{k - 1} (-1)^{2n - k} + (x^2 + 1) P(x)$$

Therefore, if

$$Q(x) = x \sum _{k = 1} ^{2n} \binom {2n} k (x^2 + 1)^{k - 1} (-1)^{2n - k} + \sum _{k = 1} ^{4n + 1} \binom {4n + 1} k (x^2 + 1)^{k - 1} x^{4n + 1 - k}$$

then $(x^2 + x + 1)^{4n + 1} - x = (x^2 + 1) Q(x)$ and $Q(x)$ clearly has integer coefficients, therefore $\int _0 ^1 Q(x) \ \textrm d x$ will clearly be a rational number.