# Global minimum for $frac{2(q - 1)(q^k + 1)}{q^{k+1} + q - 1}$, if $q geq 5$ and $k geq 1$

Mathematics Asked by Arnie Bebita-Dris on December 15, 2020

Let $$q$$ be a prime number, and let $$k$$ be an integer.

THE PROBLEM

Does the function
$$f(q,k) = frac{2(q – 1)(q^k + 1)}{q^{k+1} + q – 1}$$
have a global minimum, if $$q geq 5$$ and $$k geq 1$$?

MY ATTEMPT

I tried asking WolframAlpha, it was unable to find a global minimum for $$f(q,k)$$ in the domain $$q geq 5$$ and $$k geq 1$$.

I then computed the partial derivatives (still using WolframAlpha):

Partial derivative with respect to $$q$$
$$frac{partial}{partial q} f(q,k) = frac{2q^{k-1}bigg(q^{k+1} – k(q – 1) + qbigg)}{bigg(q^{k+1} + q – 1bigg)^2} > 0$$

Partial derivative with respect to $$k$$
$$frac{partial}{partial k} f(q,k) = -frac{2(q-1){q^k}log(q)}{bigg(q^{k+1} + q – 1bigg)^2} < 0.$$

Does this mean that we can have (say)
$$f(q,k) geq f(5,1) = frac{48}{29} approx 1.65517?$$

$$frac{2}{f(q,k)}=frac{q^{k+1}+q-1}{(q-1)(q^k+1)}=1+frac{1}{q-1}left(1-frac{1}{q^k+1}right)$$ is strictly increasing in $$k$$, so it can't have a global maximum, hence $$f(q,k)$$ can't have a global minimum.

Correct answer by metamorphy on December 15, 2020

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