Is it true that the equation $$(2k+1)(2^{4k+1}+(2k-1)^{4k+1})=a^2+(4k+1)b^2$$ has no positive integer solution ?
Let $A=(2^{4k+1}+(2k-1)^{4k+1})/(2k+1)$, if $q$ is a prime factor of $A$ such that $x^2+4k+1\equiv 0 \mod q$ has no integer solution, and $A \not\equiv 0 \mod q^2$, then $(2k+1)^2A=a^2+(4k+1)b^2$ has no integer solution.
What I know:
We have $A\equiv 3 \mod 4$,so the Jacobi symbol $J=(\frac{-(4k+1)}{A})=(\frac{-1}{A})(\frac{4k+1}{A})=-(\frac{A}{4k+1})$,
if $(\frac{A}{4k+1})=1$, then there is a prime factor of $A$, named $q$, such that $x^2+4k+1\equiv 0 \mod q$ has no integer solution.
Especially,if $4k+1=p$ is prime,then $A\equiv \frac{2+(2k-1)}{2k+1}\equiv 1 \mod p$, so $(\frac{A}{p})=1,J=-1$.
