$\frac{x^5-y^5}{x-y}=p$,give what p ,the diophantine equation is solvable

Question

for$$\frac{x^3-y^3}{x-y}=x^2+xy+y^2=p$$$p=6k+1$give p prime, On what conditions,the diophantine equation $$\frac{x^5-y^5}{x-y}=p$$ is solvable in integers.does it have a linear expression.for $$\frac{x^n-y^n}{x-y}=p$$n =7,11,13...etc,is it easy to solve.and in what conditons ,it's solvable in $Z[i]$

Answer 1

Write out $$ x^4 + x^3 y + x^2 y^2 + x y^3 + y^4. $$ Up to some large positive $N,$ the maximum number of possible values of this for $0 < y < x$ is no larger than $\sqrt N.$ As a result, it is not possible to describe the set of primes just by congruences. The quantitative aspect of Dirichlet's result on primes in arithmetic progressions is this: given fixed $a,b > 0,$ with $\gcd(a,b)=1,$ the number of primes in the progression $a t + b,$ up to $N,$ is asymptotically $$ \frac{N}{\phi(a) \log N.} $$ This is much, much larger than $\sqrt N.$