the number of ordered triplets(x,y,z) such that x,y,z are primes and (x^y) +1=z
Hint: Recall that
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
and notice that
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$
Compare your equation $x^y+1=z$ to the form $a^n+b^n$ (ie replace $a,b,n$ with numbers or variables), and see if you can find a way to generalize this for all odd $n$.