the number of ordered triplets$(x,y,z)$ such that $x,y,z$ are primes and $(x^y) +1=z$
Options $(0,1,2,\infty$ many)
i tried that since they are prime and all primes except 2 are odd.
so a $prime^{prime}$ will be odd (except$2$), and since $z$ cant be even (prime), so $x =2$
so the equation now is $(2^n)+1=z$. I have no idea how to get any further. any help would be appreciated
Best Answer
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 $n$.