Current Events > For which prime numbers p, q, and r does p^3-q^3=r and why?

(+) Yeah! (+)

And for a slightly tougher one, for which values does p^3+q^3=r ?

For which prime numbers p, n and s does p^3-n^1=s and why?

For topic title
3, 2, 19.
3^3 = 27
2^3 = 8
27-8 = 19

Since r is prime, we can assume that p > q, otherwise p^3 - q^3 is negative.

p^3 - q^3 = r
(p - q)(p^2 + pq + q^2) = r

Since r is prime, one term must be equal to r and the other must be equal to 1.

Case 1
p - q = 1 is trivial as the only adjacent primes are p = 3 and q = 2.

Evaluating p^2 + pq + q^2 gives r = 19.

Case 2
p - q = r is not possible. If p and q are odd primes, that would make r even which forces r to be 2, but p^3 - q^3 cannot be 2.

Since I'm too lazy to prove my claim about the minimum value of a difference of cubes, it suffices to look at the corresponding term and notice that p^2 + pq + q^2 = 1 is not possible as this is the sum of 3 positive integers.

I got distracted and forgot to cover p - q = r assuming q = 2. I'm sure there is a good explanation why p and r can't be twin primes but I'm even more lazy to think about that.

(2,3,19) is the only triplet for the first question. There are no triplets for the second question.