Find all triplet(s) (p, q, r) of prime numbers, that satisfy this equation:
p3 = p2 + q2 + r2
Justify your answer with valid reasoning.
The squares of primes of form (6k+-1) are always worth 1, mod 6. so RHS is worth exactly 3, mod 6.
The cubes of primes of form (6k+-1) are always worth 1 or 5, mod 6. so LHS is worth {1,5} mod 6, a contradiction.
This leaves just 2 and 3 as possibilities to be checked, leading to the solution offered by earlier posters.
Note: the only other solutions (prime or not) appear to be of the form
(n^2+1)^3-(n^2+1)^2 = (n^3+n)^2, when the partition:
(2(n^2+1)-2)^2+((n+1)*(n)*(n-1))^2 is always possible.
Edited on May 22, 2023, 4:22 am
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Posted by broll
on 2023-05-22 04:19:11 |
Possible Solution
