Let n be a natural number such that the equation a^n +b^n = c^2, ( a, b and c being prime numbers) has at least one solution.
Find the maximal possible value of n.
Source: Bilken University
1. a must be 2, because otherwise odd+odd is even
2. 2^n +b^n = c^2; obviously there are solutions where n=1
(just need a square differing from a prime by 2)
Low Hanging Fruit
3. Let n=2; no solutions (2^2 is too small to bridge the gap, since 9=2^2+5)
4. Let n=4; no solutions (Fermat: no square is the sum of 2 fourth powers)
5. Let n=3 or any odd number >3; no solutions as (a+b) is found in the factorisation of c^2 and so c.
6. Let n be even and divisible k times by 4 : a^k^4+b^k^4=c^2, see 4.
7. Let n be even and worth 0, mod (some odd number,k):a^2^k+b^2^k=c^2; see 5.
Hence the maximal value of n is 1
Edited on January 7, 2011, 2:19 am
Posted by broll
on 2011-01-07 02:04:43