20121027, 16:04  #1 
2·59 Posts 
Generalized Mersenne Primes
I believe that I remember a theorem having to do with Generalized Mersenne primes of the form: (a^n+b^n)/(a+b).
The theorem states that if (a^n+b^n)/(a+b) is prime, then n must also be prime. Does anyone know a reference that has a proof of this? Thanks. 
20121027, 18:18  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3×7×109 Posts 
For even values of n, this doesn't divide. For them, there's no denominator, but the logic below still applies.
For odd values of n: Suppose you have a composite n = k*m, with k>=3. Then, (a^{n}+b^{n})/(a+b) = (a^{km}+b^{km})/(a+b) = = (a^{k}+b^{k})/(a+b)*(a^{(k1)m}  b^{m}*a^{(k2)m} + b^{2m}*a^{(k3)m} + ... + b^{(k1)m}) where (a^{k}+b^{k})/(a+b) is > 1. QED, composite. For proper Mersenne numbers a+b = 2+(1) = 1, which makes for an interesting case  there's no denominator. When b=1, a^{n}+b^{n} is a proper Fermat number, and for b>1 and odd (prime) n, (a^{n}+b^{n})/(a+b) has the property that you sought for. Last fiddled with by Batalov on 20121027 at 18:24 
20121027, 18:28  #3  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
Code:
(15:18)>for(a=1,100, for(b=a,100, for(n=1,100, if(((a^n+b^n)/(a+b))%1.==0 &&isprime((a^n+b^n)/(a+b)) && !isprime(n), print(((a^n+b^n)/(a+b))","a","b","n);) ) ) ) Quote:


20121027, 19:14  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9156_{10} Posts 
For even values of n (and not coprime a,b), indeed, you can arrive to some convenient cancellations (a+b will have to be 2*gcd(a,b)^{n}, where n will be a power of two) and get an extended Generalized Fermat in disguise.
E.g. (6^4+26^4)/(6+26) is xGF'(3,13,4). 
20121027, 19:31  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·3·7·109 Posts 
P.S. Correction re: 2*gcd(). Not always 2, of course. 2 will frequent, because there are very many prime xGF'(a,b,4) (=(a^4+b^4)/2) with odd a and odd b. Tons. When a+b is 2^{x}, you can finish the counterexample by multiplying a and b by 2^{x1} or similar. Ugh, I need some coffee.
Last fiddled with by Batalov on 20121027 at 19:33 
20121030, 14:47  #6 
7,793 Posts 
Thanks so much for your replies. Very helpful.
Batalov, if I understand your arguments, if n is odd and composite then (a^n+b^n)/(a+b) must also be composite. Can this be extended to any composite n with an odd factor? If so, then only n that are powers of two could possibly result in a prime. I have found some counterexamples and have observed that a and b are never coprime. Can you think of an argument to support this observation about coprimes? Thanks. 
20121031, 14:16  #7  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
(a^n+b^n)/(a+b) = c^n*(d^n+e^n)/c*(d+e) =c^(n1)*(d^n+e^n)/(d+e) so if (d^n+e^n)/(d+e) is integer so is (a^n+b^n)/(a+b) but with a integer divisor >1 so it's not prime. 

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