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.

[Batalov]

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,
(aⁿ+bⁿ)/(a+b) = (a^km+b^km)/(a+b) =
= (a^k+b^k)/(a+b)*(a^(k-1)m - b^m*a^(k-2)m + b^2m*a^(k-3)m + ... + b^(k-1)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, aⁿ+bⁿ is a proper Fermat number, and for b>1 and odd (prime) n, (aⁿ+bⁿ)/(a+b) has the property that you sought for.

[science_man_88]

Originally Posted by Unregistered

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.

(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);)
                    )
                )
          )

14321,6,26,4
280097,12,69,4
4481,14,18,4
134417,24,57,4

shows that n=4 has counterexamples.

[Batalov]

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).

[Batalov]

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^x-1 or similar. Ugh, I need some coffee.

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.

[science_man_88]

Originally Posted by Unregistered

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.

well assume gcd(a,b)=c then the equation comes to:

(a^n+b^n)/(a+b) = c^n*(d^n+e^n)/c*(d+e) =c^(n-1)*(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.