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 2012-10-27, 16:04 #1 Unregistered   2×5×292 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.
 2012-10-27, 18:18 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23×7×163 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, (an+bn)/(a+b) = (akm+bkm)/(a+b) = = (ak+bk)/(a+b)*(a(k-1)m - bm*a(k-2)m + b2m*a(k-3)m + ... + b(k-1)m) where (ak+bk)/(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. Last fiddled with by Batalov on 2012-10-27 at 18:24
2012-10-27, 18:28   #3
science_man_88

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Quote:
 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.
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:
 14321,6,26,4 280097,12,69,4 4481,14,18,4 134417,24,57,4
shows that n=4 has counterexamples.

 2012-10-27, 19:14 #4 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23×7×163 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).
 2012-10-27, 19:31 #5 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 912810 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 2x, you can finish the counterexample by multiplying a and b by 2x-1 or similar. Ugh, I need some coffee. Last fiddled with by Batalov on 2012-10-27 at 19:33
 2012-10-30, 14:47 #6 Unregistered   32·241 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.
2012-10-31, 14:16   #7
science_man_88

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Jul 2009
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Quote:
 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.

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