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 Prime candidates (Posted on 2016-12-20)
For what values of integer n are n^n + 1 and (2n)^(2n) + 1 both prime numbers?

 See The Solution Submitted by Ady TZIDON No Rating

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 computer exploration | Comment 2 of 5 |
10   for n=1 to 100
20     e1=n^n+1
30     e2=(2*n)^(2*n)+1
40     if fnPrime(e1) and fnPrime(e2) then print n,e1,e2
80   next n
90   end
10000   fnOddfact(N)
10010   local K=0,P
10030   while N@2=0
10040     N=N\2
10050     K=K+1
10060   wend
10070   P=pack(N,K)
10080   return(P)
10090   '
10100   fnPrime(N)
10110   local I,X,J,Y,Q,K,T,Ans
10115   if N=2 then Ans=1:goto *EndPrime
10120   if N@2=0 then Ans=0:goto *EndPrime
10125   O=fnOddfact(N-1)
10130   Q=member(O,1)
10140   K=member(O,2)
10150   I=0
10160   repeat
10170     repeat
10180       X=fnLrand(N)
10190     until X>1
10200     J=0
10210     Y=modpow(X,Q,N)
10220     loop
10230       if or{and{J=0,Y=1},Y=N-1} then goto *ProbPrime
10240       if and{J>0,Y=1} then goto *NotPrime
10250       J=J+1
10260       if J=K then goto *NotPrime
10270       Y=(Y*Y)@N
10280     endloop
10290    *ProbPrime
10300     I=I+1
10310   until I>50
10320   Ans=1
10330   goto *EndPrime
10340   *NotPrime
10350   Ans=0
10360   *EndPrime
10370   return(Ans)
10380   '
10400   fnLrand(N)
10410   local R
10415   N=int(N)
10420   R=(int(rnd*10^(alen(N)+2)))@N
10430   return(R)
10440   '
10500   fnNxprime(X)
10510   if X@2=0 then X=X+1
10520   while fnPrime(X)=0
10530     X=X+2
10540   wend
10550   return(X)

finds only n=1 and n=2, resulting in primes 2 and 5, and 5 and 257, respectively, having tested n up to 100.

 Posted by Charlie on 2016-12-20 13:34:08

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