Puzzle: prime numbers

Show that (N^4 + 4^N) is a prime number if and only if N=1.

Comments

• Anonymous
  March 08, 2006
  Ok, let's crunch some number. The last digit of both terms N^4 and 4^N is a function of the last digit of N.
  
  Thus, if a number ends with 0 1 2 3 4 5 6 7 8 9    
  the last digit of the sum is    6 5 2 5 2 9 2 5 2 5.
  
  (i.e. 3^4=81 + 4^3=64 gives 155. Note that 5 is the last digit for every number ending with a 3: 83, 657673, and so on).
  
  Every number ending with any of the digits in the second row is not a prime.
  
  This is true for every N integer positive, with the notably exception for N=1, just because 5 (1^4+4^1) ends with, erm, 5, which is prime. Note that with N=0 the result is 1, which is not a prime.
  
  Thank you, Adi, this puzzle reminded me of the time spent learning htonl :-)

• Anonymous
  March 08, 2006
  >> Every number ending with any of the digits in the second row is not a prime.
  
  Well, if the last digit of the sum is "9" (the one you mentioned when N ends with 5) then the number is not necessarily non-prime. For example, 19, 29, 59, 79 are primes.

• Anonymous
  March 08, 2006
  Since these numbers are supposedly composite when N>1, let's try to find some factors.
  
  If N is even, then both terms are even and their sum is even, so 2 is a factor.  The formula is equal to 5 when N=1 and strictly growing so there must be another factor greater than 1 as well.
  
  If N is odd, we can work out the squares to find a factorization.
  
  4^N+N^4
  = (2^N+N^2)^2 - 2N^22^N
  = (2^N+N^2)^2 - N^22^(N+1)
  = (2^N+N^2)^2 - (N2^((N+1)/2))^2
  = (2^N+N^2+N2^((N+1)/2))(2^N+N^2-N2^((N+1)/2)))
  
  The first factor is 5 when N=1 and strictly growing.
  
  The second factor is 1 when N=1, and if you set M=2^N, then it grows like M-sqrt(M)log(M), which means it is strictly growing as well.
  
  So we've found a factorization into two factors greater than 1 when N>1.  Therefore, it's always composite.

• Anonymous
  March 09, 2006
  Excellent proof! Yes, there are several approaches to do the factorization, and this is one of them.

• Anonymous
  March 09, 2006
  Am I missing something in the notation?  
  
  4*N + N^4 is just N(4 + N^3) so we have a trivial factorization.

• Anonymous
  April 02, 2006
  Hi to human that not foolish!!!!!!!!!!!
  Mr.decio you are goof because your proof is full of wrong method...

• Anonymous
  April 04, 2006
  @Nicholas
  
  well if n is even then your factors are not integral (2^((N+1)/2) being non-integral) therefore more work needs to be done in that case to prove that it is composite.

• Anonymous
  April 04, 2008
  PingBack from http://drinksairportsblog.info/antimail-puzzle-prime-numbers/

• Anonymous
  June 13, 2008
  PingBack from http://ciara.sattelect.com/puzzleswithnumbers.html