An exponent and prime problem (Posted on 2007-01-06)

	Submitted by K Sengupta
Rating: 4.0000 (2 votes)
Solution:	(Hide)
PART A: Let F(n) = 2n+n2 For n> 3, all primes n can be written in the form, n = 6k+/-1. Therefore n^2 = 36*k^2 +/- 12k + 1 = 1 mod 3. If 2^m= 2 mod 3, then 2^(m+1) = 1 mod 3, and 2^(m+2) = 2 mod 3. As 2^1 = 2 mod 3, it follows that 2^q = 2 mod 3 for all odd q. Hence when n> 3, 2^n + n^2 = 0 mod 3, and cannot be prime. When n = 2, 2^n + n^2 = 8. When n = 3, 2^n + n^2 = 17. That is, F(n) = 2^n + n^2 is prime only when n = 3. NOTE: The problem as presented above is a modified version of a puzzle which appeared earlier in http://mathschallenge.net.

	Subject	Author	Date
	I liked it!	David Johnson	2007-01-15 00:52:54
	unique solution?	Dennis	2007-01-06 15:58:03
	re(2): needs a second attempt	elementofsurprize	2007-01-06 14:56:21
	re: needs a second attempt	K Sengupta	2007-01-06 14:52:11
	re: needs a second attempt (correction)	elementofsurprize	2007-01-06 14:51:48
	needs a second attempt	elementofsurprize	2007-01-06 14:42:54