Powerful Sum? (Posted on 2007-06-23)

	Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:	(Hide)
For m=2; 2^m + 3^m = 17, which is not a perfect power. For m> =3, m must be odd since it is a prime number. Substituting m= 2s+1, we observe that: (2^m + 3^m)/5 = {2^(2s) – 2^(2s-1)*3 + 2^(2s-2)*3^2 – 2^(2s-3)*3^3 +.....+(2^2)*3^(2s-2) – 2*3^(2s-1)) + 3^(2s)} = {2^(2s) + 2^(2s-1)*2 + 2^(2s-2)*2^2 + 2^(2s-3)*2^3)+.....+ (2^2)*2^(2s-2) + 2*2^(2s-1)) + 2^(2s)}(Mod 5) = (2s+1)*2^(2s) (Mod 5) = m*2m-1(Mod 5)……(i) (2^m + 3^m) is divisible by 5 for any given odd prime m, for (2^m + 3^m) to correspond to a perfect power, we must have (2^m + 3^m) possesing the form 5c, where c is an integer ≥ 2 Accordingly, we must have: m*2m-1(Mod 5) = 0, so that: m Mod 5 = 0, since (m, 2) = 1 for prime m>=3. This yields m =5, since all the other numbers divisible by 5 are composite. For m=5, we observe that 2^5+3^5 = 275, which is not a perfect power. Consequently, it is never the case that 2^m + 3^m is the perfect power whenever m is a positive prime number.

	Subject	Author	Date
	solution for perfect even powers	xdog	2007-06-28 21:00:43
	re(2): No Subject	xdog	2007-06-28 17:12:15
	re: No Subject	atheron	2007-06-28 14:36:40
	No Subject	xdog	2007-06-25 08:45:46