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Powerful Sum? (Posted on 2007-06-23) Difficulty: 3 of 5
Is it ever the case that 2m+3m is a perfect power, whenever m is a positive prime number?

  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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
solution for perfect even powersxdog2007-06-28 21:00:43
re(2): No Subjectxdog2007-06-28 17:12:15
re: No Subjectatheron2007-06-28 14:36:40
No Subjectxdog2007-06-25 08:45:46
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