Prove [2^(2m +1) + 1] mod 3 = 0?

Question

2^2m - 1  = (2^m +1)(2^m -1) 
hence

2^m -1 , 2^m, 2^m + 1 are sequential & 2^m not div 3

so
2^m + 1 or 2^m-1 is div 3

hence 
2^2m - 1 is div 3
2(2^2m - 1) == 2^(2m +1) -2  is div 3

so 
2^(2m +1) -2 + 3 == 2^(2m +1) + 1

Anonyme · Accepted Answer

Note that if 2^(2m + 1) + 1 ≡ 0 (mod 3), then 2^(2m + 1) + 1 is divisible by 3.

Then, we have:

2^(2m + 1) + 1
= 2*2^(2m) + 1
= 2*4^m + 1
= 2*4^m + 3 - 2
= (2*4^m - 2) + 3
= 2(4^m - 1) + 3.

Since a^n - b^n is divisible by a - b for all integers n by the factor theorem. 4^m - 1 = 4^m - 1^m is divisible by 4 - 1 = 3 for all m.

Since 2(4^m - 1) and 3 are both divisible by 3, then 2(4^m - 1) + 3 = 2^(2m + 1) + 1 is divisible by 3 as required.

I hope this helps!

? · Answer

If 2 distinctive factor elevated jointly = 0, then one or the two factors are = 0 Set each and each factor = 0 and remedy for x -7x + 3 = 0 -7x = - 3 x = 3/7 a million.2x - 10 = 0 a million.2x = 10 x = 10/a million.2 = 8.333 = 8 a million/3

Prove [2^(2m +1) + 1] mod 3 = 0?

2 réponses