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Number Theory?
Show all numbers of the form:
5·(2^n)
can be written as the sum of two squares
for all natural numbers n
2 Answers
- ?Lv 63 years agoFavorite Answer
When n = 2k (k = 0,1,2, ...)
5*(2^n)
= 5*[2^(2k)]
= (1+4)*[2^(2k)]
= 2^(2k) + 2^(2k+2)
= (2^k)^2 + [2^(k+1)]^2 : The sum of two squares
When n = 2k+1 (k = 0,1,2, ...)
5*(2^n)
= 5*[2^(2k+1)]
= 10*[2^(2k)]
= (1+9)*[2^(2k)]
= 2^(2k) + (3^2)*[2^(2k)]
= (2^k)^2 + [3*(2^k)]^2 : The sum of two squares
- ?Lv 73 years ago
n = 0, 5 = 1² + 2²
n = 1, 10 = 1² + 3² = 1² + (2+1)²
n = 2, 20 = 2² + 4² = (2*1)² + (2*2)² (terms twice those of n = 0 for 4 times that result)
n = 3, 40 = 2² + 6² = (2*1)² + (2*3)² (terms twice those of n = 1 for 4 times that result)
n = 4, 80 = 4² + 8² = (2*2)² + (2*4)² (terms twice those of n = 2 for 4 times that result)
n = 5, 160 = 4² + 12² = (2*2)² + (2*6)² (terms twice those of n = 3 for 4 times that result)
etc.
5(2^n) = ( 2^(⌊n/2⌋) )² + ( (2^(⌊n/2⌋))(2 + mod(n,2)) )², which is the sum of two squares.
