3=1+1+1
9=4+4+1
27=25+1+1
Prove:
Every power of 3 is a sum of three squares relatively prime to 3.
All powers of 3 are of the form x^2+2y^2.
x is {1,1,5,7,1,23,...} A087455 in Sloane
y is {1,2,1,4,11,10,...} A088137 in Sloane
From the same source:
The general formula for x^2 is ((1/2)*((1 - i*sqrt(2))^(n) + (1 + i*sqrt(2))^(n)))^2
The general formula for y^2 is -1/8 ((1 + i sqrt(2))^(n + 2) - (1 - i sqrt(2))^(n + 2))^2
So no divisibility by 3 in either case.
And ((1/2)*((1 - i*sqrt(2))^(n) + (1 + i*sqrt(2))^(n)))^2+ -1/8 ((1 + i sqrt(2))^(n) - (1 - i sqrt(2))^(n))^2+-1/8 ((1 + i sqrt(2))^(n) - (1 - i sqrt(2))^(n))^2 simplifies nicely to:
(i*(sqrt(2)+-i))^n (-i*(sqrt(2)+i))^n, commonly known as 3^n
Edited on January 20, 2019, 10:54 am
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Posted by broll
on 2019-01-20 10:41:26 |
Solution