For a nonnegative integer n, if the remainder is 1 when 2^n

Please check the questions when posting.

Given: \(n=integer\geq{0}\) and \(2^n=3q+1\), for some non-negative integer \(q\):

If \(n=0=even\) --> \(2^0=1\) --> remainder upon division 1 by 3 is 1 - OK;
If \(n=1=odd\) --> \(2^1=2\) --> remainder upon division 2 by 3 is 2 - not OK;
If \(n=2=even\) --> \(2^2=4\) --> remainder upon division 4 by 3 is 1 - OK;
If \(n=3=odd\) --> \(2^3=8\) --> remainder upon division 8 by 3 is 2 - not OK;
...

So we can see the pattern of reminders 1-2-1-2-.... --> given condition that the remainder is 1 when \(2^n\) is divided by 3 holds true when \(n=even\). So \(n\) must be non-negative even number: 0, 2, 4, ...

I. \(n\) is greater than zero --> not necessarily true, as \(n\) can be zero;
II. \(3^n = (-3)^n\) --> as \(n\) is even then this statement is always true;
III. \(\sqrt{2}^n=integer\) --> as \(n\) is non-negative even number then this statement is always true.

Answer: E (II and III only).