Question 699138
I can prove that even more similarly defined numbers are divisible by 3.
If you (or the teacher do not like a generalization, you can always adapt the proof to just the subset of {{{1+2^(6K+1)}}} numbers.
 
All numbers of the form {{{1+2^(2n+1)}}} are divisible by 3.
Some of them (but not all) are of the form {{{1+2^(6K+1)}}} , if {{{n=3K}}},
as for {{{1+2^(2*3+1)=1+2^7=1+2^(6*1+1)}}} (with {{{n=3}}}, or {{{K=1}}})
{{{1+2^7=1+128=129=3*43}}} .
 
{{{2^2=4=3+1}}} so {{{2^(2n)=(2^2)^n=(3+1)^n
We can expand the power of a binomial {{{(3+1)^n}}} .
{{{2^(2n)=(2^2)^n=(3+1)^n= 3^n+n*3^(n-1)+(n(n-1)/2)*3^(n-2)}}} + ... + {{{(n(n-1)/2)*3^2+n*3+1^n=3}}}[{{{3^(n-1)+n*3^(n-2)+(n(n-1)/2)*3^(n-3)}}} + ... + {{{(n(n-1)/2)*3+n}}}]+{{{1=3M+1}}} , calling the messy bracket {{{M}}}.
The conclusion so far is (in words) the even powers of 2, when divided by 3, have 1 as the remainder.
In formulas: {{{2^(2n)=3*M+1}}} , where {{{M}}}is a whole number.
{{{2^(2n+1)= 2^(2n)*2^1=2^(2n)*2=(3M+1)*2=6M+2}}}
{{{1+2^(2n+1)=1+6M+2=6M+3=3(2M+1)}}}
Since {{{1+2^(2n+1)}}} is {{{3}}} times the whole number {{{(2M+1)}}},
{{{1+2^(2n+1)}}} is divisible by {{{3}}}.