Question 909567
<pre>
What we are to prove is that for any positive integer n,

{{{2^(2n)-1}}} is divisible by 3, that is, it is a multiple of 3.

For n=1

{{{2^(2(1))-1=2^2-1 =4-1=3}}}. 3 is a multiple of 3.

So it's true for n=1 

We must prove that if you know it's true for integer n=k, 
then it will be true for n=k+1.

Assume it's true for n=k.

{{{2^(2k)-1=3m}}} for some positive integer m

Multiply both sides by {{{2^2}}}

{{{2^2*2^(2k)-2^2=2^2*3m}}} for some positive integer m
{{{2^2*2^(2k)-4=4*3m}}} for some positive integer m

{{{2^(2k+2)-4=12m}}}

Write -4 as -1-3

{{{2^(2(k+1))-1-3=12m}}}

{{{2^(2(k+1))-1=12m+3}}}

{{{2^(2(k+1))-1=3(4m+1)}}, thus {{{2^(2(k+1))-1}}} is a
multiple of 3.

Therefore since the theorem is true for n=k=1, it's true for n=k+1=2.

And since the theorem is true for n=k=2, it's true for n=k+1=3

Thus since the theorem is true for n=k=3, it's true for n=k+1=4

And this goes on forever, for we've proved it can't stop!

Edwin</pre>