Question 1044530
<pre>
Here is a proof using the binomial theorem.

{{{7^(2n)+16n-1}}}

{{{(8-1)^(2n)+16n-1}}}

Every term of the binomial expansion of (8-1)<sup>2n</sup>
except the last two terms contain a factor of 8 to a power
2 or greater and thus they are all divisible by 64.  So the
above becomes upon writing out the first two and last two
terms of the binomial expansion: 

{{{(8^(2n)-2n*8^(2n-1)+""*""*""*""+C(2n,2n-1)*8^1*(-1)^(2n-1)+(-1)^(2n))+16n-1}}}

Simplifying and using the property of combinations C(2n,2n-1)=C(2n,1)=2n

{{{(8^(2n)-2n*8^(2n-1)+""*""*""*""+C(2n,1)*8*(-1)+1)+16n-1}}}

{{{(8^(2n)-2n*8^(2n-1)*(-1)^1+""*""*""*""+2n*8*(-1)+1)+16n-1}}}

{{{(8^(2n)-2n*8^(2n-1)*(-1)^1+""*""*""*""-16n+1)+16n-1}}}

The last two terms in the binomial expansion cancel with the 
16n-1, so all the remaining terms are divisible by 64.  Thus 
the theorem is proved.

Edwin</pre>