<s>[INST] Greatest prime factor of $4^{17}-2^{28}$ I have seen the solution to this problem. 

What is the greatest prime factor of $ \ 4^{17} - 2^{28} \ $?

Answer: 7
$$ 4^{17}-2^{28} \ = \ 2^{34}-2^{28} \ = \ 2^{28} \ (2^6-1) \ = \ 2^{28} \ \cdot \ 63 \ = \ 2^{28}\ \cdot \ 3^2 \ \cdot \ 7 $$
I understand the problem up to this part: $2^{28} \ (2^6-1) \ $ .
Why did this occur? I can't explain it.
Thanks!
 [/INST] $$4^{17} = (2^2)^{17} = 2^{2 \cdot 17} = 2^{34} = 2^{28 + 6} = 2^{28} \cdot 2^6$$
Now recognize the common factor of $2^{28}$ and factor it out.
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