Trying to show the the prime factorization of $$2^{16}-1$$ without a calculator.
I know that $2^{16} - 1$ yields the prime numbers$$3*5*17*257$$ because I calculated $2^{16}-1$ on my calculator (65,535), then found that it is divisible by the prime number 5 (since the last number is 5) and so on to get my answer.
Is this possible to solve using modular arithmetic?
Using $a^2-b^2=(a-b)(a+b)$ Would $a=16$ and $b=1$? After reading for the past few hours I am unable to grasp how the difference of two squares equation is applied here.
AH! I did not realize you could apply it in that way
$2^{16}-1= (2^8-1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=(15)(17)(257)=(5)(3)(17)(257)$
I thought that $a^2-b^2$ meant that a and b had to be squared.... Thank you so much for explaining that to me! I appreciate your help greatly. This is much more simple than I had thought.
As suggested in the comments, $2^{16}-1= (2^8-1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=(15)(17)(257)=(5)(3)(17)(257)$