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 Oct 26 awarded Notable Question Nov 19 awarded Popular Question Apr 24 awarded Student Oct 23 awarded Scholar Oct 23 accepted RSA square and multiply Oct 16 comment RSA square and multiply ah. thanks for that. Oct 16 comment RSA square and multiply so if $e$ were to be another value, say... $4$. how would that add up? would we have to find the binary representation of it? square for each place, multiply by the running value? what would that gives us for something like $4$? Oct 16 comment RSA square and multiply ps: the description of d (decryption) makes perfect sense. thanks :D Oct 16 comment RSA square and multiply So if my understanding is correct... if $e = 2^{16} + 1$ , we will have 16 squarings and 1 multiplication. but, if $e = 3$ and that is equivalent to $2^1 + 1$ (like you've indicated?) then for $e = 3$ we have 1 squaring and 1 multiplication? Oct 16 comment RSA square and multiply @DrLecter I understand the second part, but that 'simple' explanation just looked a lot like lots of complicated squiggles... The mathematics of this escape me. I certainly have never seen anything as complicated as that in our lecture slides, which frankly, don't explain much of anything. Oct 16 asked RSA square and multiply