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 Oct8 accepted RSA decrypt long messages (javascript) Oct24 awarded Student Oct24 comment RSA decrypt long messages (javascript) upon some further research I did discover that the message cannot be bigger than $n$ so then I had the same thought: split the message into blocks and encrypt/decrypt separately. I have a very light knowledge of javascript though. I think this must involve creating an array but.. I'm not sure. This subject is very.. complicated. They repeatedly say it's not a programming subject, just theory, but time and time again in this assignment I'm faced with javascript issues I can't resolve because I don't deal with javascript very often! Oct24 awarded Commentator Oct24 comment RSA decrypt long messages (javascript) @CodesInChaos I cross-posted in the hopes that someone here might have the answer - since no one has answered or commented in stackoverflow). this is equal parts a progamming question (for stackoverflow) and a crypto question (for crypto.stack). so I figured it was appropriate for both. Oct24 comment RSA decrypt long messages (javascript) Hi, thanks for your answer however, I don't care about the security, I am not doing this for real life, this is a task for a university subject. doPrivate and doPublic are the only methods available to us. Oct24 asked RSA decrypt long messages (javascript) Oct16 awarded Scholar Oct16 accepted DSA generate signature and verify Oct16 comment DSA generate signature and verify alright! so $r=9$ $s=9$ $w=5$ $u1=7$ $u2=1$ and BAM! it all works out to $9$ so $v=r$! correct right? so the error was simply the calculation of the inverse all along? Oct16 comment DSA generate signature and verify ahhh well yes, they do say $y^{-u2}$... so in that case, what would $s$ be?? Oct16 comment DSA generate signature and verify actually I figured it out shortly after I posted that. but I couldn't edit any more. it was obvious hahah. BUT I think I have identified the next obvious error (besides doing the inverse wrong).... it's not $H(m) **-** x$ x $r$ , it's $H(m) **+** x$ x $r$ right?? now this isn't my fault.. $H(m) **-** x$ x $r$ is what the LECTURES say. so I think they may be wrong? Oct16 comment DSA generate signature and verify ok so we're computing $2^1 (mod$ $11)$ ... but... why? why $(mod$ $11)$ ? I'm sorry if this is a noob question but I $am$ new to crypto haha. the function for $s$ is $s = k^{-1} (H(m) - x$ x $r) mod$ $q$ right? so why should we calculate $k^{-1}$ as $k^{-1} (mod$ $q)$ ? Oct16 comment DSA generate signature and verify sorry but... how do you get $6$ from $2^{-1}$?? Oct16 revised DSA generate signature and verify deleted 43 characters in body Oct16 comment DSA generate signature and verify are you sure?? I mean, the very first question in this set is to show that g has order q as required. our lectures say to calculate this by showing that: $g^q$ mod $p$ = $1$ . which it does... $3^1$$^1$ mod $23$ = $1$ . So how can it be wrong? Oct16 comment DSA generate signature and verify hi! thank you so much for answering. I see what you're saying and I had not thought of it before. HOWEVER, this is revision I am doing and the question is on a worksheet given to the class. I didn't choose g... they gave us the value for g, p, q, x and H(m). That's why I said I couldn't come up with a suitable value for k... the only value they didn't specify in the question. Oct16 awarded Editor Oct16 revised DSA generate signature and verify added 62 characters in body Oct16 comment DSA generate signature and verify @B-Con oops I forgot to give the private key. x=5. as for the signature - I have produced multiple ones while attempting to find a value of k that 'adds up' when calculating v. I think I may be doing something wrong in the calculations because surely at least one value > 0 < 11 must work.