add numbers

content/2018-10-19-twenty-years-of-rsa-attacks.md

@@ -62,17 +62,23 @@ Since this is a big problem if you were to really use this cryptosystem, I'll be
 ```py
 >>> from Crypto.PublicKey import RSA
 >>> k1 = RSA.generate(2048)
+<_RSAobj @0x7f3d3226dfd0 n(2048),e,d,p,q,u,private>
 ```
 
 Now, normally when you generate a new key, it'd generate a new modulus. For the sake of this common modulus attack, we'll force the new key to use the same modulus. This also means we'll have to choose an exponent e other than the default choice of 65537 (see [this link][5] for documentation):
 
 ```py
 >>> N = k1.p * k1.q
+29977270253913673973269594877868500604696844309480395834898813292056864035968758602074842333119394545818563664205865827843973433118231606201251719390934610989873635763197929136439794366715495587924829697045618064595517091398323127000591150167969423793125376862942962617933168868125721044755585292104012767604921511927694421931531763256179277376290836490302585046803170658011843375751827334637689505406974645481089358325805114205957009910758378725866614617688361814922628596814445370820099034880786971816556547138716303030977389113515312289367195090368607322922710704592536914377782096784092012774047931602714559411641
 >>> e = k1.e
+65537
 >>> d = k1.d
+15565200260470091881477501931717765645013182095721628848830000114674199708256113134107524142907363428287225581416015506594787249272629252596585055146773790032720599834991872233759704632573379913049026195290680640250863651116064783079834540016568221344526961094787464713454198443832494032866744158338151738236661515444305521301583312800890473043854752775780731961801793612989845832052044110301479536119434333369042172368546513808726742737729539432085793131998509039970952524552914892677427673231515899625998973161553704772256496315467235759715665448324408858980400807019213972046972829905566822336304711418843041721957
 >>> e2 = 65539
 >>> d2 = modinv(e2, (k1.p - 1) * (k1.q - 1))
+28155004966198083605557147846430301877082565365203402029588435163682086478799751838610856433805281302245406343554098644058282620662395619703047797297929171630352487059669029554823105971149580111303390225692229359101863845359614581890498607677708812792166993283364928728648227920436362454567967968010840963546889938282011875589987758165583590886451185216017928261116297436515322115306907044332595229241201447860504794919920665520170088035323466070517987985855014612353911537010064927051269052451478774966384895845225295261610911375622081716902881447610710645142912550905885899057916649884624811336671599114611316629599
 >>> k2 = RSA.construct((N, e2, d2))
+<_RSAobj @0x7f3d31c7c5f8 n(2048),e,d,p,q,u,private>
 ```
 
 Ok, now we have two keys, `k1` and `k2`. Now I'll show how using only the public and private key of `k1` (assuming this is the pair that we got legitimately from the crypto operator), and the public key of `k2`, which is tied to the same modulus, we can find the private key of `k2`.
@@ -85,14 +91,16 @@ It turns out that `k` is extremely close to `ed/N`: `ed/N = (1 + k*phi(N)) / N =
 >>> from decimal import Decimal, getcontext
 >>> getcontext().prec = 1000
 >>> k = round(Decimal(e) * Decimal(d) / Decimal(N))
+34029
 >>> phi = (Decimal(e) * Decimal(d) - 1) / Decimal(k)
-1
+Decimal('29977270253913673973269594877868500604696844309480395834898813292056864035968758602074842333119394545818563664205865827843973433118231606201251719390934610989873635763197929136439794366715495587924829697045618064595517091398323127000591150167969423793125376862942962617933168868125721044755585292104012767604575090001864613992237960887242026855773279634028088706121371418922552125986506064146112561599205615974813154971272528592745144988174228621487749404677959591894452249599588096076892574585613962026186332366180174253118634077603697727952204486962202338916762987146793208323561031870496718547544796269555861921652')
 ```
 
 Then we can get `p + q` through the formula mentioend above:
 
 ```py
 >>> B = Decimal(N) - phi + 1
+Decimal('346421925829807939293802368937250520517556856274496340681799239089291249765321270491576943807769029506276203354532585613211864922584150104378865213010402223028176347214857274743206460295173009790370214772536128777858755035911614561414990603406404984005947717445743706054221064913595294226503135333158697489990')
 >>> C = Decimal(N)
 ```
 
@@ -100,7 +108,9 @@ Check to make sure B and C are integers. If they're not, try using a higher prec
 
 ```py
 >>> p = (B + (B * B - 4 * C).sqrt()) / Decimal(2)
+Decimal('178187650567807686297508761669341068026596182918164336679269778091413760248796912297951278062644499145975246732979455707116872915963269648808994075794761810506203681312867668286737214808081540392248516550834072470288052831951959306342657446325786002900014749794262752196461389552859745880480150585554246119623')
 >>> q = (B - (B * B - 4 * C).sqrt()) / Decimal(2)
+Decimal('168234275262000252996293607267909452490960673356332004002529460997877489516524358193625665745124530360300956621553129906094992006620880455569871137215640412521972665901989606456469245487091469398121698221702056307570702203959655255072333157080618981105932967651480953857759675360735548346022984747604451370367')
 >>> p * q == N
 True
 ```