Liber primus is an RSA code!
Is useless try to use bruteforce or other kinds of decryption, it's already been used without results.
3301 gave us hints with "all primes are sacred" and "totient function is sacred", it's clearly linked with RSA cryptography.
So, I believe that the keys are already given by 3301 but we cannot able to see or using wrong approach.
I really like to share some ideas about this point of view.
Bruteforce Software & Programs Section Edit
This is a tool I programmed to find words with a repeated letter in a specific position, or even a sequence with different letters. Note : You need Linux to run it. Setting up a bootable USB with Linux takes 10 minutes. Check this for instructions :
Tuto for Linux/Ubuntu/Backtrack ...
Launch a terminal from the same folder containing the program, write ./Prog then press enter if you want to use the 1-sequence mode, or ./Prog2 for the 2-sequence mode.
The more people on it the better results we get.http://paf.im/H0rBI
Update 1.0 : This tool will allow you to find for example words like : xxxSS, xxBBxxxx, xxxOUxxxx, xABxxxxx ... You have to enter the length of the word, the sequence you're looking for, and its position in the word. The tool will give you all the words with this combination. For this beta version you can use only 360 000 words, I will work on it in order to use the whole dictionnary. Notice that it is very optimised, it will take no time. Update 1.1 : - Some bugs now are corrected - Optimization ++ - 3 new modes 1 : Fixed length / Fixed position 2 : Free length / Fixed position 3 : Fixed length / Free position 4 : Free length / Free position Update 1.2 : - 2 new modes : 5 : One character / Fixed position / Fixed length 6 : One character / Fixed position / Free length Update 1.3 : - 8 new modes for 2 sequences - You have many combines for the searched sequences based on the length of the word containing the sequences (fixed and free), the order of sequences (fixed = one before the other, free = all possibilities for them) and their position (Fixed seq_1 at position X and seq_2 at position Y). PS : A Bruteforce code may come in the next updates PPS : Next step the Bruteforce program
The code starts being heavy, I have included the 2-sequence search option (with many modes like for the one sequence program) in a new program.
PS : You don't have to pay the guy for his program "DECPRYPTIONIST", since this one does the same thing, whatever ... (EDIT: decryptionist is much more powerful... anyway this is a quite ok free alternative)
EDIT : Instructions on how to help us with runes guessing and what we are looking for based on my conversation with Cicada_Solver and XDDD dude
For example : ᚳ ᛁ ᚱ ᚳ ᚢ ᛗ ᚠ ᛖ ᚱ ᛖ ᚾ ᚳ ᛖ - C I R C U M F E R E N C E - this 13 letter word was not encrypted at all ( all it took is to look at the gematria and replace runes with letters) . There are only 4 words that long in the entire book :
ᛇᛈᛋᚢᛚᚪᛈᚢᚳᛖᚠᛞᛉ - (EO ) P (S/Z) U L A P U (C/K) E F D X
ᛖᛞᚪᚫᛏᚩᛠᛖᛠᛉᚳᛠᛏ - E D A (AE) T O (EA)E (EA) X (C/K) (EA) T
ᛈᚳᛇᚢᛏᚳᛡᛇᛝᚾᚢᚻᚦ - P (C/K) (EO) U T (C/K) (IA/IO) (EO) (NG/ING) N U H (TH)
ᛒᛠᚠᛉᛁᛗᚢᚳᛈᚻᛝᛚᛇ - B (EA) F X I M U (C/K) P H (NG/ING) L (EO)
ᚳ ᛁ ᚱ ᚳ ᚢ ᛗ ᚠ ᛖ ᚱ ᛖ ᚾ ᚳ ᛖ ᛋ - C I R C U M F E R E N C E S - this 14 letter word was also not encrypted . There are 3 more words that long in the book :
ᛒᚷᛞᛉᛗᛒᛉᚳᛝᚦᚣᛞᚫᛠ - B G D X M B X (C/K) (NG/ING) (TH) Y D (AE) (EA)
ᛠᛁᛡᚦᛝᚾᛖᚾᚠᚩᛗᛖᚣᚪ - (EA) I (IA/IO) (TH) (NG/ING) N E N F O M E Y A
ᛏᚠᛂᚱᚹᚠᛋᚾᚹᛂᛖᛒᚢᚦ - T F J R W F S N W J E B U (TH)
The explanation from Cicada_Solver : Let's take the word "CIRCUMFERENCES". If we assume the word repeats itself in the book and you think that after encryption it gives ᛏᚠᛂᚱᚹᚠᛋᚾᚹᛂᛖᛒᚢᚦ - T F J R W F S N W J E B U (TH) then you have to guess which opetations have been applied on the word ᚳ ᛁ ᚱ ᚳ ᚢ ᛗ ᚠ ᛖ ᚱ ᛖ ᚾ ᚳ ᛖ ᛋ ( CIRCUMFERENCES) with the gematria to have the result ᛏᚠᛂᚱᚹᚠᛋᚾᚹᛂᛖᛒᚢᚦ ( T F J R W F S N W J E B U (TH) )
When you have an idea try it on other words and correct your guess since you can verify the other possibilities (if you detect a rune, then use cicada_breaker to guess the others one by one and remember each rune reduce the possibilities from n! to (n-1)!).
If everyone try this, by the end we would have lots of runes guessed (wrong and maybe right) but it is the same operation so we all think about the results and a dozen of minds are much better than one for this task. By comparing the results we can be sure about some runes.
Don't forget that an operation should be bijective and should be able to be reversed otherwise the operation of encryption/decryption could not be executed, so Cicada has necessarily used bijections to encrypt Liber Primus.
And when we have some runes guessed we can launch the bruteforce program.
Anybody that wishes to help - please download the software and try this method . The more people on it the more info for bruteforcing.
I have 29! possibilities which is about 8.841762*10^30 permutations, if we suppose that there is no other operations added on the other pages, this is just not possible to calculate even with super calculators I've got in labs (in which case it would take years before having the solution). It is a factorial-complexity so it is O(n!) for 2X variables we need 1000 years to find. (For n = 20 it takes about 800 years)
In mechanical engineering, we deal with problems like this one since the inversion of a matrix needs the calculation of determinants and the law looks like this one O(n!), we transform the system to another form having a polynomial-complexity O(n^p) then we have our solution in no time.
The Liber Primus problem is easier cuz we have words and sequences (this is its Achilles' hell), which can be guessed cuz they are comprehensible not like numbers and results of a calculation for determinant (In CFD and FEM the dimensions of the problem give us n = 10^9 easily ... so it is 10^9! operations).
I maybe repeate what I've said, but it is important to guess runes with the cicada_breaker V1 I gave you, it would reduce the possibilities to 2.092279*10^13 possibilities and this is rational to bruteforce in labs.
When I said statistical studies and probabilities, you have to take the decrypted pages, take the words reapeated and look for them in the other pages since lenght is known. What you should do is to take long words, write them in runes and copare them with long words from Liber Primus, if you figure out about the permutations applicated on the word, each time you reduce the possibilities from n! to (n-1)! and so on, and this is why I said that is ok if the runes are wrong, the calculation doesn't take time and the worst thing that could happen is having a wrong text. By the end we eliminate the "lenght-of-the-word" possibilities, and since the word is a finite space we can try all the other possibilities by hand.
The only solution to bruteforce is to cut the gematria table to 1X - 1Y permutations, 1X to guess and 1Y to bruteforce.
Guys this is a mathematical problem, we should deal with it with maths if you don't find the key fixed by cicada.
I cannot be more clear than this time.
PS : Do not forget, I assumed that there is only one opperation applied on every rune.
Dr B 67
what if they knew that whoever would decode this would have dyslexia, and so Liber Primus would automatically be decoded in their head as Libra Primary (as in the constellation Libra) libra is the scales so ''liber primus is the way'' would only mean [balances is the way] as for ''beware false paths'' Scorpions are a symbol of treachery, but libra is the symbol of justice idk... but some key words would be (balance,scorpion,claw) and key numbers would be 2.6/ 2.7/ 3.9/ 4.5/ 4.9/ 5.2/ 5.7/ 5.9/ 6.1/ 6.5/ 9.4 only got this much out of 30min OCD but you might be able to do something with it, but beware i got no clue so...yeh
if balance is the key, and this Libra is the way then during the equinox where day and night are equal follow where the Libra constellation is pointing.
Python Program for Solving Runes: Edit
A collaborative effort to get as many different cyphers put into the program: https://github.com/Be5haram/CICADA2K16/blob/master/RuneSolver.py
Important notice for Primus solvers Edit----------------------------------------------------------------------------------------------------------------------------------------
UPDATE: Two members of our community are currently working on decryption tools (one - a dictionary word search, the other - a Vigenere solver)
Pages 28 to 32 Discoveries and Decryptions Edit
Page 28 EditCharacters have been translated from Runic to Latin on page 28:
(Lines here correspond to lines on the page)
Anyone who has ideas about these words, please leave your thoughts in the comment section. Please understand, that after #, the coding system changes (# = 13 red dots).
EDIT: If you would like to use a word-searching software, please, help yourself
Actually this software is way better imho so I'm putting this one up as well :
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
EDIT: Thank you, <jacquerie> for pointing that pages were already transliterated. Please follow this link:
2016-07-14: Did some research and I was wondering about word #19 of page 28. Used several word finders on the web incl. bestwordlist and I couldn't find any match for words with at least 12 characters having a ????????X?XX structure. Can anyone confirm or disprove this?
2016-07-15: I just did the same before reading your statement. With the same result. There is no word in modern English with that structure.
I also isolated the two-letter-words and checked with the 23 modern English two-letter-words that are likely to appear in a written text. I might have made a mistake, but I think I didn't: There is no correlation that makes sense.
Furthermore: Some word structures do resemble Germanic or Romanic languages, but I do have a problem with the short words in this text. There are too many. Appart from English I don't know any language with that many short words (that are actually being used).
And why did the use the Ango-Saxon Futhorc instead of the way more popular Nordic alphabets? By the way, I'm pretty sure that the language used is NOT Old English, and no original text as well (as far as I know, they didn't have consonant gemination in AS Futhorc).
Last but not least I examined the drawings to decide whether they contain cipher/cryptic runes (Wikipedia for more information on that topic). I havn't found a system that I know of (in addition I've never heard of Futhorc cipher runes, only Futark). On the other hand you do talk about twigs and branches when dealing with runes, so maybe the trees do hold a key...
(Written by a linguist who used to professionally work with runes. So the thoughts should be somewhat qualified.)
Some interesting facts about the number 15: Edit
- The number of chapters in the book (whereas for "chapter" I mean a paragraph with a big red starting rune) is exactly 15.
- From page 41 to page 56 you can see some kind of "plants" at the left and right edges of the page. The right side plant, near the center, has 5 dots. The left side one, however, has only 3 (two of them being removed. It's easy to see if you zoom in). 3 and 5 are the prime factors of 15.
Pages 8-14 straws/saplings meanings & Discoveries Edit
Fibonacci Tree reference Edit
The saplings from page 8 and on represent the Fibonacci tree. The top of the sapling has exactly 13 branches, which is the number you'd expect from a Fibonacci tree like the one on the right. To find more: http://oeis.org/A000045/a000045.html
So the highlighted branches are: 2-3-5-6-8-10-11-13. Anything up with this? Branch 2 and branch 5 make me think of runes somehow.
There are 58 pages in total, labled 0-57. 57 is the most accurate integer that can be squared to reach a number closest to 3301. It cannot be a coincidence.
This leads me to believe that the page number is extremely relevant to solving the next stage of this puzzle. Seeing as page 57 is unencrypted, I believe that we are looking for an encryption where the page number (x) can be substituted with 57 (or 58) to equal 0, since the 57th (or possibly 58th if you count the number of pictures) runic page is unencrypted.
Calculating the number 3301 in Base58 gives "ur," which is a Sumerian city. This supports the idea of Cuneiform, and also adds the possibility of x being 58.
I will continue to post my findings.If you add up all the numbers of the pages in the runes you get 0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57 = 1653 which is the 57th triangular number, if you add up its numbers then u get 15.
If you add up the numbers in the runes individually like this 0+1+2+3+4+5+6+7+8+9+1+0+1+1+1+2+1+3+1+4+1+5+1+6+1+7+1+8+1+9+2+0+2+1+2+2+2+3+2+4+2+5+2+6+2+7+2+8+2+9+3+0+3+1+3+2+3+3+3+4+3+5+3+6+3+7+3+8+3+9+4+0+4+1+4+2+4+3+4+4+4+5+4+6+4+7+4+8+4+9+5+0+5+1+5+2+5+3+5+4+5+5+5+6+5+7 = 393 which if u add up its numbers u get 15 too. If you reverse 393 u get 393 which is not prime, but it is only divisible by 11 and 53 which are both prime (without counting 1 and itself).32 is the closest number that if you square it you get almost 1033, and 57 is the closest to 3301, which is the total number of names of pages. Also if you reverse 1653 and 393 and do 3561 - 393 you get 3168 which is 133 away from 3301, which is 1033 without counting its 0, and their two factors are prime, 7 and 19, 7+19 = 26. 133 is also a polygonal number of octagonal(7). 7 appears both when you add up the numbers in 3301/1033/133 and 133 in its polygonal. The 133rd prime is 751 which I really like because its digits appear almost everywhere in their relationships with other numbers, the 751 prime is 5701, the octal of 751 is 1357 (going up 2 by turn and all primes except 1), its duodecimal is 527 (close tho).
32 + 26 = 58 which is the total amount of runes. 32 is a hamming number and its prime factorization is 2*2*2*2*2. The 32nd prime number is 131, which adds up to 5, same as 32. The factors can be found by multiplying the first factor (1) by 2 until you get to 32 (1, 2, 4, 8, 16, 32). If you reverse 32 you get 23 which is also a prime, the 9th prime. The base8 of 26 is 32 which I think isn't a coincidence. A funny thing about 26 and 101 (the 26th prime) is that in 26 between the base8 and the base12 have a difference of 10 (octal 32, duodecimal 22) and 101 has a difference of 20 between the base12 and the base16 (duodecimal 85, hexadecimal 65). Also since 32 is the closest to 1033 in squaring there could be something, but 32 squared is 1024 which if you add its numbers you get 7, same as 1033 and 3301.
From this I think it is aiming at page 32 and maybe 26 or something like that. And by the way in this titanpad https://titanpad.com/vFCy7T5p0O with the runes and its transcriptions in a few pages there are
(13) at the end or in the middle, the pages with that are 26, 32, 56, 3 (page number 3 has it 3 times), 53, 54. Also in page 32 there is a blurry tree which then only appears in page 55, but in 55 only appears the outline and in 32 it is the interior and blurry. Interesting fact about 32 + 26, it is equal to 58 but if you add 3+2 which is 5 and put it before 2+6 which is 8 you get the same answer which is 58, which is the total amount of pages.
And I forgot to say that in the bible the Genesis 32:26 is "And he said, Let me go, for the day breaks. And he said, I will not let you go, except you bless me." which has 101 characters (the 26th prime) and 23 words (32 reversed), I think that might not just be a coincidence and that it might also be pointing at pages 32 and 26.
Also there are 32 letters in the icelandic runes (see https://en.wikipedia.org/wiki/Icelandic_orthography) and there are also 26 letters in the alphabet, I don´t think this is a coincidence either. When Q, V and X are excluded, there are 32 letters in the Polish alphabet too (https://en.wikipedia.org/wiki/Polish_alphabet) maybe this is just to decipher the runes in the pages or something.
I want to add to the Pabster's edit that 32th page is linked to the 2016 puzzle image, both contain the tree.
Nothing new for long time. I guess it is over this time.
- These are just ideas.
- This is regarding the amount and number of pages of runes.
- 58 pages named 0-57.
- 3301/57 = 57.9122807018 which makes it the closest to its root
- 1033/32 = 32.28125 which also makes it the closest root
- Maybe something to do with pg 32?
- 5 + 8 = 13
- 3 + 2 = 5
- 5 + 7 = 12
- All add up to 30
- Maybe something?
- The factors of 57 are 3 and 19, both primes.
- Factors of 32 are 2, 4, 8 and 16. Nothing special I think.
- 3 + 19 = 22
- 2 + 4 + 8 + 16 = 30
- 30 + 30 (from the above section) - 22 = 38
- 3 + 8 = 11 prime
- 1 + 1 = 2 also prime
- Maybe something?
- Any ideas or do you think this isn't anything, just add up to this or dismiss it.
So, I was thinking about it and i think their may be a correlation with each individual page adding up to a certain number and then either being, or creating two prime factors. Each page must lead to certain numbers that, when arranged correctly are an encryption key of some sort possibly in order from front to back, but it wouldn't be a long shot to say that its just another encrypted puzzle. It wouldn't be a longshot to say that it spells out another link in the deep to visit. Hopefully we'll get more information but this is just my take after mulling over it for a while. But i think its just another code to crack. I think once we figure out each pages hidden numbers we'll figure out whether it is or isn't another riddle, then we'll be well on our way to the end of this one, if they're even still participating...
3301 gave a hash of a new onion link, saying that we should find it; the hash was broken into 5 lines; putting one after another and passing the full string (128 bytes = 1024 bits) to "hash-identifier" returns it is likely sha512 or Whirlpool; 3301 uses sha1, so I would guess it is not Whirlpool;
knowing it was a long shot, I created sha1 hashes of each outgessed data; then I took the first character of each line, added "onion" to the end and generated the sha512 hash of it; then the same with the second character of each line and so on (made a script to do it ;-)); none of the sha512 hashes matched the hash given by 3301;
bruteforcing the sha512 hash with hashcat would take several months, if not years, even with the community working together;
----- EDIT_3: PLEASE. It's been over 2 years so FORGET THE OUTGUESS DATA. You can replicate same kind of results very easily by creating similar pages and trying to embed outguess data on them. The steganography artefacts Outguess creates are very visible and apparent in the image, especially at the amount you suppose it would yield. The data is just palin random.
You can reverse engineer this effect by trying to embed the same amount of binary data on a similar image to the pages. The outguess data is a coincidence and the similar headers it gives is due to the fact that all the pages are alike with a lot of whitespace. I've gone this route. The solution lies elsewhere and even if you don't believe me, you can replicate my test results very easily in 2-3 minutes. Put your efforts to something useful.
Page 57 QR Scan Edit
<a href="tel:0421812877725">0421812877725</a>. (scanned from mayfly, most likely redlaser glitch)
digits are most likely glitch with red laser. Saved just in case.
Regarding pages 49, 50, and 51... Edit
Note: I'm new here. I just noticed a few things when reading through this. I hope it can help.
"Where 4F corresponds to 255": RGB color values range from 0-255. Also mentioned that 4F is maximum value seen in the pages. Could it be reference to colors
--A note on this. If 255 refers to the RGB color system, 4F is equal to 79 in this system. 79 is a prime number.
Also, sexagesimal number systems were developed by the Babylonians - this is closely related to 360 days/year. If I counted correctly, the 255th day in a non-leap year is October 13th. Leap years it is October 14th. Does this have anything to do with it?
I didn't see this covered anywhere, so I put it here. Should this be a comment? I dunno. -Gopher
4F corresponds to 255 if you think in base 60. It starts from 0? to 4F, so we have something to 255. ASCII? Colours? Maybe it is a black and white image.
Final page possible clue? Edit
I haven't been following this for long but I thought about the possibility that the final page might contain important information so I opened it in notepad. I'm sorry if it's not much but I found a copyright in it and am wondering if it could potentially be a clue. It said "copyright Artifex Software" so maybe they set it up or something.
Artifex Software sites:
Artifex is the software they used to create the PDFs, nothing special there.
Regarding 3301 referring to Babylon and Ur Edit
After some research I noticed that the characters from the rune pages looked strikingly like Babylonian text. I'll look into that and see if I can decrypt something from the document.
(17 13 55 1) 17*60+13=1033 55*60+1=3301http://it.stlawu.edu/~dmelvill/mesomath/Bignums.html
Another instance of 15 Edit
Probably a coincidence, but of note.
Research on apostrophe symbol on page 35 Edit
Since apostrophes' usage are limited in English, I found this:
PE'T in the beginning of a sentence on 4th line.
I guess this is IT'S. And then calculated the shifting, I got this numbers (3,2,1).
Could this be a hint on the encryption method?
Some observations on the OutGuess data Edit
- The length of header and footer increases as the page number increases.
- All header/footer overlapping part are the same.
- Each file has the same length, roughly 58KB (and there are 58 pages)
Some thoughts: because the file is unreadable, and header length/ footer length is changing. The only thing I could come up with is XOR. We know that a XOR a = 0 and 0 XOR a = a. Finally all evenly overlapped headers/footers will be canceled to 0. If this still not going to give us any hint, I will quit... (Because the last possibility is that this whole shit was created by some cipherpunks who prefer it will never be decrypted :(
Outguess often outputs false positive outputs.
Liber Primus in gematria values Edit
Here is link to unsolved pages of Liber Primus in gematria values. http://pastebin.com/NrLjVjdq
Low occurrence of two-same-rune bigrams Edit
After a lot of analyzing of runes in unsolved pages of LP only conclusion we figured out is that LP is not coded by simple short key vigenere cipher (or monoalphabetic cipher).
Analysis of two grams (pairs of runes) is the only deviation from randomness of other analysis results. It shows strange and unexpected low number of rune pairs with two same runes. Somehow key/cipher produces very few same rune pairs in ciphertext. Same runes 2-grams happen but are very rare.
Its almost impossible that such low occurrence (86 same rune 2-grams in ~13.000 runes) would happen randomly using standard polyalhabetic cipher with most stream shift keys. Algorithm or stream key for some unknown reason cause this. This is so far only solid clue towards cipher/key found in runes.
And we can reproduce similar results with made up enciphering algorithms created to deliberately cause same thing. Such algorithms include previous rune gematria position or value when calculating shift for enciphering. Note that cipher doesnt totally exclude 2 rune repetition, it only makes it happen ~5 times less likely as any other 2 different rune 2-gram.
In the 58 pages there are 89 (prime) repeating pairs, and their Gematria values sum to 4337 (twin prime). They appear evenly spread throughout the pages, with some clusters on certain pages. When placed in order they appear in the text there is a suggestive "W,W,W" 7 repeats from the end... For more info see:
Frequency analysis of runes in Liber Primus by chapters Edit
Nothing unexpected, distribution of runes is close to random distribution. Google doc sheet
Hcaltroms version: http://imgur.com/D7wpFZG
IMPORTANT UPDATE: we found patterns in Frequency analysis, for more info: http://uncovering-cicada.wikia.com/wiki/Frequency_Analysis_(Liber_Primus)
3 ways how to apply key to runes in cipher text Edit
Often when encrypting / decrypting, you want to combine a message-rune with an encryption key to give a cipher-rune. The encryption key can be expressed as another rune. Therefore, it is informative to investigate possible ways that this can be done. Presented below are some common examples. The runes have two numbers that can represent them, their position (pos) in the Gematria Primus or their rune-prime-equivalent (pri). So there are four possible combinations, (where r is any rune)...
Spredsheet for decoding Liber PrimusEdit
Guide on how to use the spreadsheet:
- Sheet2 is the Gematria converted to shifts. To create these shifts we use the Prime Counting Function ,Pi(p), on the prime numbers associated with each letter in the Gematria.
- page_x is the xth page of the Liber Primus. There are 58 pages total, beginning with a count of 0 we have pages page_0 to page_57
(*) There are many "hidden" pages in the spreadsheet for organization, right click on the bottom and select "unhide" imgur.com/asLy62L
(**) The visible pages are those that begin each distinct section of the Liber Primus. The current theory is that each section uses a different key, so these pages are the best to start.
- Top row is the original cipher text of the page. imgur.com/wZVmpq9
- Rows 3-32 are the shifts from the cipher text in row 1 imgur.com/UasnRcL
- Rows 35-64 are the english letters corresponding to the shifts in 4. imgur.com/RQANPSJ
- Rows 66 - 70 are the ones to use. imgur.com/RHV0bYJ
The "Count" row denotes the index corresponding to the letter position on the page. The "Modifier" row is the Encryption Key - use this to try various key sequences. The Modifier (Key) is applied to the cipher text to obtain the "Decoded" row.
See page_56 and page_57 to see how these were decrypted. Page 56 uses the "Count" row to look up the index of the prime corresponding to Pi(count) and subtract it from 59, mod 29.
Ether8unny CICKADA tool Edit
As seen on TV!
Repeat of "DJU BEI" in Liber Primt Edit
We found that string " DJU BEI" or in runes "-ᛞᛄᚢ-ᛒᛖᛁ" repeats two times in Liber Primus. On 27.jpg and 55.jpg. Both times next character after DJU BEI is "AE" "ᚫ". But on page 27 its mid sentence while on page 55 it appears on next page in new paragraph.
This could mean that same key and same plaintext was used on that string or it could be just coincidence.
Links that can be usable investigating this: