2229:"Because a^2 ≡ (n − a)^2 (mod n), the list of squares (mod n) is symmetrical around n/2, and the list only needs to go that high." This, as far as I am aware, is not provably true. I do know, though, that if a and n are relatively prime, this is absolutely true, and I have proven it myself. To avoid confusion, I changed it to say "Because a^2 ≡ (n − a)^2 (mod n) while a and n are relatively prime, ..." I add the citation needed because it is not immediately obvious, and there are citations that can be added. I have written up a proof in LaTex and compiled it to a PDF. Should I add this? If so, how should I cite it?
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1013:, only visible by admins). To answer your questions about article translation, most articles on different language projects are independently written, but there are some ad hoc translation projects. Please feel free to leave any other questions you have on my talk page, and remember to sign your talk page comments with four tildes (~~~~).
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and integer factoring are equivalent. It is clear that a polynomial time algorithm for integer factoring implies a polynomial time for QR. But how does the opposite direction work, i.e., how can we obtain a polynomial time algorithm for integer factoring from a polynomial time algorithm for QR? Could
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Modulo an odd prime number p there are (p + 1)/2 residues (including 0) and (p − 1)/2 nonresidues. In this case, it is customary to consider 0 as a special case and work within the multiplicative group of nonzero elements of the field Z/pZ. (In other words, every congruence class except zero (mod p)
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I think there may be a mistake in the section "Composite modulus not a prime power". The first sentence ends "... and the product of a residue and a nonresidue is a nonresidue (or zero)." But for example, modulo 6 the residue 3 times the nonresidue 5 is the residue 3. I think the statement would
579:
What's the point in writing "field of even characteristic" instead of "characteristic 2" ? AFAIK, the characteristic of a field is always prime, except if its 0, which is even, but for which the statement of every element being a square is wrong: char(Q)=0 i.e. even, but not every element of Q is a
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Yikes, definitely an error! I think I'll fix it by saying that it is hard to generalize the other cases (multiplication by a nonresidue) and give examples. I *think* that it works differently for even and odd moduli, but I don't know any reference off the top of my head and will exchew original
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for taking modular square roots can do composite modula and does not need factorisation of the modula. It's possible but not easy to take the two different square roots of 5, for instance, so that p*q can be factored. Therefore, does
Kunerth's algorithm show that taking the square root of a
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It is suspected to be outside of all three of the complexity classes P, NP-Complete, and co-NP-Complete. If it could be proved that it is in either NP-Complete or co-NP-Complete, that would imply NP = co-NP. That would be a very surprising result, and therefore integer factorization is widely
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This shows equivalency. Nageh is being too censurous reviewing the contribution. This contribution would absolutely have gotten into the wikipedia, without any legal problems, not too long ago. This is actually holding back a better understanding of the equivalency claim with the undoing.
1333:. Under prime modulus it says the convention is to exclude zero, and under composite modulus not a prime power it says that some authors include gcd = 1 as part of the definition, and has a footnote to one of the authors, and says that the wiki article does not follow that convention.
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That square root computation reduces to factoring is described in the beginning of section "Complexity of finding square roots". That factoring reduces to finding square roots is described in the indented paragraph of subsection "Composite modulus". Other than that, please provide
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Thanks, EJ. It's done. Another question (the documentation here is overwhelming, I know I've seen the answer somewhere but I can't remember where). There are a bunch of references at the bottome to foreign languges, most of which I don't know. How/when do articles get translated?
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Hi, I was intrigued when the article (and source) states that no-one has provided a direct proof that there are more residues than non-residues among the numbers 1,2..(p-1)/2. I think I've managed to do this - can anyone suggest where I should get this checked out?
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The definition in the article and in the appropriate article in the german wikipedia don't seem to match: while here we have "q is a QR (mod n) if it is congruent to a square (mod n)", the german article says "q is a QR (mod n) if it is congruent to a square (mod n)
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Yes, it is an intriguing claim, I've seen it in a couple of references: Landau and (?) Lemmermyer at least. I'm not an academic, I'm not sure what is the best way to get your proof looked at by an expert. 2 things come to mind: 1) try submitting it to the
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Dear
Virginia-American, welcome to Knowledge. Your rewrite is very nice. It is applaudable that you are seeking approval of other editors before making such a sweeping change, but generally you do not have to worry about notifying people, just
205:-problem is in NP, but not that it is NP-hard. As far as I can see, factoring directly solves the decision problem mentioned here, thus factoring would solve an NP-hard problem and be NP-hard itself. It is open, whether factoring is NP-hard.
1229:
QRs (mod n) For example, there are 11 QRs (mod 35) ". Now, 35=5*7, (5-1)(7-1)/4 = 6, not 11. There are 11 QRs (mod 35) according to the definition on this page, but 6 (relatively prime) QRs (mod 35) according to the german wiki definition.
1877:? In any case, I think this fact (about number of quadratic residues being about n/2 for primes n and much less for composite n) is worth mentioning in the article, because IIRC it has applications in primality testing and cryptography.
638:. First of all, the claim is being made for algorithms finding square roots (rather than the QR decision problem), and the reduction is randomized, so it gives a probabilistic poly-time algorithm for factoring, not a deterministic one.
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Gauss defines quadratic residue in the
Disquisitiones, art. 95 and explcilty includes 0. In article 96 he says that since 0 is always a residue he will exclude it from the discussion because doing so makes the theorems easier to
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My point is: A quick internet search revealed me that QRP is proved in UP ∩ coUP. If one can show that factorization is not in UP or coUP, doesn't it mean that QRP is easier than factorization, without implying P = NP? Thanks!
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The text states "if the modulus n is prime the
Legendre symbol (a|n) can be quickly computed using a variation of Euclid's algorithm." It would be nice if it were described, somewhere in Knowledge, how to do this quick thing.
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For the sake of organization, I think the table of quadratic residues should be at the bottom of everything else and not immediately follow the introduction. The table should be more like an appendix. Just a thought. —
2908:. But that symbol is not defined anywhere in the article. (And I have never seen that symbol used in the context of number theory.) So: I hope someone knowledgeable will modify this formula so that it is both
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I put it there for ease of refrence while looking at the theorems and examples. I think at the bottom the readers would be scrolling excessively. Is there some way to have the table open in a separate window?
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I know a bit of cryptography too, and there is just a claim in the article that there is equivalency. It is not shown, as my contribution in the article showed, that an equivalent square (that contained
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I'd love to see a copy, though I'm not sure what the best way to get me one would be. You can't put original research in the actual wiki, I dunno what the rules are for putting it on your own talk page
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Both problems are in the polynomial hierarchy (in fact, in NP ∩ coNP). Thus if P = NP, then they are solvable in polynomial time, and in particular, they are polynomial-time reducible to each other. —
2016:. The fact that the subgroup of squares in a direct sum is exactly the direct sum of the subgroups of squares in the summands is completely trivial, it follows from the definition of a direct sum. —
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I will also add stuff about how different the theory is depending on whether the moduls is prime or not (eg, one can solve x ≡ a (mod p), but mod a composite it is basically factoring the modulus.
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Ah, yes, you're right. Anyway, regarding the misunderstanding, I was so used to read statements like "if this can be shown it would imply P = NP" that I just read it again. ;) Thanks and cheers,
1743:.) Obviously, the number of squares in a direct sum is the product of their number in each summand, and exactly half of the elements are squares in a cyclic group of even order. Therefore, if
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composite modulus is possible without being able to factor the modulus. And that Rabin's assertion that the modular square root problem is equivalent to integer factorisation is not true.
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Hardy & Wright define quadratic residues as a subset of the numbers 1...p-1 (mod p). Ireland & Rosen explicilty include gcd(a, m) = 1 as part of the definition of a residue mod m.
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If there is variation in the definition we probably ought to introduce those variants somewhere in the article, attribute each one to a source, and describe why each one is useful.
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Instead of a link to "Distribution of quadratic residues" I will add a new section. Keeping the links to acoustics and crypto, and covering Jacobi-Dirichlet and Polya-vinogradov.
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Thanks for the explanation. There's just one part I'm unsure of: "Obviously, the number of squares in a direct sum is the product of their number in each summand"—is this by the
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This is a gross error, as it suggests that there'd be an easy way to exclude many potential prime factors for a given number. Apperently, even Lehmer fell for that error:
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I always wondered about an obvious discrepancy in
Schneier, "Appl. Crypt.", 2nd ed. p.251, where he says: "If n is the product of two primes, p and q, there are exactly
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934:. At any rate, it is sufficient to leave a note here on the article talk page, as anybody who cares about the article is supposed to have it on their watchlist. --
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is given with the input instance, the problem remains NP-complete (although currently I still don't understand it, I will look for
Manders and Adleman's paper). --
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Nageh undid a contribution I made to this article showing just how an algorithm for a square root of a composite modulus is equivalent to integer factorization.
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From the article: "The law of quadratic reciprocity says something about quadratic residues and primes." Surely we can do better? 02:12, 19 October 2005 (UTC)
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An intriguing fact about these two theorems is that all known proofs rely on analysis; no-one has ever published a simple or direct proof of either statement.'
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there will be a solution for any number that divides kn evenly. Let k' = kn/c, n' = c, which leads to x^2 = k'n' + a = k'c + a, and a is a residue of c.
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states that " is not as hard as factorization, but is thought to be quite hard." QRP cannot be harder than factorization, but I am not aware that it is
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I glanced over your rewrite as well, and it's well-written and comprehensive. Welcome and I hope your future contributions are as great as this one!
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I Disagree over equivalency between composite root square root algorithm and integer factorization being shown in the modular square root article.
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I guess the intended meaning was "we do not know any reduction of factoring to QRP, and chances are that none exists". Nobody knows how to actually
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Right, I was wrong. I didn't notice, that the reference to Garey & Johnson refers to this section. In the mean time, I thought, that a simple
2494:"...some authors add to the definition that a quadratic residue q must not only be a square but must also be relatively prime to the modulus n.
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But nothing in that section is labeled a "theorem". This would be easier to understand if the two relevant formulas were labeled "theorem".
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Along this line, it might be helpful to amateurs like myself to give an alternate defintion of "quadratic residue" in terms of algebra:
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that it is strictly easier, as this would imply P ≠ NP (inter alia). I reworded it so that it does not ostensibly make any such claim. —
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Is there something more that is known under less general settings? (Something like: the number of residues is at most n/4, when n is ) --
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in UP ∩ coUP. (This is in fact fairly easy to see using the facts that factorization is unique, and primality is polynomial time.) —
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2009:{\displaystyle (\mathbb {Z} /n\mathbb {Z} )^{\times }\simeq \bigoplus _{i<k}(\mathbb {Z} /p_{i}^{e_{i}}\mathbb {Z} )^{\times }}
1572:{\displaystyle (\mathbb {Z} /n\mathbb {Z} )^{\times }\simeq \bigoplus _{i<k}(\mathbb {Z} /p_{i}^{e_{i}}\mathbb {Z} )^{\times }}
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Following this convention, the multiplicative inverse of a residue is a residue, and the inverse of a nonresidue is a nonresidue.
2040:(Sorry for the long delay; I forgot about this!) You're right, the fact is indeed trivial. :) Thanks again for the explanation,
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Nope, there's not - otherwise vandals would be able to identify pages to vandalize that wouldn't get noticed (although there is
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where presumably residues are used as some kind of source of randome numbers. But the linked article says nothing about that.
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variables. So, if you have two possible roots modulo each prime, you get an exponential amount of possible roots modulo
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This is inconsistent. I'd just revert it, but I don't want to get into an edit war over something fairly peripheral.
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Although it makes things tidier, this article does not insist that residues must be coprime to the modulus."
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Kunerth's modular square root algorithm can do composite modula and does not need factorisation of the modula
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Undo by Nageh over new material for square root of composite modulus is equivalent to integer factorization
2547:- however, the Jacobi symbol is multiplicative. That means, if there's a residue modulo a composite, it is
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So the misunderstanding is resolved, but as an aside, (the decision version of) integer factorization
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The second statement could probably also be generalized, but I haven't studied it enough to be sure.
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be true provided the residue is coprime to the modulus n, but the section ends by ruling that out:
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2789:. Is there a published source, such as a peer-reviewed academic publication, for this material?
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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Now I scanned through the proof, I recognize, where my intuition was wrong. I considered
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Thanks for the kind words, Dcoetzee. Is there any way to tell who is watching a page?
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Both definitions are possible, depending on what is more useful in a given context. —
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I am thinking that the 1st statement could be replaced by the more general statement
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a is a residue of b iff there is a solution for x^2 = kb + a, using only integers.
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may be harder, but I think you could do that by an extension of the same method. —
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914:. Is there anything required, customary or courteous to do before I copy it over?
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How can every integer be a quadratic residue modulo 2? This is sheer nonsense!
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is divisible by 8. Counting also residues which share a common factor with
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Added a section about counting, also has a link with a nice derivation.
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They don't, AFAICS. The anon's change was misguided, though harmless. —
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I'm reverting the edit; relative primality has nothing to do with it.
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has a multiplicative inverse. This is not true for composite moduli.)
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Aaaahhh, sorry, I misread P = NP instead of P ≠ NP. Time to go home.
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Composite modulus not a prime powerThe basic fact in this case is
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On the other hand, if we want to know if there is a solution for
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Distribution of quadratic residues when p is congruent to 3mod4
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distinct prime factors, then the number of residues coprime to
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Question concerning the equivalence of QR and integer factoring
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Other editors here can probably help. Maybe you should ask at
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Which is the proper definition? Is, say, 5*5=25 a QR of 35?
337:{\displaystyle \exists x\leq c\,(x^{2}\equiv q{\pmod {n}})}
1635:{\displaystyle (\mathbb {Z} /p^{e}\mathbb {Z} )^{\times }}
469:. The NP-hardness result is derived from a reduction from
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I'm too lazy to dig up a reference, there may be some in
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Really? I wasn't aware of the implication. Why is that?
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I had said that every integer is a QR (mod 2); he said
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2983:Knowledge level-5 vital articles in Mathematics
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619:anybody give a hint or provide a reference?
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2781:Knowledge works with material that can be
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537:(at least, that's my new intuition). --
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598:The text claims that QR for composite
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1122:Move the Table of quadratic residues?
839:This is convincing. Thanks a lot. --
266:{\displaystyle \langle q,n,c\rangle }
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558:Indeed, that's my intuition too. --
106:This article is within the scope of
2089:Quadratic_residue#Composite_modulus
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49:It is of interest to the following
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235:: IOW, the set of all triples
231:You are wrong. The keyword is
1:
2926:00:47, 17 December 2015 (UTC)
2823:20:25, 16 December 2012 (UTC)
2799:17:10, 16 December 2012 (UTC)
2776:15:29, 16 December 2012 (UTC)
2695:01:45, 16 February 2012 (UTC)
2631:16:55, 14 February 2012 (UTC)
2600:American Mathematical Monthly
2592:14:58, 14 February 2012 (UTC)
2429:
2279:
2074:02:38, 19 November 2012 (UTC)
2024:10:56, 27 February 2009 (UTC)
1887:23:20, 26 February 2009 (UTC)
1865:11:09, 26 February 2009 (UTC)
1388:03:46, 26 February 2009 (UTC)
976:18:18, 29 February 2008 (UTC)
963:18:07, 29 February 2008 (UTC)
944:14:08, 29 February 2008 (UTC)
924:19:57, 27 February 2008 (UTC)
905:17:50, 26 February 2008 (UTC)
881:02:52, 26 February 2008 (UTC)
749:
702:
314:
181:I doubt, that this paragraph
120:and see a list of open tasks.
2993:C-Class mathematics articles
2949:19:41, 13 January 2023 (UTC)
2785:by reference to independent
2571:16:57, 21 October 2016 (UTC)
2209:18:07, 26 January 2010 (UTC)
2195:18:00, 26 January 2010 (UTC)
2179:17:50, 26 January 2010 (UTC)
2164:17:49, 26 January 2010 (UTC)
2148:17:40, 26 January 2010 (UTC)
2136:17:03, 26 January 2010 (UTC)
2122:16:51, 26 January 2010 (UTC)
2105:16:36, 26 January 2010 (UTC)
1683:{\displaystyle (p-1)p^{e-1}}
1343:21:31, 4 February 2009 (UTC)
1321:20:50, 3 February 2009 (UTC)
1308:19:59, 3 February 2009 (UTC)
1272:10:48, 3 February 2009 (UTC)
1257:10:17, 3 February 2009 (UTC)
1222:{\displaystyle (p-1)(q-1)/4}
1157:12:51, 6 November 2008 (UTC)
1141:18:57, 5 November 2008 (UTC)
1116:10:37, 6 December 2013 (UTC)
864:The text provides a link to
849:14:52, 5 December 2007 (UTC)
831:11:44, 5 December 2007 (UTC)
655:. Thus if you pick a random
629:10:10, 5 December 2007 (UTC)
2656:If there is a solution for
2539:17:38, 28 August 2011 (UTC)
2518:12:45, 26 August 2011 (UTC)
2083:as integer factorization???
1642:is a cyclic group of order
189:less than some given limit
3014:
1057:12:08, 10 April 2008 (UTC)
233:less than some given limit
2881:, this sentence appears:
2867:11:27, 8 March 2013 (UTC)
2844:01:30, 8 March 2013 (UTC)
2050:16:26, 4 March 2009 (UTC)
1875:Chinese remainder theorem
1042:19:07, 9 April 2008 (UTC)
1018:04:22, 1 March 2008 (UTC)
1004:01:18, 1 March 2008 (UTC)
667:which is different from ±
563:17:26, 16 June 2006 (UTC)
542:09:10, 16 June 2006 (UTC)
449:06:27, 16 June 2006 (UTC)
353:18:49, 14 June 2006 (UTC)
226:11:02, 13 June 2006 (UTC)
145:
78:
57:
2481:12:01, 15 May 2010 (UTC)
2241:19:24, 14 May 2010 (UTC)
1579:. Moreover, for a prime
1096:21:03, 7 June 2008 (UTC)
647:is odd, and it is not a
589:20:32, 21 May 2007 (UTC)
473:which actually requires
152:project's priority scale
2896:: In the next section,
2225:a^2 ≡ (n − a)^2 (mod n)
1849:, depending on whether
1724:{\displaystyle 2^{e-2}}
1285:Elemenary Number Theory
109:WikiProject Mathematics
2963:C-Class vital articles
2754:
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2010:
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1011:Special:UnwatchedPages
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2733:{\displaystyle r^{2}}
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2011:
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1397:looks as follows. If
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526:{\displaystyle \ell }
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210:Integer_factorization
36:level-5 vital article
2879:Dirichlet's formulas
2849:See the articles on
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477:to be a product of
2904:) uses the symbol
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2674:Regards, Charlie
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1349:Number of residues
1236:Thanks, -manfred
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636:Rabin cryptosystem
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363:many-one reduction
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193:, this problem is
101:Mathematics portal
45:content assessment
2873:Some improvements
2859:Virginia-American
2753:{\displaystyle m}
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2623:Virginia-American
2573:
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2531:Virginia-American
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2473:Virginia-American
1940:
1503:
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1335:Virginia-American
1300:Virginia-American
1260:
1243:comment added by
1149:Virginia-American
1106:comment added by
1088:Virginia-American
1076:I went on to say
1034:Virginia-American
1006:
992:Virginia-American
990:comment added by
955:Virginia-American
916:Virginia-American
907:
893:Virginia-American
891:comment added by
873:Virginia-American
611:{\displaystyle n}
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2877:In the section
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792:
789:
759:
754:
745:
742:
739:
736:
712:
707:
696:
692:
688:
683:
679:
670:
666:
662:
658:
654:
650:
649:perfect power
646:
642:
641:
637:
633:
632:
631:
630:
626:
622:
605:
593:
591:
590:
587:
583:
574:
564:
561:
557:
556:
555:
554:
553:
552:
551:
550:
543:
540:
536:
520:
495:
491:
476:
472:
468:
464:
460:
459:
458:
457:
456:
455:
450:
447:
443:
439:
420:
410:
407:
404:
401:
398:
384:
381:
378:
364:
360:
359:
358:
357:
354:
351:
348:is known. --
347:
324:
319:
310:
307:
302:
298:
289:
286:
283:
257:
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245:
234:
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206:
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196:
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176:
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168:
153:
149:
148:High-priority
143:
140:
139:
136:
119:
115:
111:
110:
102:
96:
91:
89:
86:
82:
81:
77:
73:High‑priority
71:
68:
65:
61:
56:
52:
46:
38:
37:
27:
23:
18:
17:
2934:
2913:
2909:
2897:
2893:
2892:
2889:
2885:
2883:
2878:
2876:
2832:
2766:
2762:
2711:
2708:
2705:
2677:— Preceding
2673:
2670:
2667:
2664:
2661:
2658:
2655:
2652:
2649:
2646:
2643:
2640:
2580:
2557:— Preceding
2552:
2548:
2542:
2529:
2500:— Preceding
2496:
2493:
2489:
2471:
2468:
2464:
2461:
2458:
2360:
2248:
2244:
2233:Raptortech97
2231:
2228:
2184:
2152:
2111:
2095:. Comments?
2092:
2086:
2080:
2060:
1854:
1850:
1752:
1748:
1744:
1736:
1732:
1731:elements if
1695:
1691:
1584:
1580:
1459:
1394:
1377:
1371:
1367:
1363:
1359:
1355:
1353:
1330:
1284:
1235:
1232:
1173:
1168:
1165:
1144:
1125:
1102:— Preceding
1099:
1085:
1075:
1069:
1066:
1031:
909:
884:
870:
863:
855:
818:
668:
664:
660:
656:
652:
644:
597:
578:
534:
474:
467:Blum integer
462:
441:
345:
232:
220:
207:
202:
200:
190:
186:
180:
172:
147:
107:
51:WikiProjects
34:
2791:Deltahedron
2361:Therefore,
2081:not as hard
1239:—Preceding
1063:Consistency
986:—Preceding
887:—Preceding
671:. That is,
195:NP-complete
123:Mathematics
114:mathematics
70:Mathematics
2957:Categories
2042:Shreevatsa
1879:Shreevatsa
1739:≥ 3. (See
1694:is odd or
1380:Shreevatsa
1366:even) or (
273:such that
2683:Wkcharlie
2526:research.
2506:Wkcharlie
1283:Landau's
1129:cosmotron
1028:Perfect ?
866:acoustics
169:Vagueness
39:is rated
2808:reliable
2783:verified
2691:contribs
2679:unsigned
2559:unsigned
2514:contribs
2502:unsigned
2087:Section
1735:= 2 and
1370:+ 1)/2 (
1362:/2 + 1 (
1318:Dcoetzee
1253:contribs
1241:unsigned
1137:contribs
1104:unsigned
1015:Dcoetzee
1000:contribs
988:unsigned
973:Dcoetzee
901:contribs
889:unsigned
416:divides
203:decision
2941:Endo999
2836:Gaohoyt
2811:sources
2768:Endo999
2250:Proof:
1462:, then
932:be bold
150:on the
41:C-class
2093:easier
1293:state.
841:Desi73
729:, but
621:Desi73
47:scale.
2815:Nageh
2604:ArXiv
2246:Huh?
2201:Nageh
2171:Nageh
2156:Nageh
2128:Nageh
2112:prove
2097:Nageh
1587:≥ 1,
1374:odd).
821:. --
471:3-SAT
208:From
197:; ...
28:This
2945:talk
2922:talk
2918:Daqu
2912:and
2910:true
2894:ALSO
2863:talk
2857:. -
2853:and
2840:talk
2819:talk
2795:talk
2772:talk
2687:talk
2627:talk
2588:talk
2567:talk
2535:talk
2510:talk
2477:talk
2462:QED
2237:talk
2205:talk
2189:Emil
2175:talk
2160:talk
2142:Emil
2132:talk
2116:Emil
2101:talk
2079:QRP
2070:talk
2046:talk
2018:Emil
1949:<
1883:talk
1859:Emil
1747:has
1583:and
1512:<
1419:<
1384:talk
1339:talk
1304:talk
1266:Emil
1249:talk
1245:Mz65
1153:talk
1133:talk
1112:talk
1092:talk
1053:talk
1038:talk
996:talk
959:talk
940:talk
920:talk
897:talk
877:talk
845:talk
827:talk
625:talk
586:Talk
539:GrGr
446:GrGr
402:<
223:GrGr
142:High
2555:.
2549:not
2435:mod
2285:mod
1799:or
1755:is
1690:if
1169:and
784:gcd
755:mod
708:mod
582:MFH
320:mod
2959::
2947:)
2924:)
2902:ij
2865:)
2842:)
2821:)
2813:.
2797:)
2774:)
2693:)
2689:•
2629:)
2590:)
2569:)
2537:)
2516:)
2512:•
2479:)
2413:−
2404:≡
2388:−
2318:−
2273:−
2270:≡
2264:−
2239:)
2207:)
2192:J.
2185:is
2177:)
2162:)
2145:J.
2134:)
2119:J.
2103:)
2072:)
2048:)
2021:J.
2002:×
1942:⨁
1938:≃
1933:×
1885:)
1862:J.
1807:φ
1763:φ
1714:−
1673:−
1656:−
1628:×
1565:×
1505:⨁
1501:≃
1496:×
1412:∏
1386:)
1341:)
1306:)
1269:J.
1255:)
1251:•
1203:−
1188:−
1155:)
1139:)
1135:|
1131:(
1114:)
1094:)
1055:)
1049:EJ
1040:)
1002:)
998:•
961:)
942:)
936:EJ
922:)
903:)
899:•
879:)
847:)
829:)
823:EJ
799:±
743:±
689:≡
627:)
560:EJ
521:ℓ
492:ℓ
485:Θ
396:∃
391:‖
388:⟩
376:⟨
350:EJ
308:≡
287:≤
281:∃
261:⟩
243:⟨
212::
2943:(
2920:(
2916:.
2906:∧
2884:"
2861:(
2838:(
2817:(
2793:(
2770:(
2748:m
2726:2
2722:r
2685:(
2625:(
2615:.
2606:.
2586:(
2565:(
2533:(
2508:(
2475:(
2443:)
2440:n
2432:(
2424:2
2420:)
2416:a
2410:n
2407:(
2399:2
2395:)
2391:a
2385:(
2382:=
2377:2
2373:a
2342:2
2338:a
2334:=
2329:2
2325:)
2321:a
2315:(
2293:)
2290:n
2282:(
2276:a
2267:a
2261:n
2235:(
2203:(
2173:(
2158:(
2130:(
2099:(
2068:(
2044:(
1998:)
1993:Z
1985:i
1981:e
1975:i
1971:p
1966:/
1961:Z
1957:(
1952:k
1946:i
1929:)
1924:Z
1920:n
1916:/
1911:Z
1907:(
1881:(
1855:n
1851:n
1835:1
1832:+
1829:k
1825:2
1820:/
1816:)
1813:n
1810:(
1785:k
1781:2
1776:/
1772:)
1769:n
1766:(
1753:n
1749:k
1745:n
1737:e
1733:p
1717:2
1711:e
1707:2
1696:e
1692:p
1676:1
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1666:p
1662:)
1659:1
1653:p
1650:(
1624:)
1619:Z
1613:e
1609:p
1604:/
1599:Z
1595:(
1585:e
1581:p
1561:)
1556:Z
1548:i
1544:e
1538:i
1534:p
1529:/
1524:Z
1520:(
1515:k
1509:i
1492:)
1487:Z
1483:n
1479:/
1474:Z
1470:(
1460:n
1442:i
1438:e
1432:i
1428:p
1422:k
1416:i
1408:=
1405:n
1395:n
1382:(
1372:n
1368:n
1364:n
1360:n
1356:n
1337:(
1302:(
1247:(
1217:4
1213:/
1209:)
1206:1
1200:q
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819:n
805:)
802:b
796:a
793:,
790:n
787:(
763:)
760:n
752:(
746:a
740:≢
737:b
716:)
713:n
705:(
697:2
693:a
684:2
680:b
669:a
665:b
661:a
657:a
653:n
645:n
623:(
606:n
584::
535:n
501:)
496:3
488:(
475:n
463:n
442:n
424:}
421:n
411:p
408::
405:c
399:p
385:c
382:,
379:n
373:{
346:n
332:)
328:)
325:n
317:(
311:q
303:2
299:x
295:(
290:c
284:x
258:c
255:,
252:n
249:,
246:q
191:c
187:x
154:.
53::
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