638:. Express LM≡0 (mod 7) in terms of a and b, then we have: (i) if (a-b)≡0 mod 3, then (a+3b)(a-b)≡0 mod 7; (ii) if (a+b)≡0 mod 3, then (a-3b)(a+b)≡0 mod 7; (iii) if 2b≡0 mod 3, then 4ab≡0 mod 7. (i) is the same as: either (a-b)≡0 mod 21, or { (a-b)≡0 mod 3 and (a+3b)≡0 mod 7 }, and the thing inside the curly brackets can be rewritten as ((a-4b)≡0 mod 3 and (a-4b)≡0 mod 7), i.e. (a-4b)≡0 mod 21. Similarly, (ii) means either (a+b)≡0 mod 21, or (a+4b)≡0 mod 21. (iii) is the same as b≡0 mod 3 and ab≡0 mod 7, hence either b≡0 mod 21, or (b≡0 mod 3 and a≡0 mod 7). Now we can see that (7|p)
84:
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6^3=221-5≡-5). The condition (#2) is true for p=13, as a=1, b=2, 7|(a-4b), but it should not. By the way, according to the section titled Other theorems, 7 is a cubic residue mod p iff LM≡0 (mod 7), which is correct both for p=19 (L=7, M=1, LM≡0) and for p=13 (L=-5, M=1, LM≢0). Since 7 is also cubic residue modulo 19 (a=4, b=1), 73 (a=5, b=4), 181 (a=13, b=2), 313 (a=11, b=8), 367 (a=2, b=11)..., the right condition seems to be
682:
7|(a±4b), or when the modulus is 13, 19, 31, 73, 103, 139, …. In this case, 7 is truly a cubic residue only if 3|(a±4b), or modulus = 19, 73, …. Other moduli of this type (13, 31, 103, 139, …) make 7 an “Euler pseudo cubic residue” that is a nonresidue (I complained about modulus=13 satisfying (#2), but 13 is just one of these guys). 157 is the smallest non 7|(a±4b) type modulus that makes 7 a cubic residue.
694:(mod 13), resp. The coefficient of M seems to be μ=9r/(2u+1), which can be easily calculated if you first determine u and r such that 3u+1≡r^2(3u-3), u≠0, 1, -1/2, -1/3. For q=11, (u,r)=(3,±3), (9,±5), (10,±2) satisfy the condition, each giving μ=±4. Similarly, μ=±1 for q=13. For q=17, μ=±3,±8, and we could write this as LM(L-3M)(L+3M)(L-8M)(L+8M)≡0; and so on. For now, I’ll just comment out (11|p)
22:
693:
in this article are LM(L-3M)(L+3M)≡0 (mod 11) and LM(L-2M)(L+2M)≡0 (mod 13), respectively. These are exactly what
Lemmermeyer has in his book (p. 212), but again, I think he is wrong. I haven’t checked this carefully yet, but the correct expressions seem LM(L-4M)(L+4M)≡0 (mod 11) and LM(L-M)(L+M)≡0
479:
And that’s the current version as of writing this (June 17, 2013). Though (#2) happens to be correct for p=19, it is worse than (#1). For example, 7 is a cubic nonresidue mod 13, because (I think) only ±5 is non-trivial cubic residues modulo 13 (2^3=13-5≡-5, 3^3=26+1≡1, 4^3=65-1≡-1, 5^3=130-5≡-5,
380:
I agree that the current presentation is too abrupt. In a more gentle version the notion of "cubic residue" should be defined separately, and there should be a better explanation of the term "reciprocity" in this context. The lede mentions cubic equations, a term that does not recur in the article,
681:
Actually, the 7|(b±2a) at the very end seems meaningless, because it is equivalent to 7|(4b±a) — i.e. if 4b±a≡0 mod 7, then a≡∓4b, hence 7|(b±2a); conversely, if b±2a≡0 mod 7, then b≡∓2a, hence 7|(4b±a). So (#2) and (#2') are equivalent, and Euler’s criterion often gets a false positive when
626:
residue for a modulus that satisfies (#2). The first mistake of
Lemmermeyer is, he misquoted if as iff. The second mistake is, he swapped a and b. Those two mistakes totally messed things up. Anyway, now we have the correct version of Euler’s conjecture for q=7, and this conjecture is
621:
In translation I replaced p with b, q with a, so that the expressions are compatible with ours. As you can see, he simply states that if 7 is a cubic residue modulo 3b^2+a^2, then a and b satisfy (#2). He does not say the converse is also true. Namely, 7 can be a cubic
983:
I think that there is no generalized way to calculate whether m is n-th power residue mod d (i.e. x^n == m mod d has solutions) for n = 5, 7, etc. i.e. n is not divisor of 24, since there is no algebraic solution for a generalized algebraic equation with degree n for n:
331:
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The above is untrue for p=19. Namely, 7 is a cubic residue mod 19, because 4^3=64≡7 (mod 19), when 19=4^2+3*1^2, a=4 and b=1; for which (#0) does not hold. Since it is unlikely that Euler was wrong for p=19 (the second smallest
238:
491:…much simpler than (#0)-(#2). For the time being, I will just comment out this part for q=7, because (#0) is wrong and (#1)-(#3) are OR. I think someone needs to check Euler’s original work for this. —
764:(To make matters worse, there is also no explanation of its meaning, and no link to any explanation. But please do not think that a link to an explanation would have been a good idea: It would not be.)
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In the text it is written "...cubic reciprocity is most naturally expressed...". Is there some other definitions, since it seems I need one "not naturally expressed".
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Sometime before 1748 Euler made the first conjectures about the cubic residuacity of small integers, but they were not published until 1849, after his death.
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I think that there is no generalized way to calculate whether m is n-th power residue mod d (i.e. x^n == m mod d has solutions) for n = 3, 4 but not 5, 7
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989:, thus for n=8 there is also a way, therefore, I think that there is a way if and only if n is divisor of 24 (possibly 48 instead of 24). ——
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This article does not clearly state an algorithm to find the cubic residue to satisfy x^3=p mod q. Is there any algorithm available?
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I rewrote this based on my article on quartic reciprocity. I think the abruptness has been removed. I gave it a B+ rating.
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467:(#1) is indeed better than (#0), or so it seems to me at least, but still incorrect for p=19, etc. On 23 June 2012,
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4, however, for n=6 it is only needed to test n=2 and n=3, thus there is still a way, and there is an article of
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Is the writer truly unaware that the word "primes" could refer to the primes of the unique factorization domain
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Euler’s original version is basically (#2), except he does not say “if and only if”. He only says:
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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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or 21|(a±4b)”. When I first wrote it here, I accidentally dropped 21|(a±b). With that fixed—
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326:{\displaystyle \alpha ^{(P-1)/3}\equiv \left({\frac {\alpha }{\pi }}\right)_{3}\mod \pi }
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There is nothing wrong with using the word "residuacity" except for the fact that
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another editor, Maxal, changed this condition again, perhaps accidentally, to:
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Let p=a^2+3b^2 be a rational prime ≡1 (mod 3). According to
Lemmermeyer’s
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writing in
Knowledge is for the purpose of making things clear to readers
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21|b, or (3|b and 7|a), or 21|(a±b), or 7|(4b±a), or 7|(b±2a) … (#2)
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Virginia-American tried to fix this problem, by replacing (#0) with:
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mathematically correct too, provided that 3b^2+a^2 is a prime : -->
419:, p. 223, Euler guessed that 7 was a cubic residue modulo p iff
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which would be somewhat analogous to Euler's criterion for the
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There's nothing wrong with using the word "residuacity" except
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Another thing. Currently the expressions we have about (11|p)
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Tractatus de numerorum doctrina capita sedecim, quae supersunt
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702:. I’ll update the article more properly when I have time. —
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A footnote says: “an apparent misprint has been corrected”.
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7. First of all, (#3) should be “21|b, or (3|b and 7|a),
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21|b, or (3|b and 7|a), or 21|(a±b), or 7|(a±4b) … (#2')
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21|b, or (3|b and 7|a), or 21|(a±b), or 21|(a±4b) … (#3)
642:, iff LM≡0 (mod 7), iff (i) or (ii) or (iii), iff (#3).
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will hold too, and an apparently even weaker condition
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So I found the original Latin text on the
Internet:
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99.9999% of readers will have no idea what it means
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109:and see a list of open tasks.
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937:{\displaystyle \mathbb {Z} }
822:{\displaystyle \mathbb {Z} }
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406:23:15, 5 December 2008 (UTC)
355:06:26, 27 October 2006 (UTC)
345:18:45, 24 October 2006 (UTC)
944:that was just mentioned???
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411:Euler’s conjecture for q=7
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375:01:25, 16 April 2008 (UTC)
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Text is available under the Creative Commons Attribution-ShareAlike License. Additional terms may apply.
