Talk:Cubic reciprocity

Suspected error

I strongly suspect that

$\alpha ^{(P-1)/3}\equiv \left({\frac {\alpha }{\pi }}\right)_{3}$

should be replaced by

$\alpha ^{(P-1)/3}\equiv \left({\frac {\alpha }{\pi }}\right)_{3}\mod \pi$

which would be somewhat analogous to Euler's criterion for the Legendre symbol. DRLB 18:45, 24 October 2006 (UTC)

Yes indeed, fixed now: thanks. Richard Pinch 06:26, 27 October 2006 (UTC)

Natural? Maybe. Understandable? Less so

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". Vavlap (talk) 01:25, 16 April 2008 (UTC)

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, making this only more mysterious.  --Lambiam 04:49, 24 April 2008 (UTC)

Rewrite

I rewrote this based on my article on quartic reciprocity. I think the abruptness has been removed. I gave it a B+ rating. Virginia-American (talk) 23:15, 5 December 2008 (UTC)

Euler’s conjecture for q=7

Let p=a^2+3b^2 be a rational prime ≡1 (mod 3). According to Lemmermeyer’s Reciprocity Laws: from Euler to Eisenstein, p. 223, Euler guessed that 7 was a cubic residue modulo p iff
21|a, or (3|a and 7|b), or 21|(a±b), or 7|(4a±b), or 7|(a±2b) … (#0)

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 p > 7), probably Lemmermeyer misquoted Euler. When this was first quoted on Wikipedia, on 5 December 2008,[1] Virginia-American tried to fix this problem, by replacing (#0) with:
21|b, or (3|b and 7|a), or 21|(a±b), or 7|(4a±b), or 7|(a±2b) … (#1)

A footnote says: “an apparent misprint has been corrected”.[2]

(#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,[3] another editor, Maxal, changed this condition again, perhaps accidentally, to:
21|b, or (3|b and 7|a), or 21|(a±b), or 7|(a±4b), or 7|(2a±b) … (#2)

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, 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
21|b, or (3|b and 7|a), or 21|(a±b), or 21|(a±4b) … (#3)
…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. — Gyopi (talk) 11:48, 17 June 2013 (UTC)

So I found the original Latin text on the Internet: Tractatus de numerorum doctrina capita sedecim, quae supersunt,[4] chapter 11.[5] Euler’s original version is basically (#2), except he does not say “if and only if”. He only says:

Ut 7 sit residuum divisorque 3pp+qq, debet esse vel p=3m et q=7n, vel p±q=21n, vel 4p±q=7n, vel p=21m, vel p±2q=7n.
[Translation: In order that 7 is a residue and a divisor is 3b^2+a^2, it must be that either b=3m and a=7n, or (b±a)=21n, or (4b±a)=7n, or b=21m, or (b±2a)=7n.]

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 nonresidue 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 mathematically correct too, provided that 3b^2+a^2 is a prime > 7. First of all, (#3) should be “21|b, or (3|b and 7|a), or 21|(a±b), or 21|(a±4b)”. When I first wrote it here, I accidentally dropped 21|(a±b). With that fixed—

Proof of (#3). 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)₃, iff LM≡0 (mod 7), iff (i) or (ii) or (iii), iff (#3).

Proof of Euler’s conjecture (#2). If

21|b, or (3|b and 7|a), or 21|(a±b), or 21|(a±4b) … (#3)

holds, then a weaker condition

21|b, or (3|b and 7|a), or 21|(a±b), or 7|(a±4b) … (#2')

will hold too, and an apparently even weaker condition

21|b, or (3|b and 7|a), or 21|(a±b), or 7|(4b±a), or 7|(b±2a) … (#2)

will of course hold.

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 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.

Another thing. Currently the expressions we have about (11|p)₃ and (13|p)₃ 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 (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)₃ and (13|p)₃. I’ll update the article more properly when I have time. — Gyopi (talk) 12:23, 19 June 2013 (UTC)

algorithm to find the cubic residue

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? Jackzhp (talk) 15:45, 20 January 2018 (UTC)

There's nothing wrong with using the word "residuacity" except

The first sentence in the section History is as follows:

"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."

There is nothing wrong with using the word "residuacity" except for the fact that 99.9999% of readers will have no idea what it means.

(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.)

Please remember that writing in Wikipedia is for the purpose of making things clear to readers. 2601:200:C000:1A0:D02F:D749:A7D7:6091 (talk) 20:13, 21 June 2021 (UTC)

Sloppy writing

The section Facts and terminology contains this passage:

"$\mathbb {Z} [\omega ]$ is a unique factorization domain. The primes fall into three classes:

• 3 is a special case:

$3=-\omega ^{2}(1-\omega )^{2}.$"

Is the writer truly unaware that the word "primes" could refer to the primes of the unique factorization domain $\mathbb {Z} [\omega ]$ that was just mentioned??? 2601:200:C000:1A0:D02F:D749:A7D7:6091 (talk) 20:22, 21 June 2021 (UTC)

References

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

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>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 octic reciprocity, 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). ——220.132.54.182 (talk) 09:09, 29 December 2021 (UTC) — Preceding unsigned comment added by 118.163.215.24 (talk) — Preceding unsigned comment added by 61.221.60.160 (talk)