Talk:Cubic reciprocity

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Field: Number theory

Suspected error

I strongly suspect that

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

should be replaced by

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

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)