mersenneforum.org Lucas-Lehmer Test proof
 Register FAQ Search Today's Posts Mark Forums Read

 2005-12-07, 19:22 #1 alpertron     Aug 2002 Buenos Aires, Argentina 22·337 Posts Lucas-Lehmer Test proof I was reading more carefully to the Lucas-Lehmer Test proof presented in the Mersennewiki and I found that it is not a proof at all. After the introduction there is the sentence "We are assuming that $S_{p-1}$ is divisible by $2^p-1$. " If we compute S1 = 4, ..., until Sp-1 = 0 we know that it is prime because we are assuming it (?????)
2005-12-07, 22:05   #2
philmoore

"Phil"
Sep 2002
Tracktown, U.S.A.

45F16 Posts

Quote:
 Originally Posted by alpertron I was reading more carefully to the Lucas-Lehmer Test proof presented in the Mersennewiki and I found that it is not a proof at all. After the introduction there is the sentence "We are assuming that $S_{p-1}$ is divisible by $2^p-1$. " If we compute S1 = 4, ..., until Sp-1 = 0 we know that it is prime because we are assuming it (?????)

There are two parts to the proof. One part is to show that if $2^p-1$ is prime, then $S_{p-1}$ is divisible by $2^p-1$. The other part of the proof is to show that if $S_{p-1}$ is divisible by $2^p-1$, then $2^p-1$ is prime. Because this second part guarantees that candidates passing the Lucas-Lehmer test are actually prime, and because it assumes very little background in number theory, I placed it first. Notice that the conclusion of this paragraph is that $2^p-1$ is prime. It guarantees that composite numbers will always fail the Lucas-Lehmer test, but doesn't necessarily prove that all prime numbers pass it. The proof that all prime numbers pass the L-L test is the point of the third paragraph. It does require a little knowledge of quadratic reciprocity, and was placed last for that reason.

 2005-12-08, 17:06 #3 alpertron     Aug 2002 Buenos Aires, Argentina 101010001002 Posts So the proof is OK, but the paragraphs are reversed. I think the last paragraph should be first, so after that we can safely assume that Sp-1 because it was proved in the previous paragraph.
 2005-12-08, 22:05 #4 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 21378 Posts Perhaps the second and third paragraphs should be labeled "Proof of Sufficiency" and "Proof of Necessity". There is no logical reason that either paragraph must come before the other because the two proofs are logically independent; i.e, neither depends upon the results of the other. However, each proof does make use of the set of numbers $a+b\sqrt 3 mod Q$, and if you find a compelling pedagogical reason to switch the paragraphs, you certainly have my permission.
 2005-12-09, 20:42 #5 alpertron     Aug 2002 Buenos Aires, Argentina 22×337 Posts I exchanged the proofs and also replaced in Rosen's proof the letter Q by F, because the variable Q was used in Bruce's proof as 2p - 1, so it is clear that they are different values.
 2005-12-09, 21:44 #6 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts Thanks, the main problem that I see is that the properties of the numbers $a+b\sqrt 3 mod Q$ are not discussed until the sufficiency proof, but these properties actually get used in the necessity proof. Also the phrase "From the previous proof" in the second sentence of the sufficiency proof needs to be removed. The sufficiency proof does not depend at all on the necessity proof, and this phrase is extremely misleading. Sufficiency means that if the L-L test is satisfied for some prime exponent p, then 2p - 1 is prime. Necessity means that if 2p - 1 is prime, then the L-L test is satisfied. Each proof is completely independent of the other. Unfortunately, the necessity proof now seems very unclear, because much of the notation that is used there does not get introduced until the second proof. If I get time over school break, I may try to fix this.
 2005-12-09, 23:42 #7 alpertron     Aug 2002 Buenos Aires, Argentina 22×337 Posts I still fail to see why the proof of sufficiency does not depend on the other proof, since it starts assuming the "output" of the proof of necessity: $S_{p-1}$ is multiple of $2^p-1$. It appears that the text should be reworked a bit. I will continue reading it.
 2005-12-10, 15:43 #8 T.Rex     Feb 2004 France 22·229 Posts Proof of LLT by Ribenboim Alpertron, Have you tried to read Ribenboim's proof of LLT ? It appears in page 77 and 78 (but requires to read many pages before ...). I prefer this proof, though it is longer. Tony
 2005-12-10, 18:35 #9 alpertron     Aug 2002 Buenos Aires, Argentina 101010001002 Posts I added some steps to the proof and now it is "crystal clear" to me.
 2005-12-11, 01:10 #10 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 21378 Posts Suppose one proved that if a, b, and c are the lengths of the sides of a right triangle with c the side opposite the right angle, then a2+b2=c2. Now suppose one proved that if a, b, and c are the lengths of the sides of any triangle and a2+b2=c2, then the angle opposite the side of length c must be a right angle. Even though the second proof starts by assuming the formula proven in the first proof, one cannot say in general that the second proof depends on the first. Each of the theorems assumes as input the output of the other theorem. The two theorems are converses, and in general, a theorem may be true even though the converse theorem may be false. Of course, in some cases, the proof of a converse may be dependent on the truth of the original theorem, and Euclid does in fact prove the converse to the Pythagorean theorem by SSS congruence using the Pythagorean theorem itself (Book I, theorems 47 and 48), but I challenge you to show me where the proof of sufficiency of the Lucas-Lehmer tests depends upon the proof of necessity. The fact that there is some common notation between the two proofs may have confused you, but look at the sufficiency proof alone: it simply says that IF Sp-1 is divisible by 2p-1, THEN 2p-1 is prime. The statement that Sp-1 is divisible by 2p-1 is the supposition of the theorem, and in no way is justified by the necessity piece, which assumed that 2p-1 is prime anyway. Here we cannot assume this, it is what must be proven.
 2005-12-11, 21:15 #11 alpertron     Aug 2002 Buenos Aires, Argentina 25048 Posts Arggghhh!!! I was wrong, the fact that $S_{p-1}$ is divisible by $2^p-1$, is the hypotesis of the proof. I changed the wording again. Please let me know if there are errors.

 Similar Threads Thread Thread Starter Forum Replies Last Post Trilo Miscellaneous Math 25 2018-03-11 23:20 carpetpool Miscellaneous Math 2 2017-07-30 09:21 science_man_88 Miscellaneous Math 48 2010-07-14 23:33 kurtulmehtap Math 13 2009-10-02 12:12 storm5510 Math 22 2009-09-24 22:32

All times are UTC. The time now is 17:04.

Fri Apr 23 17:04:24 UTC 2021 up 15 days, 11:45, 0 users, load averages: 1.66, 1.94, 1.98