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-   -   "New" primality test/check (https://www.mersenneforum.org/showthread.php?t=22838)

 10metreh 2017-12-31 19:25

[QUOTE=gophne;475627]Hi science_man_88

LHS of == sign

2^n-1....let n=5

Then LHS result is 31

RHS of == sign

(n+1)/2 mod (n+2).......substitute for n=5

(5+1)/2 mod (5+2)

RHS result is =3[/QUOTE]

"mod" isn't an operator. You can't just apply it to one side of an equation. It isn't the same as the "%" operator in many programming languages, though it is somewhat similar.

When we say that a ≡ b (mod n), what we mean is that n divides a-b exactly (or, equivalently, that a and b leave the same remainder on division by n). So, for example, 1 ≡ 11 (mod 5), because 5 divides 1-11 = -10 exactly.

When n=5, indeed 2^n-1 = 31 and (n+1)/2 = 3.
These are not equal, but 2^n-1 ≡ (n+1)/2 (mod n+2) is still true, because 7 divides 31-3 = 28 exactly, and 31 ≡ 3 (mod 7).

The "%" operator gives the remainder you get when you divide one number by another.
Note that if b = a%n, then a = kn+b for some integer k: that's what we mean by a remainder. Then a-b = kn, so a ≡ b (mod n). This is how "mod" and "%" are related.
For positive integers, a ≡ b (mod n) is equivalent to a%n = b%n. (You need to be careful with negative numbers because programming languages may define a%n to have the same sign as a.)

 CRGreathouse 2017-12-31 19:36

[QUOTE=gophne;475595]Could you please run the two algorithms, the one I posted (on gophne thread #73 I think) and the one that awmayer "reduced" a bit further on in the thread. If you could that present to all your results. If the two algorithms gives the same result/s then I will accept that the two algorithms are indeed the same.[/QUOTE]

I can’t take on unpaid projects at this time, sorry.

 gophne 2017-12-31 19:41

[QUOTE=10metreh;475632]"mod" isn't an operator. You can't just apply it to one side of an equation. It isn't the same as the "%" operator in many programming languages, though it is somewhat similar.

When we say that a ≡ b (mod n), what we mean is that n divides a-b exactly (or, equivalently, that a and b leave the same remainder on division by n). So, for example, 1 ≡ 11 (mod 5), because 5 divides 1-11 = -10 exactly.

When n=5, indeed 2^n-1 = 31 and (n+1)/2 = 3.
These are not equal, but 2^n-1 ≡ (n+1)/2 (mod n+2) is still true, because 7 divides 31-3 = 28 exactly, and 31 ≡ 3 (mod 7).

The "%" operator gives the remainder you get when you divide one number by another.
Note that if b = a%n, then a = kn+b for some integer k: that's what we mean by a remainder. Then a-b = kn, so a ≡ b (mod n). This is how "mod" and "%" are related.
For positive integers, a ≡ b (mod n) is equivalent to a%n = b%n. (You need to be careful with negative numbers because programming languages may define a%n to have the same sign as a.)[/QUOTE]
[B]Hi 10metreh[/B]

No disagreement here: 31 mod 7 ≡ 3

However the logical [B]==[/B] determinant means [B]IS IDENTICAL.[/B] 31[B]<>[/B]3

Consider the following:

x==31 and x==31 mod 7....THEY ARE NOT THE SAME

 gophne 2017-12-31 19:45

[QUOTE=CRGreathouse;475636]I can’t take on unpaid projects at this time, sorry.[/QUOTE]
Hi CRGreathouse

You seem to have no problem with gratis comments :)

I know I am dumb, but please help me to understand and stop me from being a nuisance by posting the results for all to see and to analyse the outcomes. Ppl wil have a lot of faith in your results.

 CRGreathouse 2017-12-31 20:06

[QUOTE=gophne;475638]You seem to have no problem with gratis comments :)

I know I am dumb, but please help me to understand and stop me from being a nuisance by posting the results for all to see and to analyse the outcomes. Ppl wil have a lot of faith in your results.[/QUOTE]

In fact I may have to cut back the gratis comments to make time for other commitments. But for the moment I can spare a bit of time.

I’ll be happy to carry out the analysis however you like, and write whatever supporting code is needed, as soon as the check clears in my bank account. Until then I recommend further study which will (in addition to enhancing your life) make it easier for you to understand the equivalence. If interested, forum members are usually more than happy to recommend resources.

 gophne 2017-12-31 20:35

[QUOTE=CRGreathouse;475647]In fact I may have to cut back the gratis comments to make time for other commitments. But for the moment I can spare a bit of time.

I’ll be happy to carry out the analysis however you like, and write whatever supporting code is needed, as soon as the check clears in my bank account. Until then I recommend further study which will (in addition to enhancing your life) make it easier for you to understand the equivalence. If interested, forum members are usually more than happy to recommend resources.[/QUOTE]
Hi CRGreathouse

I respect that you have other tasks on the Site, it is just that I had thought that if you had done the analysis, that would have been much more authoritative.

 science_man_88 2017-12-31 20:41

[QUOTE=gophne;475652]Hi CRGreathouse

I respect that you have other tasks on the Site, it is just that I had thought that if you had done the analysis, that would have been much more authoritative.

Look up proof by authority...

 VBCurtis 2017-12-31 22:25

[QUOTE=chalsall;475522]That doesn't surprise me at all.[/QUOTE]

:tu:

 George M 2017-12-31 22:39

X!=Y means X does not equal Y.... right?

I read the title on a mathematical symbols keyboard. Wait, I am not replying to the right post, am I? If so, then just ignore this.

 gophne 2017-12-31 23:00

Hi George M

[B] !=[/B] does mean "[I]not equal to[/I]" in some computer code.

 gophne 2018-01-01 06:54

RETRACTION

Hi Everybody

I acknowledge that "my" algorithm is a "clone" of Fermat, after the answer to runs of the two algorithms by [B]10metreh, post #22[/B], in the [B]OMG, I cannot spam anymore in the forum Feedback where my question was answered!!!!!!!!!!111111[/B]

[I]The results posted prove conclusively that "my" algorithm is a clone/copy of Fermat's.[/I]

I apologise for not being to graps this earlier. Egg all over my face and many frustrated contributers.

All I can do now is to provide the background work that I had used to derive "my" algorithm, if anybody might be interested. I did not use Fermat.

Oh my gosh!!!

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