Question

I need help with this algebra problem, you will definitely have to use the product log.

Answer 1

\[\Large a^p-a=np\] I need to solve for p

Answer 2

True, but I thought it would be fun to do it without using WA just as an exercise. I want to figure out the steps with anyone who's interested.

Answer 3

Why am I interested in this? Well for all prime numbers there is an n and a you can plug in to get p.

Answer 4

Haha yes, I just thought it would be interesting to solve for the prime number in it. =P

Answer 5

this is fermat theorem , i thought i posted this comment :O

Answer 6

it says
a^p=a mod p for any p prime s.t GCD(a,p)=1

Answer 7

I guess alternatively, how do I get wolfram alpha to graph this: http://www.wolframalpha.com/input/?i=f%28x%2Cy%29%3D%28-y+W%28-x%5E%28-x%2Fy%29ln%28x%29%2Fy%29-xlnx%29%2F%28ylnx%29

Answer 8

gcd(a,p) = 1 is not needed

Answer 9

w is wronsckin ?

Answer 10

but looks Kainui wants the continuous version

Answer 11

W is product log function, you want to abuse wronskians als here ha!

Answer 12

@dan815 @UnkleRhaukus might know plotting this in matlab

Answer 13

Since this is true for any integer a, we can just set it equal to 2. And I believe this will map all the prime numbers uniquely 1-to-1 to another integer. \[\Large \frac{2^p-2}{p}=n\] This seems pretty cool.

Answer 14

ganesh gcd is needed else it would be
a^p=0 mod p

Answer 15

that work kai , but note
n,a are changeable

Answer 16

u can change 2 by any other prime xD
but its means nothing

Answer 17

eh , im not expert with w function though i wud like to learn it

Answer 18

Here are two sites that really helped me, there are some good exercises on there! It's really fun, I'll help you if you have questions. =D

http://www.had2know.com/academics/lambert-w-function-calculator.html
https://luckytoilet.wordpress.com/tag/product-log/

Answer 19

True, we can make it anything, but I figured I'd make it 2 since it's the smallest possible one to make computation easier.

Answer 20

\(a,n\) are real numbers?

Answer 21

No, they are natural numbers.

Answer 22

To anyone curious looking for patterns, I've listed the primes on the left and to the right are different values of n for consecutive integers starting at 2. \[\Large \frac{a^p-a}{p}=n\]

So if you look in the third row we have p=5, and it lists for a=2, n=6; for a=3, n=48; for a=4, n=204; etc...
```
2: 1, 3, 6, 10, 15, 21, 28, 36,
3: 2, 8, 20, 40, 70, 112, 168, 240,
5: 6, 48, 204, 624, 1554, 3360, 6552, 11808,
7: 18, 312, 2340, 11160, 39990, 117648, 299592, 683280,
11: 186, 16104, 381300, 4438920, 32981550, 179756976, 780903144, 2852823600,
13: 630, 122640, 5162220, 93900240, 1004668770, 7453000800, 42288908760, 195528140640,
17: 7710, 7596480, 1010580540, 44878791360, 995685849690, 13684147881600, 132458812569720, 981010688215680,
19: 27594, 61171656, 14467258260, 1003867701480, 32071565263710, 599941851861744, 7585009898729256, 71097458824894312,
23: 364722, 4093181688, 3059510616420, 518301258916440, 34336096654504472, 1189945536525257216,
29: 18512790, 2366564736720, 9938978487990060, 6422914307692954624,
31: 69273666, 19924948267224, 148764065110560896,
37: 3714566310, 12169835294351280,
41: 53634713550, 889585277491970432,
43: 204560302842, 7633882962663652352,
47: 2994414645858,
53: 169947155749830,
57: 2528336632909752, ```

Answer 23

So what are you looking for? finding \(p\) given \(a\) and \(n\) (which is not always exist) or finding \(n\) given \(a\) and \(p\) which is what you did on the table?

Answer 24

What I'm saying is we have this function: \[\Large f(n) = \frac{-W(\frac{1}{\sqrt[n]{4}}\ln \frac{1}{\sqrt[n]{2}})}{\ln2}-\frac{2}{n}\] and on the domain where n is only natural numbers, the only values in the range of f(n) that are integers are all prime numbers.

Answer 25

To put another way to make it apparent how cool this is!
\[domain \ f = \mathbb{N} \]\[(range \ f) \cap \mathbb{N} = set\ of\ all\ primes\]

Answer 26

Basically the way I know it is just like it says on this site: https://www.math.hmc.edu/funfacts/ffiles/10007.5.shtml

Answer 27

i see , thanks for the new thing to think about xD