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Chuck
Daedalian Member
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 Posted: Fri Mar 22, 2013 9:03 pm Post subject: 1 |
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Everyone knows that if 2^n-1 is prime, n is prime. But 2^n-1 is just an exmaple of the more general x^n-(x-1)^n where x = 2. If x^n-(x-1)^n is prime, is n always prime? I can't find any counter-examples. For higher x I get:
| Code: |
x 3^x-2^x primes
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2 5
3 19
5 211
17 129009091
29 68629840493971
31 617671248800299
53 19383245658672820642055731
59 14130386091162273752461387579
101 1546132562196033990574082188840405015112916155251
277 1454077510067338869372316944847370699315973030...
...8977340746977842962031155489999650314740218...
...74144975537134405477448628043552826333261491
x 4^x-3^x primes
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2 7
3 37
7 14197
17 17050729021
59 332306984815842876487217260305275077
x 5^x-4^x primes
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3 61
43 1136791005963704961126617632861
59 173472015290681763212224222187425603741981
191 3186183822264904553072710640625561630875233107881...
...6472270207782250106896363274089867800367051529...
...351065966102374800998198276889145001421 |
It's similar for higher x. I've never read about this anywhere that I can remember. |
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Thok
Oh, foe, the cursed teeth!
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| Posted: Fri Mar 22, 2013 9:56 pm Post subject: 2 |
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| More generally, if a^n-b^n is prime, n must be prime. Proof: if q|n, then a^q-b^q|a^n-b^n, and 1<a^q-b^q<a^n-b^n. (Details left to the reader.) |
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Zag
Unintentionally offensive old coot
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| Posted: Fri Mar 22, 2013 10:26 pm Post subject: 3 |
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| Thok wrote: |
| More generally, if a^n-b^n is prime, n must be prime. Proof: if q|n, then a^q-b^q|a^n-b^n, and 1<a^q-b^q<a^n-b^n. (Details left to the reader.) |
Well, this reader isn't clever enough.
I assume that this notation: if q|n means "if q is a divisor of n." Right? (Also, q and n have to be integers.)
Wouldn't that imply that Chuck's less general case is all that is useful if you were trying to generate primes? That is, even when n is a prime, it would be useless to try a^n-b^n, hoping it will also be prime, when |a-b| != 1, because a^n-b^n will always be divisible by |a-b|. Right, I see now that you'll always be able to fill the space a^n-b^n with |a-b| x 1 bricks. |
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Trojan Horse
Daedalian Member
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| Posted: Fri Mar 22, 2013 11:34 pm Post subject: 4 |
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| Zag wrote: |
| I assume that this notation: if q|n means "if q is a divisor of n." Right? (Also, q and n have to be integers.) |
Correct.
| Zag wrote: |
| Wouldn't that imply that Chuck's less general case is all that is useful if you were trying to generate primes? That is, even when n is a prime, it would be useless to try a^n-b^n, hoping it will also be prime, when |a-b| != 1, because a^n-b^n will always be divisible by |a-b|. Right, I see now that you'll always be able to fill the space a^n-b^n with |a-b| x 1 bricks. |
Correct again. |
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Chuck
Daedalian Member
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| Posted: Sat Mar 23, 2013 1:56 am Post subject: 5 |
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| If a^n+b^n is prime, does n have to be a power of 2? I can't find a counter-example. |
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Thok
Oh, foe, the cursed teeth!
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| Posted: Sat Mar 23, 2013 5:39 am Post subject: 6 |
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| Chuck wrote: |
| If a^n+b^n is prime, does n have to be a power of 2? I can't find a counter-example. |
Yes. If n=ep, where p is odd, there's a standard factorization of the polynomial x^p+y^p, that gives a factorization of a^n+b^n by setting x=a^e, y=b^e. |
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