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Topic: Prime power plus one (Read 1952 times) |
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NickH
Senior Riddler
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Prime power plus one
« on: Jan 22nd, 2003, 6:36pm » |
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Show that p^n + 1 = m^r,
where p > 1 is prime, and r > 1 is not a multiple of p,
has no solution for n, m in positive integers.
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Nick's Mathematical Puzzles
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SWF
Uberpuzzler
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Re: Prime power plus one
« Reply #1 on: Jan 23rd, 2003, 7:37pm » |
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Assume the two are equal, then
\sum_{k=0}^{r-1}m^k)
Since (m-1) is a factor of pn, it must be that: m=1 mod p. That means each term in the summation is 1 mod p, so the summation is r mod p, and therefore so is (m-1) times the summation. But pn is zero mod p, and r is not a multiple of p which is a contradiction.
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NickH
Senior Riddler
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Re: Prime power plus one
« Reply #2 on: Jan 23rd, 2003, 7:54pm » |
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That was my solution, SWF!
See puzzle 39 on my puzzle site, below.
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Nick's Mathematical Puzzles
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Icarus
wu::riddles Moderator
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Boldly going where even angels fear to tread.
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Re: Prime power plus one
« Reply #3 on: Jan 23rd, 2003, 8:09pm » |
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on Jan 23rd, 2003, 7:37pm, SWF wrote:| so the summation is r mod p, and therefore so is (m-1) times the summation. |
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I don't follow this. Yes the sum = r mod p, but how do you get from there to (m-1)[sum] = r mod p ?
Unless m = 2, (m-1) | pn means p | (m-1), and so (m-1)(anything) = 0 mod p.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
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Re: Prime power plus one
« Reply #4 on: Jan 23rd, 2003, 8:15pm » |
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31+1 = 22
71+1 = 23
311+1 = 25
1271+1 = 27
Haven't found a solution for n != 1 yet.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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NickH
Senior Riddler
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Posts: 341
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Re: Prime power plus one
« Reply #5 on: Jan 23rd, 2003, 8:17pm » |
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Since p is a prime, both factors must be powers of p.
You're right -- SWF didn't explicitly state that in his proof.
<edit>
Thanks! I forgot to specify n,m > 1.
</edit>
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| « Last Edit: Jan 23rd, 2003, 8:24pm by NickH » |
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Nick's Mathematical Puzzles
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SWF
Uberpuzzler
Posts: 879
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Re: Prime power plus one
« Reply #6 on: Jan 26th, 2003, 9:43am » |
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Yes, that "proof" was quite reckless, and and I will try to salvage something from it. The (m-1) term in the factored expression must equal either 1 or be a multiple of p. If it is a multiple of p, the sum equals either r mod p or 1. The sum can't be one since r>1. The sum must also be a multiple of p which is a contradiction.
A more serious omission from the proof, which is also missing from the proof on NickH's web site, is the case when m-1=1; m=2. In this case, m is not 1 mod p and the sum is not necessarily r mod p. It still remains to show that pn=2r-1 has no solution. Obviously p can't be 2 because that would make one side odd and the other even. Since r>1, 2r is a mulitiple of 4 and 2r-1 is 3 mod 4. Since p is odd, for pn to also be 3 mod 4, n must be odd, because (2N+1)2M=(4N2+4N+1)M=1 mod 4.
With n determined to be odd, the 1 in the assumed equality can be moved back to the left side and the pn+1 term factored:
\sum^{n-1}_{k=0}(-p)^k)
The summation has an odd number of odd terms, so it is odd. It is also greater than 1. This is not possible because it needs to be a factor of 2r.
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NickH
Senior Riddler
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Re: Prime power plus one
« Reply #7 on: Jan 26th, 2003, 10:33am » |
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Oops -- thank you for pointing that out, SWF! I have temporarily excluded the case m = 2 on my site, until I can update the proof.
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