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Old 2013-10-03, 05:29   #34
Jayder
 
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Clearly, Richard Guy should have bet no less than $1000. Who does he think he is, offering to split his original friendly-bet of $20? I pass by 20-dollar bills every day without even bothering to pick them up. Chump change!
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Old 2013-10-03, 05:34   #35
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Quote:
Originally Posted by Jayder View Post
Clearly, Richard Guy should have bet no less than $1000. Who does he think he is, offering to split his original friendly-bet of $20? I pass by 20-dollar bills every day without even bothering to pick them up. Chump change!
Would a small amount from Erdös or Knuth have been chump change too?
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Old 2013-10-03, 05:59   #36
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Originally Posted by only_human View Post
Would a small amount from Erdös or Knuth have been chump change too?
If I would have a $10 cheque (why is this underlined in red?) from any of those two, first I would put it in a frame over my bed, to look at it every evening and morning. Then second, I would put it on ebay for bidding, starting from, say, half a million. Third, I will watch the bets going to 6 or 7 digits... Then fourth, I would say "sorry guys, it is not for sale, it was a joke"... (well, no character, as someone just tried to tell me... hehe )

(edit: fifth: I would tell my boss "funk you! even Erdos/Knuth paid me 10 bucks!" )

Last fiddled with by LaurV on 2013-10-03 at 06:03
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Old 2013-10-03, 07:53   #37
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Originally Posted by Prime95 View Post
"Save Richard Guy $20" project.
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Old 2013-10-03, 08:02   #38
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It is anecdotal that Knuth has a fixed price for valid errata for his books and that most of his checks (for whatever amount, $2.56 or something?) are never cached. One must have quite limited imagination that people actually work to get that money, literally. For some people it would have paid for 2 or say 3 minutes of their daytime job.
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Old 2013-10-03, 08:26   #39
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Quote:
Originally Posted by Batalov View Post
It is anecdotal that Knuth has a fixed price for valid errata for his books and that most of his checks (for whatever amount, $2.56 or something?) are never cached. One must have quite limited imagination that people actually work to get that money, literally. For some people it would have paid for 2 or say 3 minutes of their daytime job.
Knuth always used to pay $2.56 for each reported bug and 32¢ for each accepted enhancement. He gave up doing so some years back because so many people posted images of his checks and fraudsters tried to use the bank details.

My check for $2.88 from DEK is framed and hanging on the wall near the reproductions of the RSA-129 and RSA-140 counterparts.

Paul

(Actually, most checks from Knuth are cached AFAIK. Very few were cashed )
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Old 2013-10-03, 08:30   #40
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Of course, Serge. That is why I posed my question because I was trying to understand if the poster really thought that commensurate pay was intended. Humor falls flat so often on the attenuated media.

Richard Guy, of multiple editions of Unsolved Problems in Number Theory, would be a particularly good example of the rewards not intended to be drudge work remuneration.

Here is a math problem that offered a Goose as a reward.
Open problems with monetary rewards [mathoverflow.net]
Quote:
You may also be interested in the following picture of Mazur awarding Per Enflo a live goose as promised for the solution of the approximation problem. http://en.wikipedia.org/wiki/File:MazurGes.jpg
http://en.wikipedia.org/wiki/Per_Enflo
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In 1972, Per Enflo constructed a separable Banach space that lacks the approximation property and a Schauder basis. In 1972, Mazur awarded a live goose to Enflo in a ceremony at the Stefan Banach Center in Warsaw; the "goose reward" ceremony was broadcast throughout Poland.
http://en.wikipedia.org/wiki/Scottish_cafe

Last fiddled with by only_human on 2013-10-03 at 08:45
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Old 2013-10-03, 08:59   #41
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If it wasn't clear, I was being sarcastic. Actually, I was poking fun at the post before mine.
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Old 2013-10-03, 09:52   #42
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how to get rich...
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Old 2013-10-03, 11:57   #43
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Originally Posted by LaurV View Post
Would a moderator please move this thread to the crank sub-forum?
It has become devoid of information content.
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Old 2013-10-03, 11:59   #44
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We actually want to look at the expression F(X)-(sqrt(F(X)-1)-n)2 (note the exponent) and determine if it is a perfect square rather than find its factors. The value for n = 0 is trivially the square 1, so we start at n = 1 and increase it. In general, we can eliminate non-squares by sieving. Any square must be congruent to 0, 1, or 4 mod 8, for example, so any values of F(X)-(sqrt(F(X)-1)-n)2 which are congruent to 2, 3, 5, 6, or 7 mod 8 need not be further considered. This immediately eliminates all values of n congruent to 2 mod 4. Next we consider that any square must be congruent to 0 or 1 mod 3. This eliminates all values of n congruent to 1 mod 3. (Or we could say congruent to 0, 1, 4, or 7 mod 9 to eliminate more values of n.) In general, we set up a table for each prime p that we use for sieving that allows us to infer from n's congruence class mod p whether the above expression is a non-square or possibly a square. After we sieve with a reasonable number of different p's (Lehmer used all p's less than or equal to 127 in his Delay Line Sieve, later increased to 157) we expect that the few remaining n values are good candidates for generating perfect squares in the above expression. We can then test those candidates for each remaining n by taking the integer square root of F(X)-(sqrt(F(X)-1)-n)2, squaring that, and seeing if it is equal to the original expression.

I have coded such a sieve in C, but to optimize it, the sieving routines should be done in assembly language. It would be a good learning exercise, though, to code it in a higher level language. The integer square-root function is only called occasionally, and the C routine in GMP or MPIR would probably work just fine. Or you could just write the candidate n values to a list (there shouldn't be very many) and check to see if the resulting expression is a square using PARI or some other computational package. Try it with F5. You are sieving values of F5 - (65536 - n)2 = 1 + 131072*n - n2 by all small primes p to eliminate all values of n for which this expression is not a perfect square mod p. If you set up an array to check all n values from 1 to say, 20,000, you should find that the only value of n which gives you a perfect square will be 3272 = 65536 - 62264.

There is more to my method than this, I suggested working in general on trying to find sums of squares representations of N, where N could be a Fermat number, or a composite factor of a Fermat number, or either of these two possibilities multiplied by a product of distinct prime factors which all have representations of sums of squares. Or N could be any other number whose factors are all equal to 1 mod 4. The possibilities of trying many multipliers takes this from being a deterministic method of order n1/2 to a probabilistic method of order n1/4. The method of sieving was around for centuries before the Lehmers used an electronic device to automate and speed up the sieving. In general, we sieve on N - (sqrt(N) - n)2 where sqrt(N) is the integer truncation of the square-root of N.
This method is totally dominated by faster methods. It is pointless.
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