Prime Numbers Formula: 1800s Math Discovery

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SUMMARY

The discussion centers around a polynomial formula from the 1800s that purportedly predicts prime numbers accurately for values of n from 1 to 79 but fails beyond that range. The formula's limitations are highlighted, indicating that while it can generate primes up to a certain point, it becomes unreliable thereafter. Participants mention the possibility of constructing polynomials that yield primes if the primes are already known, emphasizing the inherent limitations of such approaches. The conversation also references a potential connection to Wittgenstein's "Philosophical Investigations," although this link remains uncertain.

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Mathematicians, educators, and students interested in the historical context of prime number theory and the limitations of polynomial predictions in mathematics.

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I was told by a math teacher I met recently that there is a formula that a mathematician in the 1800's came up with that accurately predicted all of the primes up to a certain point, but after that point began to miss a few primes, and after awhile, wasn't useful at all. Does anyone have any information on that?
 
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There is a polynomial in N that gives primes for something like n=1 through 79, but then falls apart. I can't remember what it is at the moment, but I'll try to find it if nobody else posts anything
 
For some reason I'm recalling that it actually appears in Wittgenstein's Philosophical Investigations, but I'm not sure if that's right...
 
http://www.jstor.org/view/07468342/di020779/02p0348s/0

This site seems to have good information
 
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You can, of course, construct polynomials that will give you all the primes up to any arbitrary point, if you already know what they are!
 
The positive solutions to the following system of equations are precisely the primes. But if you look closely you'll see that it's cheating you...

0 = wz + h + j − q
0 = (gk + 2g + k + 1)(h + j) + h − z
0 = 16(k + 1)3(k + 2)(n + 1)2 + 1 − f2
0 = 2n + p + q + z − e
0 = e3(e + 2)(a + 1)2 + 1 − o2
0 = (a2 − 1)y2 + 1 − x2
0 = 16r2y4(a2 − 1) + 1 − u2
0 = n + l + v − y
0 = (a2 − 1)l2 + 1 − m2
0 = ai + k + 1 − l − i
0 = ((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2
0 = p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m
0 = q + y(a − p − 1) + s(2ap + 2p − p2 − 2p − 2) − x
0 = z + pl(a − p) + t(2ap − p2 − 1) − pm.
 
Dragonfall said:
But if you look closely you'll see that it's cheating you...

Could you explain this?
 

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