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i mean yeah if you trivially set the function equal to a prime constant i guess it won't change _but why would you do that? why would you do any of that?_
This result can be extended even for multivariate polynomials, though the proof is of course much more complicated. But a polynomial can still generate prime numbers for the appropriate meaning of "generate." A polynomial can't _only_ have prime values, but it can have every prime value. In fact, Ribenboim's polynomial in 10 variables not only takes on every prime value, but _every_ positive value of the polynomial is prime. It just also has nonpositive values. And that result can probably be improved.
Shouldn't you state, obvious though it may seem, "non-constant polynomials"? Since f(x) = 2 always generates a prime number; and such exceptions can mess up the logic of seemingly bullet proof proofs.
On a tangent, following up on the example polynomial f(x) which produces 40 prime numbers for x from 1 to 40, in fact you can in principle construct a polynomial that passes through any given finite set of points. So for example, if you have a finite set of n points (1,2) , (2,3), (3,5), (4,7), etc, where for x=k you have the kth prime, then you can construct a polynomial which passes through all those points and thus get a polynomial which generates the first n primes by plugging in x from 1 to n. (To see a method for constructing such a polynomial, look up Lagrange interpolating polynomials which are the unique polynomials of lowest degree that pass through a given set of points.)
Doesn’t this proof only work assuming the polynomial has integer coefficients? I mean it is possible, though a bit subtle, for the polynomial to have rational coefficients, yet still produce integers. Ex: Triangular T(n)=1/2*n^2+1/2*n. This would invalidate some step where we factor p+p*stuff. The stuff could actually be a rational number with a factor of p in the denominator
You can extend it to include rational coefficients, since if you take the lcm of the coefficients’ denominators, let’s call it k, and multiply your prime generating polynomial f(x) by k, you get a polynomial g(x)=kf(x) that generates prime numbers multiplied by k and has integer coefficients. So you do the same thing and prove that g(b+mkp) is kp(1+stuff). Now divide both sides by k to get f(b+mkp) = p(1+stuff), which is not prime.
I already knew that polinomial that generates primes up to f(40) so my mind instantly tried to prove this case by case on a_0. If |a_0| > 1, then a_0 is clearly a factor. If a_0 = 0, then you can factor out x. I wonder if there is a simple enough argument for |a_0| = 1
Polynomial answers can either be rational, irrational, and imaginary. In the end, if it comes down to a "prime number," it'll most likely have its square root taken anyway in a, let's say, x² = 17 situation. Prime numbers simply become irrational more than half the time. If it's an imaginary prime number, complex numbers in general don't have "primes" to begin with, not making it prime. If it's rational, a prime number can't be created from taking a square root.
The proof seems to be valid only for a finite polynomial. An infinite polynomial would still work. Which is perhaps kind of obvious. But what is the convergence of the optimal prime number generator? Eg how many terms are needed in the polynomial to produce 1000 primes.
An infinite polynomial is not a polynomial, it is a power series. Edit: Actually not quite, I'm wrong. The prime counting function approximation is not representable with a power series (though maybe it's still possible?) I'm pretty sure there exists a power series for the inverse of the prime counting function (which would be the nth prime function, and thus all values of the function is prime). I suspect this because there's a power series that converges to the prime counting function which uses the zeros of the Riemann zeta function. I imagine there's some way to invert it, to make a power series that """"""generates"""""" all primes.
Seems like a very convoluted proof. Isn't it enough to observe that substituting in the constant term into any polynomial results in a composite number?
I feel like in order to get a prime number generator we'd need to do something different to the integer we add to a polynomial so that it isn't fixed, and that way the overall function may avoid hitting a composite every so often. I realise you can't vary the integer in a way that directly relates to the variable (Say 'n'), for example adding 'one more than n' or 'seven less than n' or 'one more than three lots of n' or whatever, as that will just cancel down into a very slightly different polynomial. My idea is that the integer added could be the nth prime number, which brings some psuedo-randomness to what gets added, whilst still using n to decide what the added integer should be for each step. I tried (n*n) + n + the nth prime, and it worked for the first seven terms. I can't help wondering if tweaking the degree of that type of expression, the individual exponents, and/or exactly which prime is being added (eg. 'n-2'th prime or 'n+1'th prime etc) might yield a prime number generator somewhere along the lines. Of course for it to be any use you'd probably want it to only use for input those primes that had already been found earlier by the same function, and ideally if it could find all of the primes that would be truly something.
@@YunxiaoChu If you are implying that you disagree with me then give me some constructive criticism to work with. Or was that just a sigh of pleasure at reading an interesting mathematical hypothesis?
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y=0x+3
i mean yeah if you trivially set the function equal to a prime constant i guess it won't change
_but why would you do that? why would you do any of that?_
@@matthewbertrand4139this was to show that he made a mistake in his proof. otherwise he would have caught this exception.
ok but that is not a polynomial
highest degree of x has a coefficient of zero so it's not a polynomial
@QwertzIsTired y=c is a constant function
May I say, "very nice proof". Easy to understand.
Thank you!
This result can be extended even for multivariate polynomials, though the proof is of course much more complicated.
But a polynomial can still generate prime numbers for the appropriate meaning of "generate." A polynomial can't _only_ have prime values, but it can have every prime value. In fact, Ribenboim's polynomial in 10 variables not only takes on every prime value, but _every_ positive value of the polynomial is prime. It just also has nonpositive values. And that result can probably be improved.
f(x) = x can generate all the primes. We just have to filter through the range of f to find them.
could you cite a source for this polynomial? I couldn't find anything for it online.
Wow this such a elegant proof I’ve only seen much harder proofs thanks for this
no
I feel like he could have made it a lot simpler by just setting b=0.
Shouldn't you state, obvious though it may seem, "non-constant polynomials"? Since f(x) = 2 always generates a prime number; and such exceptions can mess up the logic of seemingly bullet proof proofs.
On a tangent, following up on the example polynomial f(x) which produces 40 prime numbers for x from 1 to 40, in fact you can in principle construct a polynomial that passes through any given finite set of points. So for example, if you have a finite set of n points (1,2) , (2,3), (3,5), (4,7), etc, where for x=k you have the kth prime, then you can construct a polynomial which passes through all those points and thus get a polynomial which generates the first n primes by plugging in x from 1 to n. (To see a method for constructing such a polynomial, look up Lagrange interpolating polynomials which are the unique polynomials of lowest degree that pass through a given set of points.)
Doesn’t this proof only work assuming the polynomial has integer coefficients? I mean it is possible, though a bit subtle, for the polynomial to have rational coefficients, yet still produce integers. Ex: Triangular T(n)=1/2*n^2+1/2*n. This would invalidate some step where we factor p+p*stuff. The stuff could actually be a rational number with a factor of p in the denominator
On second thought, maybe this could be fixed if the polynomial is written with binomial coefficients like a*(n choose 3)+b*(n choose 2) +c*n+d
You can extend it to include rational coefficients, since if you take the lcm of the coefficients’ denominators, let’s call it k, and multiply your prime generating polynomial f(x) by k, you get a polynomial g(x)=kf(x) that generates prime numbers multiplied by k and has integer coefficients. So you do the same thing and prove that g(b+mkp) is kp(1+stuff). Now divide both sides by k to get f(b+mkp) = p(1+stuff), which is not prime.
I already knew that polinomial that generates primes up to f(40) so my mind instantly tried to prove this case by case on a_0.
If |a_0| > 1, then a_0 is clearly a factor. If a_0 = 0, then you can factor out x.
I wonder if there is a simple enough argument for |a_0| = 1
If P(n)=k is prime then P(n+km) is divisible by k
P(n+km)=P(n) mod k , P(n)=0 mod k so P(n+km)=0 mod k.
Polynomial answers can either be rational, irrational, and imaginary. In the end, if it comes down to a "prime number," it'll most likely have its square root taken anyway in a, let's say, x² = 17 situation. Prime numbers simply become irrational more than half the time. If it's an imaginary prime number, complex numbers in general don't have "primes" to begin with, not making it prime. If it's rational, a prime number can't be created from taking a square root.
there is a sense in which complex numbers have primes, actually en.wikipedia.org/wiki/Gaussian_integer#Gaussian_primes
It's not the only way to genrate numbers out of a polynomial. First, doing an operation and then setting to a value.
The proof seems to be valid only for a finite polynomial. An infinite polynomial would still work. Which is perhaps kind of obvious. But what is the convergence of the optimal prime number generator? Eg how many terms are needed in the polynomial to produce 1000 primes.
An infinite polynomial is not a polynomial, it is a power series.
Edit: Actually not quite, I'm wrong. The prime counting function approximation is not representable with a power series (though maybe it's still possible?)
I'm pretty sure there exists a power series for the inverse of the prime counting function (which would be the nth prime function, and thus all values of the function is prime). I suspect this because there's a power series that converges to the prime counting function which uses the zeros of the Riemann zeta function. I imagine there's some way to invert it, to make a power series that """"""generates"""""" all primes.
Seems like a very convoluted proof.
Isn't it enough to observe that substituting in the constant term into any polynomial results in a composite number?
Nice proof and love the Nintendo music in the background!
Thank you!
Can you make a similar argument by substituting N*A0 into the polynomial?
ao < 0 ?
What about a degree 0 polynomial? for example f(x) = 7
That's the one exception!
I feel like in order to get a prime number generator we'd need to do something different to the integer we add to a polynomial so that it isn't fixed, and that way the overall function may avoid hitting a composite every so often. I realise you can't vary the integer in a way that directly relates to the variable (Say 'n'), for example adding 'one more than n' or 'seven less than n' or 'one more than three lots of n' or whatever, as that will just cancel down into a very slightly different polynomial. My idea is that the integer added could be the nth prime number, which brings some psuedo-randomness to what gets added, whilst still using n to decide what the added integer should be for each step. I tried (n*n) + n + the nth prime, and it worked for the first seven terms. I can't help wondering if tweaking the degree of that type of expression, the individual exponents, and/or exactly which prime is being added (eg. 'n-2'th prime or 'n+1'th prime etc) might yield a prime number generator somewhere along the lines. Of course for it to be any use you'd probably want it to only use for input those primes that had already been found earlier by the same function, and ideally if it could find all of the primes that would be truly something.
Hmm
@@YunxiaoChu If you are implying that you disagree with me then give me some constructive criticism to work with.
Or was that just a sigh of pleasure at reading an interesting mathematical hypothesis?
@@MrDannyDetail just find it interesting
@@YunxiaoChu Oh fair enough. I guess I'm just slightly worried that I may have overlooked some obvious reason why my idea wouldn't work.
y=2 would work
Yeah, constant functions are the only exception
y=x+1 , y(6) =7
too confusing
Which part?