Sunday, January 28, 2024

Certifying a Number is in a set A using Polynomials

 (This post was done with the help of Max Burkes and Larry Washington.)


During this post N={0,1,2,} and  N+={1,2,3,}.

Recall: Hilbert's 10th problem was to (in todays terms) find an algorithm that would, on input a polynomial p(x1,,xn)Z[x], determine if there are integers a1,,an such that p(a1,,an)=0.

From the combined work of Martin Davis, Yuri Matiyasevich, Hillary Putnam, and Julia Robinson it was shown that there is no such algorithm.  I have a survey on the work done since then, see here
The following is a corollary of their work:


Main Theorem:
Let AN+ be an r.e. set. There is a polynomial p(y0,y1,,yn)Z[y0,y1,,yn] such that

(xA) iff (a1,,anN)[(p(x,a1,,an)=0)(x>0)]}.

Random Thoughts:

1) Actual examples of polynomials p are of the form

p1(y0,y1,,yn)2+p2(y0,y1,,yn)2++pm(y0,y1,,yn)2

as a way of saying that we want a1,,an such that the following are all true simultaneously:

p1(x,a1,,an)=0, p2(x,a1,,an)=0, , pm(x,a1,,an)=0,

2) The condition x>0 can be phrased

(z1,z2,z3,z4)[x1z12z22z32z42=0].

This phrasing uses that every natural number is the sum of 4 squares.


The Main theorem gives a ways to certify that xA: Find a1,,anZ
such that p(x,a1,,an)=0.

Can we really find such a1,,an?

A High School student, Max Burkes, working with my math colleague Larry Washington, worked on the problem of finding a1,,an.

Not much is known on this type of problem. We will see why soon. Here is a list of what is known.

1) Jones, Sato, Wada, Wiens (see here) obtained a 26-variable polynomial q(x1,,x26)Z[x1,,x26] such that

xPRIMES iff (a1,a26N)[(q(a1,,a26=x)(x>0)].

To obtain a polynomial that fits the main theorem take

(x,x1,,x26,z1,z2,z3,z4)=(xq(x1,,x26))2+(xz12+z22+z32+z42)2.

Jones et al. wrote the polynomial q using as variables a,,z which is cute since thats all of the letters in the English Alphabet. See their paper pointed to above, or see Max's paper here.

2) Nachiketa Gupta, in his Masters Thesis, (see here) tried to obtain the the 26 numbers\ (a_1,\ldots,a_{26}\) such that q(a1,,a26)=2 where q is the polynomial that Jones et al. came up with.  Nachiketa Gupta found  22 of them.  The other 4 are, like the odds of getting a Royal Fizzbin, astronomical.  Could todays computers (21 years later) or AI or Quantum or Quantum AI obtain those four numbers?  No, the numbers are just to big.

3) There is a 19-variable polynomial p from the Main Theorem for the set

{(x,y,k):xk=y}.

See Max's paper here Page 2 and 3, equations 1 to 13. The polynomial p is the sum of squares of those equations. So for example r(x,y,z)=1 becomes (r(x,y,z)1)2.

Max Burkes found the needed numbers to prove 11=1 and 22=4. The numbers for the 22=4 are quite large, though they can be written down (as he did).

 
Some Random Thoughts:

1) It is good to know some of these values, but we really can't go much further.

2) Open Question: Can we obtain polynomials for primes and other r.e. sets so that the numbers used are not that large. Tangible goals:

(a) Get a complete verification-via-polynomials that 2 is prime.

(b) The numbers to verify that 23=8.

3) In a 1974 book about progress on Hilbert's problems (I reviewed it in this book rev col, see here) there is a chapter on Hilbert's 10 problem by Davis-Matiyasevich-Robinson that notes the following. Using the polynomial for primes, there is a constant c such that, for all primes p there is a computation that shows p is prime in c operations. The article did not mention that the operations are on enormous numbers.

OPEN: Is there some way to verify a prime with a constant number of operations using numbers that are not quite so enormous.




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