A talk given at Huawei Company (Oct. 31, 2019)and Shenzhen Univ. (Nov. 28, 2019)

Computability and Undecidability

Zhi-Wei Sun

Nanjing UniversityNanjing 210093, P. R. China

zwsun@nju.edu.cn

http://math.nju.edu.cn/∼zwsun

Nov. 28, 2019

Abstract

In this talk we introduce the theory of computability and theundecidability results on Hilbert’s Tenth Problem. We will alsomention some algorithms concerning primes.

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Part I. The Theory of Computability

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Primitive recursive functions

Let N = {0, 1, 2, . . .} and call each n ∈ N a natural number.

Three Basic Functions:

Zero Function: O(x) = 0 (for all x ∈ N).

Successor Function: S(x) = x + 1.

Projection Function: Ink(x1, . . . , xn) = xk (1 6 k 6 n)

Composition:

f (x1, . . . , xn) = g(h1(x1, . . . , xn), . . . , hm(x1, . . . , xn))

Primitive Recursion:

f (x1, . . . , xn, 0) =g(x1, . . . , xn),

f (x1, . . . , xn, y + 1) =h(x1, . . . , xn, y , f (x1, . . . , xn, y))

Primitive recursive functions are the basic functions and thoseobtained from the basic functions by applying composition andprimitive recursion a finite number of times.

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Acknermann’s function

It is easy to see that all primitive recursive functions arecomputable by our intuition.

Skolem’s Claim (1924): All intuitively computablenumber-theoretic functions are primitive recursive functions.

In 1928 Ackermann showed that the following Ackermannfunction A(m, n) is computable but not primitively recursive.

A(0, n) =n + 1,

A(m + 1, 0) =A(m, 1)

A(m + 1, n + 1) =A(m,A(m + 1, n)).

For example,

A(1, 2) =A(0,A(1, 1)) = A(0,A(0,A(1, 0)))

=A(0,A(0,A(0, 1))) = A(0,A(0, 2))

=A(0, 3) = 4.5 / 39

Partial recursive functions

µ-operator:

f (x1, . . . , xn) = µy(g(x1, . . . , xn, y) = 0)

means that f (x1, . . . , xn) is the least natural number y such thatg(x1, . . . , xn, y) = 0. If g(x1, . . . , xn, y) 6= 0 for all y ∈ N, thenf (x1, . . . , xn) is undefined.

Partial recursive functions are the basic functions and thoseobtained from the basic functions by applying composition andµ-operator a finite number of times.

If a partial recursive function f (x1, . . . , xn) is defined for allx1, . . . , xn ∈ N, then f is called a total recursive function.

All primitive functions as well as the Ackermann function A(m, n)are total recursive functions.

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Church’s Thesis

For any partial recursive function f , it is easy to see that iff (x1, . . . , xn) is defined then the value f (x1, . . . , xn) is effectivelycomputable.

In 1936 A. Turing introduced the notion of Turing machine whichis an abstract machine that manipulates symbols on a strip of tapeaccording to a table of rules (i.e., a program). A functionf (x1, . . . , xn) is Turing computable if there is a program accordingto which the Turing machine with initial inputs x1, . . . , xn finallystops and yields the value f (x1, . . . , xn) as outut if f (x1, . . . , xn) isdefined, and never stops if f (x1, . . . , xn) is undefined.

Partial recursive functions and Turing computable functions wereproved to be equivalent.

Church’s Thesis (1936). If a function f into N with naturalnumber variables is effectively computable by intuition, then itmust be a partial recursive function (or a Turing computablefunction).

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Recursively enumerable sets

A subset A of N is said to be an r.e. (recursively enumerable) set(or a semi-decidable set) if the function

f (x) =

{1 if x ∈ A,

undefined if x ∈ N \ A.

is a partial recursive function.

If A = Dom(f ) for some partial recursive function f , then we mayrevise the program computing f (x) by letting the output be 1 iff (x) is computed, and thus A is an r.e. set.

In view of the above,

A ⊆ N is an r.e. set

⇐⇒ A = Dom(f ) for some partial recursive function f .

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Recursively enumerable setsIf A is an r.e. set containing an element a, and the program Pcomputes the above function f , then the function

g(x , y) =

{x if the program P computes f (x) within y steps,

a otherwise

is a partial recursive function with Ran(g) = A.

If A is the range of a partial recursive function h(x1, . . . , xn), thenthe function

f (x) =

{1 if x ∈ Ran(h) = A,

undefined otherwise,

is a partial recursive function (we may seek for x1, . . . , xn withh(x1, . . . , xn) equal to a given x ∈ A), and thus A is an r.e. set.

So, a nonempty A ⊆ N is an r.e. set if and only if A = Ran(f ) forsome partial recursive function f .

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r.e. sets and recursive sets

Enumeration Theorem. There is a partial recursive functionϕ(m, n) such that

ϕ0, ϕ1, ϕ2, . . .

list all the partial recursive functions of one variable. where ϕm isgiven by

ϕm(n) = ϕ(m, n) (n = 0, 1, 2, . . .).

A set A ⊆ N is called decidable or recursive, if the characteristicfunction

χA(x) =

{1 if x ∈ A,

0 if x ∈ N \ A.

is Turing computable (or recursive).

A set A ⊆ N is recursive if and only if both A and N \ A are r.e.sets.

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Halting Problem is undecidable

Theorem. The set K = {x ∈ N : x ∈ Dom(ϕx)} is a nonrecursiver.e. set.

Proof. As the function ϕx(x) = ϕ(x , x) is a partial recursivefunction, we see that K is an r.e. set.

Suppose that K is recursive. Then the function

f (x) =

{ϕx(x) + 1 if x ∈ Dom(ϕx),

0 otherwise,

is totally recursive, thus for some m ∈ N we have ϕm = f andhence

f (m) = ϕm(m) 6= ϕm(m) + 1

which leads a contradiction.

Let Px be a Turing program computing ϕx . Whether a Turingmachine with input x and program Px finally stops, is anundecidable problem which is called the halting problem.

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Part II. Solution to Hilbert’s Tenth Problem

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Hilbert’s Tenth Problem

In 1900, at the Paris conference of ICM, D. Hilbert presented 23famous mathematical problems. He formulated his tenth problemas follows:

Given a Diophantine equation with any number of unknownquantities and with rational integral numerical coefficients: To

devise a process according to which it can be

determined in a finite number of operations whether

the equation is solvable in rational integers.

In modern language, Hilbert’s Tenth Problem (HTP) asked for aneffective algorithm to test whether an arbitrary polynomial equation

P(z1, . . . , zn) = 0

(with integer coefficients) has solutions over the ring Z of theintegers.

However, at that time the exact meaning of algorithm was notknown.

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Diophantine equations over N and ZThroughout this talk, variables always range over Z.

Let P(z1, . . . , zn) ∈ Z[z1, . . . , zn]. Then

∃z1 . . . ∃zn[P(z1, . . . , zn) = 0]

⇐⇒ ∃x1 > 0 . . . ∃xn > 0

[ ∏ε1,...,εn∈{±1}

P(ε1x1, . . . , εnxn) = 0

].

On the other hand, by Lagrange’s four-square theorem (eachm ∈ N can be written as the sum of four squares), we have

∃x1 > 0 . . . ∃xn > 0[P(x1, . . . , xn) = 0]

⇐⇒ ∃u1∃v1∃y1∃z1 . . . ∃un∃vn∃yn∃zn

[P(u21 + v2

1 + y21 + z2

1 , . . . , u2n + v2

n + y2n + z2

n ) = 0]

So HTP has the following equivalent form (HTP over N): Is therean algorithm to decide for any polynomial P(x1, . . . , xn) withinteger coefficients whether the Diophantine equationP(x1, . . . , xn) = 0 has solutions with x1, . . . , xn ∈ N ?

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Diophantine relations and Diophantine sets

A relation R(a1, . . . , am) with a1, . . . , am ∈ N is said to beDiophantine if there is a polynomial P(t1, . . . , tm, x1, . . . , xn) withinteger coefficients such that

R(a1, . . . , am) ⇐⇒ ∃x1 > 0 . . . ∃xn > 0[P(a1, . . . , am, x1, . . . , xn) = 0].

A set A ⊆ N is Diophantine if and only if the predicate a ∈ A isDiophantine.

It is easy to see that any Diophantine set A is an r.e. set. In fact,for a given element a ∈ A we may search for the natural numbersolutions of the related Diophantine equation. If it has a solution,then we will find one and let the computer stop and give theoutput 1. If it has no solution, the computer will never stop.

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Davis Daring Hypothesis

In 1944 E. L. Post thought that HTP begs for an unsolvabilityproof, i.e., HTP might be undecidable.

In 1949 Martin Davis used Godel’s coding idea to obtain that anyr.e. set A ⊆ N has the following Davis normal form

a ∈ A ⇐⇒ ∃x > 0∀0 6 y 6 x∃z1 > 0 . . . ∃zn > 0

[P(a, x , y , z1, . . . , zn) = 0],

where a is a natural number and P is a polynomial with integercoefficients.

Davis Daring Hypothesis. Any r.e. set A ⊆ N is Diophantine.

Under this hypothesis, for the nonrecursive r.e. setK = {x ∈ N : x ∈ Dom(ϕx)} there is a polynomialP(x , x1, . . . , xn) such that for any a ∈ N we have

a ∈ K ⇐⇒ ∃x1 > 0 . . . ∃xn > 0[P(a, x1, . . . , xn) = 0].

Thus Davis Daring Hypothesis implies that HTP over N isundecidable.

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Systems of Diophantine equations

A system of finitely many Diophantine equations is equivalent to asingle Diophantine equation. In fact, ifPi (z1, . . . , zn) ∈ Z[z1, . . . , zn] for all i = 1, . . . , k, then

P1(z1, . . . , zn) = 0 & . . . & Pk(z1, . . . , zn) = 0

⇐⇒ P21 (z1, . . . , zn) + . . .+ P2

k (z1, . . . , zn) = 0.

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The Davis-Putnam-Robinson Theorem

Theorem (M. Davis, H. Putnam, J. Robinson, Annals of Math.1961) Any r.e. set is exponential Diophantine. Thus there is noalgorithm to decide for any given exponential Diophantine equationwhether it has solutions over N.

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Julia Robinson’s Hypothesis

A. Tarski conjectured in 1948 that {2n : n ∈ N} is not aDiophantine set. His PhD student J. Robinson did not succeed inproving this with serious efforts.

JR Hypothesis (J. Robinson, 1950). There is a Diophantinerelation R(a, b) with a, b ∈ N such that

R(a, b)⇒ b < aa

and∀k > 0∃a > 0∃b > 0[R(a, b) & ak < b].

Under this hypothesis, J. Robinson showed that the exponentialrelation ab = c is Diophantine and hence all r.e. sets areDiophantine. So, the JR Hypothesis implies the negative solutionof HTP.

J. Robinson tried to prove her JR Hypothesis but got no success.This made her depressed and doubt her Hypothesis.

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Matiyasevich’s Theorem

Recall that the Fibonacci sequence (Fn)n>0 defined by

F0 = 0, F1 = 1, and Fn+1 = Fn + Fn−1 (n = 1, 2, 3, . . .)

increases exponentially.In 1970 Yu. Matiyasevich, a 23-year-old Russian, confirmed the JRHypothesis by showing that the relation y = F2x (with x , y ∈ N) isDiophantine! It follows the exponential relation ab = c (witha, b, c ∈ N, a > 1 and c > 0) is Diophantine, i.e. there exists apolynomial P(a, b, c , x1, . . . , xn) with integer coefficients such that

ab = c ⇐⇒ ∃x1 > 0 . . . ∃xn > 0[P(a, b, c, x1, . . . , xn) = 0].

This, together with the Davis-Putnam-Robinson work in 1961, ledMatiyasevich finally confirm Davis Daring Hypothesis.

Matiyasevich’s Theorem (or MDPR Theorem) (1970).Recursively enumerable sets coincide with Diophantine sets. ThusHTP has a negative solution!

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Part III. Reduction of Natural Number Unknowns

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Small ν with ∃ν over N undecidable

For a set S ⊆ Z we let ∃n over S denote the set of formulas

∃x1 ∈ S . . . ∃xn ∈ S [P(x1, . . . , xn) = 0]

with P(x1, . . . , xn) ∈ Z[x1, . . . , xn].

Any nonrecursive r.e. set A has a Diophantine representation:

a ∈ A ⇐⇒ ∃x1 > 0 . . . ∃xn > 0[P(x1, . . . , xn) = 0].

It is interesting to find the least ν ∈ Z+ = {1, 2, 3, . . .} such that∃ν over N is undecidable.

ν < 200 (Matiyasevich, Summer of 1970)ν 6 35 (J. Robinson, 1970)ν 6 24 (Matiyasevich and Robinson, 1970)ν 6 14 (Matiyasevich and Robinson, 1970)ν 6 13 (Matiyasevich and Robinson, 1973 [Acta Arith. 27(1975)])ν 6 9 (Matiyasevich, 1975; details in Jones [J. Symbolic Logic,1982])

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The 9 Unknowns Theorem

The above ideas, together with some other works in the 1975paper of Matiyasevich and Robinson, led Mtijasevich obtain thefollowing celebrated theorem.

Matiyasevich’s 9 Unknowns Theorem: ∃9 over N is undecidable!

The detailed proof of this theorem appeared in Jones [J. SymbolicLogic, 1982].

Up to now, no one has shown that ∃ν over N is undecidable forsome ν < 9, although A.Baker, Matiyasevich and Robinson allbelieved that ∃3 over N might be undecidable.

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∃ over Z is decidable

Matiyasevich and Robinson [Acta Arith. 27(1975)]: Ifa0, a1, . . . , an and z are integers with a0z 6= 0 and∑n

i=0 aizn−i = 0, then

|z |n 6 |a0zn| 6n∑

i=1

|ai ||z |n−i 6n∑

i=1

|ai ||z |n−1

and hence

|z | 6n∑

i=1

|ai |.

Thus ∃ over N and ∃ over Z are decidable (in polynomial time).

It is not known whether ∃2 over Z is decidable. But A. Bakerproved in 1968 that if P(x , y) ∈ Z[x , y ] is homogenous, irreducibleand of degree at least three then for any m ∈ Z there is an effectivealgorithm to determine whether P(x , y) = m for some x , y ∈ Z.

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Relative resultsFor any m ∈ Z, by Lagrange’s four-square theorem

m > 0 ⇐⇒ ∃z1∃z2∃z3∃z4[m = z21 + z2

2 + z23 + z2

4 ].

Thus

∃n over N is undecidable⇒ ∃4n over Z is undecidable.

By the Gauss-Legendre theorem on sums of three squares,

N \ {x2 + y2 + z2 : x , y , z ∈ Z} = {4k(8l + 7) : k , l ∈ N}.

If n ∈ N, then 4n + 1 = (2x)2 + (2y)2 + (2z + 1)2 for somex , y , z ∈ Z, and hence n = x2 + y2 + z2 + z . Thus, for any m ∈ Z,

m > 0 ⇐⇒ ∃x∃y∃z [m = x2 + y2 + z2 + z ].

It follows that

∃n over N is undecidable⇒ ∃3n over Z is undecidable.

Thus ∃27 over Z is undecidable by the 9 unknowns theorem, aspointed out by S.P. Tung in [Japan J. Math., 11(1985)].

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A new relation-combining theorem

Tung (1985) asked whether ∃ν over Z is undecidable for someν < 27.

New Relation-Combining Theorem (Z.-W. Sun [Z. Math. LogikGrundlag. Math. 38(1992)]): Let A1, . . . ,Ak ,B,C1, . . . ,Cn,D,Ebe integers with D 6= 0. Then

A1, . . . ,Ak ∈ � ∧ B 6= 0 ∧ C1, . . . ,Cn > 0 ∧ D | E

⇐⇒ ∃z1 . . . ∃zn+2[P(A1, . . . ,Ak ,B,C1, . . . ,Cn,D,E , z1, . . . , zn+2) = 0],

where P is a suitable polynomial with integer coefficients.

This implies that

∃n over N is undecidable⇒ ∃2n+2 over Z is undecidable.

So ∃20 over Z is undecidable by the 9 unknowns theorem.

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∃11 over Z is undecidableIn 1992, I announced that ∃11 over Z is undecidable.

To achieve this goal, unlike others I did not simply use the relativeresult, instead I adapt the deep proof of the 9 unknowns theoremand made suitable variants so that we can use integer variablesinstead of natural number variables.

My starting point is the use of Lucas sequences with integerindices instead of the usual natural number indices. I publishedthis initial step in Sci. China Ser. A 35(1992).

The whole proof of the undecidability of ∃11 over Z is verysophisticated. It appeared in my PhD thesis in 1992. During1992-2016, despite that many mathematicians wanted to see mydetailed proof, I did not write an English version of that, since Iwas frequently busy with my new discoveries.

After 25 years have passed, I finally spent time to write an Englishpaper which contains the undecidability of ∃11 over Z as well asmy new discoveries related to HTP. The preprint is now publiclyavailable from http://arxiv.org/abs/1704.03504 27 / 39

Part III. Some Algorithms involving Primes

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Initial Results on DiscriminantsTheorem 1 (L. K. Arnold, S. J. Benkoski and B. J. McCabe[Amer. Math. Monthly 92 (1985), no. 4]). Let n > 4 be aninteger. Then the least positive integer m (denoted by D(n)) suchthat 11, 22, . . . , n2 are distinct modulo m, is

min{m > 2n : m = p or m = 2p with p an odd prime}.

Remark. The range of D(n) does not contain those primesp = 2q + 1 with q an odd prime.

The theorem arose from consideration of a problem in computersimulation. The problem may be described as developing a methodto determine quickly the square roots of a long sequence ofintegers. For a long sequence 12, 22, . . . , n2 of squares, you mayuse the function A(x) =

√x (1 6 x 6 n2) to get the square roots.

This needs an array of size 1× n2. If 12, 22, . . . , n2 are distinctmodulo m, then letting A(r) =

√x where r ∈ {1, . . . ,m} and

r ≡ x (mod m). The modulo procedure is , in general, muchfaster than the square root procedure. 29 / 39

Generate all primes in a combinatorial manner

Theorem 1 (Sun, Feb. 29, 2012) (i) For n ∈ Z+ let S(n) denotethe smallest integer m > 1 such that those 2k(k − 1) mod m fork = 1, . . . , n are pairwise distinct. Then S(n) is the least primegreater than 2n − 2.

(ii) For n ∈ Z+ let T (n) denote the least integer m > 1 such thatthose k(k − 1) mod m with 1 6 k 6 n are pairwise distinct. Thenwe have

T (n) = min{m > 2n − 1 : m is a prime or a positive power of 2}.

Remark. (a) The range of S(n) is exactly the set of all primes!(b) I proved that the least positive integer m such that those(k2

)= k(k − 1)/2 (k = 1, . . . , n) are pairwise distinct modulo m, is

just the least power of two not smaller than n.

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Another theorem

Theorem 2 (Sun, March 2012) (i) Let d ∈ {2, 3} and n ∈ Z+.Then the smallest positive integer m such that thosek(dk − 1) (k = 1, . . . , n) are pairwise distinct modulo m, is theleast power of d not smaller than n.

(ii) Let n ∈ {4, 5, . . .}. Then the least positive integer m such that18k(3k − 1) (k = 1, . . . , n) are pairwise distinct modulo m, is justthe least prime p > 3n with p ≡ 1 (mod 3).

Remark. We are also able to prove some other similar results; forexample, for each n > 5 the least m ∈ Z+ such that those18k(3k + 1) (k = 1, . . . , n) are pairwise distinct modulo m, is justthe first prime p ≡ −1 (mod 3) after 3n.

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Alternating sums of primes

Let pn be the nth prime and define

sn = pn − pn−1 + · · ·+ (−1)n−1p1.

Note that

s2n =n∑

k=1

(p2k − p2k−1) > 0, s2n+1 =n∑

k=1

(p2k+1 − p2k) + p1 > 0.

Here are values of s1, . . . , s16:

2, 1, 4, 3, 8, 5, 12, 7, 16, 13, 18, 19, 22, 21, 26, 27.

The sequence 0, s1, s2, . . . were first introduced by N.J.A. Sloaneand J.H. Conway (see A008347 at OEIS).

It is not difficult to show that those sn (n = 1, 2, 3, . . .) arepairwise distinct.

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An amazing recurrence for primes

The following surprising conjecture on recurrence for primes allowsus to compute pn+1 in terms of p1, . . . , pn.

Conjecture (Sun, March 28, 2012). For any positive integern 6= 1, 2, 4, 9, the (n + 1)th prime pn+1 is the least positive integerm such that

2s21 , . . . , 2s2n

are pairwise distinct modulo m.

Remark. I have verified the conjecture for n 6 2× 105, and provedthat 2s21 , . . . , 2s2n are indeed pairwise distinct modulo pn+1.

A Related Conjecture (Sun, March 26, 2012). The least integerm > 1 such that 2S2

k (k = 1, . . . , n) are pairwise distinct modulo

m is a prime smaller than n2 unless n | 6, where Sk =∑k

j=1 pj .

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Conjecture on alternating sums of consecutive primes

Conjecture (Sun, April 2-3, 2012). For any positive integer m,there are consecutive primes pk , . . . , pn (k 6 n) not exceeding2m + 2.2

√m such that

m = pn − pn−1 + · · ·+ (−1)n−kpk .

Examples.

10 = 17− 13 + 11− 7 + 5− 3;

20 = 41− 39 + 37− 31 + 29− 23 + 19− 17 + 13− 11;

2382 = p652 − p651 + · · ·+ p44 − p43,

p652 = 4871 = b2 · 2382 + 2.2√

2382c.

The conjecture has been verified for m up to 109. Most knownresults on primes are about local properties of primes, not aboutrelations of primes.

Prize. I would like to offer 1000 US dollars for the first proof.34 / 39

How to solve x2 ≡ a (mod p)?Let p be an odd prime and a be any quadratic residue modulo p.How to solve the congruence x2 ≡ a (mod p) quickly?

Tonelli-Shanks Algorithm. Knowing a quadratic nonresidued ∈ Z mod p, one can solve x2 ≡ a (mod p) in polynomial time:

Write p − 1 = 2st with s, t ∈ Z+ and 2 - t, and find even integers

m1, . . . ,ms with (admi )2s−i t ≡ 1 (mod p) for all i = 1, . . . , s in the

following way: m1 := 0, and after those m1, . . . ,mi (with1 6 i < s) have been chosen we select mi+1 ∈ {mi ,mi + 2i} such

that (admi+1)2s−i−1t ≡ 1 (mod p). Note that ((admi )2

s−i−1t)2 ≡ 1

(mod p) and hence (admi )2s−i−1t ≡ ±1 (mod p). If

(admi )2s−i−1t ≡ −1 (mod p), then

(admi+2i )2s−1−i t ≡ −d2s−1t = −d (p−1)/2 ≡ 1 (mod p).

As (adms )t ≡ 1 (mod p), we have x2 ≡ a (mod p) withx = ±a(t+1)/2(d t)ms/2.

Remark. In 1985 R. Schoof gave a polynomial (depending on a)algorithm to solve the congruence x2 ≡ a (mod p) with a given. 35 / 39

Find a quadratic nonresidue modulo a prime

However, there is no known deterministic, polynomial timealgorithm for finding a quadratic nonresidue d modulo the oddprime p. Under the Extended Riemann Hypothesis for algebraicfields, it can be shown that there is a positive quadratic nonresidued < 2 log2 p; and so an exhaustive search to this limit succeeds infinding a quadratic nonresidue in polynomial time.

As Fibonacci numbers grow exponentially, our following conjectureis particularly interesting since it implies that we can find squareroots for quadratic residues modulo any odd prime p indeterministic, polynomial time.

Conjecture (Z.-W. Sun, 2014) For any integer n > 4, there is aFibonacci number f < n/2 with x2 ≡ f (mod n) for no integer x .

Remark. This can be reduced to the case with n = p prime. Wehave verified that for any prime 3 < p < 3× 109 there is aFibonacci number Fk < p/2 with (Fk

p ) = −1.

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The partition function p(n)

A partition of a positive integer n is a way to write n as a sum ofpositive integers with the order of addends ignored. The partitionfunction p(n) denotes the total number of partitions of n.

Example. p(5) = 7 since

5 =1 + 4 = 2 + 3 = 1 + 1 + 3 = 1 + 2 + 2

=1 + 1 + 1 + 2 = 1 + 1 + 1 + 1 + 1.

Hardy-Ramanujan Formula:

p(n) ∼ eπ√

2n/3

4√

3nas n→ +∞,

and hence

log p(n) ∼ c√

n with c = π

√2

3.

Note that p(n) grows eventually faster than any polynomial in nbut slightly slower than 2n. Mathematica could compute p(n)quickly.

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On primes related to p(n)Conjecture 1 (Sun, 2014-02-27). Let n be any positive integer.Then one of the n numbers

p(n) + 1, p(n) + 2, . . . , p(n) + n

is prime.

Conjecture 2 (Sun, 2014-02-28). Let n > 1 be an integer. Thenp(n) + p(k)− 1 is prime for some 0 < k < n.

Conjecture 3 (Sun, 2014-03-13). Let n > 3 be an integer. Thenp(n + k) + 1 is prime for some k = 1, . . . , n.

Conjecture 4 (Sun, 2014-03-12). Let n > 1 be an integer. Thenthere exists a number k ∈ {1, . . . , n − 1} such thatkp(n)(p(n)− 1) + 1 is prime. Also, we may replacekp(n)(p(n)− 1) + 1 by p(k)p(n)(p(n)− 1) + 1 orp(k)p(n)(p(n) + 1)− 1.

These conjectures might be helpful in finding large primes forthe use of RSA.

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References

For main sources of my work mentioned here, you may look at:

1. Z.-W. Sun, Reduction of unknowns in Diophantinerepresentations, Sci. China Math. 35(1992), 257–269.

2. Z.-W. Sun, A new relation-combining theorem and itsapplication, Z. Math. Logik Grundlag. Math. 38(1992), 209-212.

3. Z.-W. Sun, Further results on Hilbert’s tenth problem,arXiv:1704.03504, http://arxiv.org/abs/1704.03504.

4. Z.-W. Sun, On functions taking only prime values, J. NumberTheory 133 (2013), 2794–2812.

5. Z.-W. Sun, Problems on combinatorial properties of primes, in:M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory:Plowing and Starring through High Wave Forms, Proc. 7thChina-Japan Seminar (Fukuoka, Oct. 28–Nov. 1, 2013), Ser.Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp.169-187.

Thank you! 39 / 39