A 60,000 DIGIT PRIME NUMBER OF THE FORM x 2 + x Introduction Stark-Heegner Theorem. Let d > 0 be a square-free integer then Q( d) has

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1 A 60,000 DIGIT PRIME NUMBER OF THE FORM x + x + 4. Introduction.. Euler s olynomial. Euler observed that f(x) = x + x + 4 takes on rime values for 0 x 39. Even after this oint f(x) takes on a high frequency of rimes. For instance among f(), f(),... f(0 6 ), 6,080 are rime comared to the sequence,, where there are 78,498 rimes.... Stark-Heegner Theorem. Let d > 0 be a square-free integer then Q( d) has class number d {,, 3, 7,, 9,, 67, 63}. Originally conjectued by Gauss, this was essentially roved by Karl Heegner albeit with gas in 95. It was comletely roved by Harold Stark in 967. The numbers are known as Heegner numbers. In 93 Rabinowitz roved that n + n + A gives rimes for 0 n A if and only if its discriminant 4A is minus a Heegner number. A yields A so A is maximal.,, 3 are not of the required form since 4A = h A = h +. So the Heegner 4 numbers that work 7,, 9,, 67, 63, yield rime generating functions corresonding to A =, 3, 5,, 7, 4. These are called the Lucky Numbers of Euler... Hardy and Littlewood Conjecture. Hardy and Littlewood had a number of recise conjectures about rime distributions. The first conjecture: the rime k-tules conjecture (rime k-tule: (, + a,... + a k ) is a sequence of rimes such that a k is as least as ossible.) imlies that for any ositive integer m, A such that x + x + A is rime for 0 x m.

2 Note that we aren t saying that m = A so this does not contradict Rabinowitz. So with a large enough choice of A, Euler s olynomial can be beaten..3. Hardy and Littlewood s Conjecture F. Consider f(x) = ax + bx + c. If f(x) for some x Z, then = b 4ac, the discriminant of f(x) must be a square modulo. Thus if is not a square modulo for many rimes, we exect f(x) to take on many rime values asymtotically. Hardy & Littlewood formalized this in conjecture F..3.. Conjecture (F). Let a > 0, b, c be integers such that gcd(a, b, c) =, = b 4ac is not a square and a + b, c are not both even. Then there are infinitely many rimes of the form f(x), and π f (n) ɛc f Li(n), where, ɛ = Li(n) = n dx log x when a + b, otherwise, and C f = > (a,b) > a ( ) ( ) The roducts in the exression for C f are taken over the rimes only, and denotes the Legendre symbol. Note here that ɛc f is what really determines the density of rime values assumed by f, since Li(n) is a function of n only. The larger ɛc f is, the higher the asymtotic density of rime values for any quadratic olynomial of discriminant.

3 For olynomials of the form f A (x) = x + x + A, π fa (n) C( )L A (n), where C( ) = 3 ( ) Here = 4A. No one has found a olynomial of the form f A (x) that reresents distinct rimes for more than the first 40 values of x. So Euler s olynomial f 4 (x) = x + x + 4 holds the record. For Euler s olynomial C( 63) = H.C. Williams, M. Jacobson, G.W. Fung have looked at finding quadratic olynomials which have a high density of rime values. For instance x + x has C( ) = So we exect this olynomial to assume more rimes asymtotically than f 4. It starts off slowly, only 4 rimes for x 00 comared to 87, but for x 0 7 it assumes 570 rime values comared to only 0897 by Euler s olynomial..4. Primality Testing. The best general rimality roving method not based on factorizations is the Ellitic Curve Primality Proving method (ECPP). The largest rime roved by ECPP has 6,64 digits by Francois Moran in 0. The Lucas-Lehmer-Riesel test is a rimality test for numbers of the form N = k n, with n > k. It is the fastest deterministic algorithm known for numbers of that form. The Brillhart-Lehmer-Selfridge test is the fastest deterministic algorithm for numbers of the form N = k n +. The largest known rime of the form x + is , with,558,647 digits. The latter two methods require knowledge of artial rime factorizations of N + and N resectively 3

4 .5. Brillhart-Lehmer-Selfridge Theorem. Suose that N > is odd and write N = F R where F is even and the rime factorization of F is known. Suose also that ( ) /3 N () F > () For each rime, dividing F, there is an integer a i, so that a N i (mod N) and N gcd(a i i, N) =, (3) If we write R = F q + r, where r < F, then either q = 0 or r 8q is not a erfect square, then N is rime..6. Aroach. To find large rimes of the form f(x) = x + x + 4, find olynomials g(x) so that f(g(x)) is reducible. Doing a comuter search reveals the choice g(x) = 40x 3 + 4x + 4x +, for which f(g(x)) = (40x + x + )(40x 4 + 8x 3 + 3x + 84x + 4) Let h(x) = 40x + x + and i(x) = 40x 4 + 8x 3 + 3x + 84x + 4. Thus for any given choice of x, we have N = f(g(x)) with h(x) = F and i(x) = R. The goal: Find a choice of x (with known rime factorization), for which 40x + x + is rime (use B-L-S theorem) and for which f(g(x)) is rime (again use B-L-S theorem). Thus there is a simultaneous rimality requirement which increases the number of candidate values of x we must search.. Strategy.. How Many Numbers Do We Need to Check? We are looking for a 0,000 digit rime (corresonding to h(x)) and a 60,000 digit rime corresonding to f(g(x)). By The Prime Number Theorem, the density of rimes close to an integer N is aroximately equal to ln(n). If we assume the same density of rimes within the values of Euler s Polynomial as within the set of all integers, we multily the robabilities of finding a 0,000 digit rime and a 4

5 60,000 digit rime. ( ) ( ) ln( ) ln( ) 6, 34, 34, 060 If we were to test M numbers, the chance of finding at least one rime air would be ((N )/N) M where N = 6, 34, 34, 060. If we set M = 3N, then ((N )/N) 3N ( ) % e.. Pre-Sieving with Primorials. The rimorial n# is the roduct of rimes less than or equal to n. f(g(x)) = 600x x x x x + 6x +. By choosing x = k. n#, h(x) and f(g(x)) will not be divisible by any rime less than n. The density of numbers n divisible by is. thus for each otential rime divisor eliminated, the number of otential rimes decreases by where is the divisor eliminated. 3, # Choose as a factor of x. Thus the number of numbers to check should be ( ) (3 6, 34, 34, 060) rime < 3, Merten s theorem states that rime x ( ) e γ ln(x) With this aroximation there are 59,48,3 numbers to check after re-sieve. ( ) 3, #.3. Sieving. In f(g(x)) we had x = k, thus sieving was designed to eliminate values of k for which f(g(x)) was a multile of a rime. To do this, look for roots of the olynomials, h(x) and f(g(x)), in F. 5

6 Factor h(x) = 40x + x + mod and comute its roots in F. Then, eliminate the choices of k for which the corresonding x value is a root. Reeat this rocess on f(g(x)). The number of numbers left after sieving u to is aroximately (59, 48, 3) rime x ( ) By Merten s Theorem aroximately, 9,94,04 numbers would remain after sieving..4. Comutations. The comutations were done on a cluster of 00 nodes, each running tests on grous of 900 numbers. Run Fermat seudo rimality tests to find robable rimes base 3 on OenPFGW. (3 x mod x). This generates a list of 0,000 digit seudo rimes. Periodically check the corresonding 60,000 digit numbers for seudo rimality base 3. Test the 60,000 digit seudo rime for rimality. The total amount of CPU time used was: 5 days for sieving 535 days for 0,000 digit seudo-rimality tests 4 days 60,000 digit seudo-rimality tests. 3. Result 3.. Theorem(Justin Debenedetto and Jeremy Rouse). Let f(x) = x +x+4 and g(x) = 40x 3 + 4x + 4x +. If we set x = # then f(g(x)) is a 60,000 digit rime number. 6