Significant improvement of Fermat's congruent numbere discriminant theorem and ten new algorithms

Zcy Congyao College of Information Engineering, Hunan University Yu Wei Department of Statistics and Financial Engineering,Ningbo University Tang Xiaoning Beijing Haitian Starting Point Technology Service Co. LTD Abstract In this paper, the sufficient and necessary condition theorem of Fermat, Guan Xun GUI is improved greatly.Ten new algorithms are proposed for the first time,which make it possible to solve many extremely difficult problems of the congruent number. 1)Algorithm 1st is for prime p≡7(mod 8),such as 127,191,etc.of course,it is also effective for composite numbers with N111 structure,such as 119. 2)Algorithm 2st is for prime p≡5(mod 8),such as 157,173,269etc.of course,it is also effective for composite numbers with 11N1 structure,such as 85. 3)Algorithm 3 is for composite numbers with type 112N,such as 2*307 or 2*411. 4)Algorithm 4 is for composite numbers with type N121,such as 2*151 5)Algorithm 5 is for composite numbers with type 1k1N,such as 5*161. 6)Algorithm 6 is for composite numbers with type k11N,such as 3*101,when N=1,It can also be calculated as 127,etc. 7)Algorithm 7 is for composite numbers with type N1k1,such as 79 *5,when N=1,It can also be calculated as 157, etc. 8)Algorithm 8 is for composite numbers with type Nk11, such as 241*5 9)Algorithm 9 is for composite numbers with type 1kN1, such as 3*257. 10)Algorithm 10 is for numbers with type 1N11,this is Fermat's discriminant theorem,but our algorithm is more expensive than Fermat-xungui Guan. 11)Algorithm Ast,can be calculated all congruent number,the disadvantage is that it is too slow. Algorithm 1 to 10 all achieve n times of operation, up to n4 effect, this is the biggest progress.

Key words: congruent number,elliptic curve,number theory,group

0.historical background It has been more than a thousand years since the Persian mathematician Karaji put forward the problem of congruent number in the 10th century[1] Up to now,there are mainly the following theorems Theorem 1)If Prime p≡5,7(mod 8),then p is congruent number(Stephens)[2] Theorem 2)1952,Heegner & Birch,It is proved that p≡3(mod 4),then 2p are congruent number[2] Theorem 3)If Prime p,q and n=pq≡5,or 6,or 7(mod 8),then n is congruent number. (Heegner,Birch,Stephens) [2] Alter,Kurtz & Kubota Further conjecture n=pq≡5,or 6,or 7(mod 8),then n is congruent number (no proof yet) Theorem 4)If Prime p≡3(mod 8),then p is not congruent number(1855,Genocchi [2] Theorem 5)If Prime p≡5(mod 8),then 2p is not congruent number(1874,Genocchi)[2] Theorem 6) If Prime p≡9(mod 16),then 2p is not congruent number (1915,Bastieni) [2]

Theorem 7)Let the coefficients of the expansion of BSD modular function are an, bn, There n is a positive integer without square factor,

When an≠0,then n is not congruent number,when bn≠0,then 2n is not. congruent number. (Cotes-wiles,1979) [3] Theorem 8)(Feng Keqin 1996,Li Delang,Tianye 2000,Zhao Chunlai 2001)[4] For any integer k, we have infinitely many non congruent numbers,which indeed contain k odd primes, meanwhile there are also infinitely many congruent numbers, which indeed contain k odd primes (existence theorem) Theorem 9)Let A be the congruent number if and only if (Guan xungui,2009) Equations x2+Ay2 = z2 x2−Ay2 = w2 There are integers a,b,v, Let n =|6 | or 4ab( ) hold,Necessary and sufficient conditions for abbreviation g_1,g_2[5]

Theorem 10)Let p0≡5,7(mod 8)(separately p0≡3(mod4),pi ≡1(mod8),i=1 …k, p0, pi is odd prime,

and satisfied (pi/p0) = −1), i=1, … ,k,and (pi/pj) = 1, 1≤j≤i-1 then [6] n=p0p1 … pk(separately n=2p0p1 … pk) is congruent number.(Tian Ye,2012) Theorem 11)If Prime p,q≡3 (mod 8),then pq is not congruent number (2019,Zhou congyao,Yu wei,Tang xiaoning) [7] Theorem 12)If Prime p,q≡3(mod 8),then 2pq is not congruent number (2019,Zhou congyao,Yu wei,Tang xiaoning) [8] Theorem 13)If Prime p,q≡5(mod 8),then 2pq is not congruent number (2019,Zhou congyao,Yu wei,Tang xiaoning) [8] Theorem 14)For positive integers without square factor,once there is no solution in N-1 position of the combination factor, i.e. N-1 criterion,then A is not congruent number. (2019,Zhou congyao,Yu wei,Tang xiaoning)[9] There are several problems about the congruent number (The third problem -- the nature of the solution -- not discussed here,and the interested parties can refer to[11]): 1. Looking for a rule, The proof of certainty whether a positive integers are the congruent number (not completely solved).If anyone solves the BSD conjecture, which is equivalent to this problem, he will get a $1 million bonus [3]. 2.If n∈N is congruent number, how to find the solution(there is no effective algorithm, even a slightly better method),The calculation of the congruent number may be a hundred times more difficult than the discrimination. Zagier is the originator of the calculation of the congruent number.Don Zagier He is recognized as a genius in today's world, At the age of 16,he received both bachelor's and master's degrees in mathematics at MIT, At the age of 19, he received a doctorate in mathematics from Oxford,Cambridge,UK, At present, he is the director of algebra and number theory in German Institute of mathematics.It's just that he worked very hard to work out the three sides of the right triangle corresponding to 157 (or the three sides with the smallest denominator)[3]. He worked out that the solution of the congruent number 157 is a number whose s and t are all 24 decimal digits, Don't underestimate it,Because the algorithm of calculating the congruent number is a double loop,That is,the calculation amount of 1048,It's a quantum computer One second 1020 times of addition, subtraction, multiplication, division and root (it's actually impossible), but it's also necessary 1019 years. Zagier There is no published algorithm,and we do not know what method to use.We find two ways to find the solution of 157. In addition to the above theorem, there are three conjectures 1)BSD conjecture,2)AKK conjecture,3)Tunnell conjecture,Please refer to[3] [7] [8] [9] [11] [72]

1．The necessary and sufficient condition theorem of Fermat

Fermat's discriminant theorem: A is congruent number if only if equations x2+Ay2 = z2 x2−Ay2 = w2 Existence of integer solutions with y ≠ 0 x,y,z,w (theorem 9 in the previous section is a step ahead of this theorem.)

2．Preparatory theorem

2.1 Preparatory theorem 1:

Let f(x)= +…+ Is an algebraic equation with integral coefficients,

T(f,m) express the number of solutions of f(x)≡0 (mod m), here m= … , is prime number

Then congruence equation f(x)≡0 (mod m) (1)

and Congruence Equations f(x)≡0 (mod ) (i=1,…k) (2) equivalent and T(f,m)=T(f, )T(f, )…T(f, ) [10] (p182,theorem 1)

That is to say,once one of the equations of (2) has no solution,then the global congruence equation (1) has no solution. 2.2 Preparatory theorem 2:

Let a ,there abc without square factor,a,b,c are not all positive or all negative, then if only if ab(mod |c|), ≡ bc(mod |a| and a (mod |b|) all have solutions, then quadratic equation a there are positive integer solutions.

[10] (p282 theorem 1) 2.3 Preparatory theorem 3: If A is congruent number, there are at least two types of solutions (s,t),(S,T), the latter is Fermat type (1,A,1,1). 2.4 Some identities: 1) The general solution of x2 + y2 = 2z2 is x=|u2 − v2 − 2uv|, y=|u2 − v2 + 2uv|, z=u2 + v2 2) The general solution of x2 + 2y2 = z2 is x=|u2 − 2v2|, y=2uv,z=u2 + 2v2 question 9 of p102 exercise 2 of reference material 10, where (u,v)=1,u,v one is odd and one is even 2 2 2 2 2 2 2 3)The general solution of x + ky = z is x=|k1u − k2v |,y=2uv,z=k1u + k2v here

k=k1 ∗ k2, k has no square factor. 2 2 2 2 2 2 2 4) The general solution of kx + y = z is x=2uv,y=|k1u − k2v |,z=k1u + k2v

there k=k1 ∗ k2,there is no square factor. Here 3) and 4) is one thing,but the solution of the congruent number is very different,because the order is different, the type of the congruent number is different.

When k1 = 1, k2 = 1,it is the Pythagorean right triangle; [10] 当k1 = 1, k2 = 2, It is the identity 2) , thousands of other cases 3) and 4) are first discovered and applied to the calculation of congruent number.

3. Symbols and definitions Definition 1: the solution and the type of solution of the congruent number Letsoution of congruent number is (s,t).here s>t is positive integer,(s,t)=1,s,t one odd and one even. According to the definition A×m2 = s − t ts(s + t), let m = xyzw, here x,y,z,w are the maximum square factors of (s-t), t, s and (s + t), this is x2||(s-t), y2||t, z2||s, w2||(s+t), let a=(s-t)/ x2,b=t/y2,c=s/z2,d=(s+t)/w2,then A=abcd is congruent number, four tuples (a, b,c, d) are called the types of solutions; st(s+t)(s-t) Is the area of a right triangle,m2 Is the largest square factor of the area of a right triangle. We call (s, t) the solution of the congruent number, and four tuples (a,b,c, d) is the type of solution. we call (a, b,c,d) = (1,A,1,1) is the Fermat type, abbreviated as F=(1a11) Obviously, s-t, t, s,s+t are mutually prime. There are only two cases of parity: case a: ooeo; case B: oeoo For type of a,b,c,d The result of all modulo 8 is called, It is abbreviated as (a'b'c'd), Its complete set is the M8 set. From (s,t) to (s-t,t,s,s+t) remove the maximum square factor of these four variables to get (a,b,c,d); On the contrary, there is no effective algorithm

Definition 2: Factor combination number Y (A) ,and factor combination set C of positive integer A The element of factor combination C1,C2,…,Cn,… Here A is a positive integer without square factor,and the factor combination of A is as follows: For composite A,each odd prime factor is multiplied by 4,and if it contains 2,it is multiplied by 2 Example 1: A=5 then C={(5111),(1511)(1151)(1115)}={C1..C4},Y(5)=4 Example 2: A=6 then C={(3211),(1611)(1231)(1213), (3121),(1321),(1161),(1123)}={C1..C8},Y(6)=8 Example 3: A=15 thenC={(15,1,1,1),(3,5,1,1),(3,1,5,1)(3,1,1,5), (5,3,1,1),(1,15,1,1),(1,3,5,1),(1,3,1,5), (5,1,3,1),(1,5,3,1),(1,1,15,1),(1,1,3,5), (5,1,1,3),(1,5,1,3),(1,1,5,3),(1,1,1,15)}={C1..C16},Y(15)=16 We have an algorithm to automatically generate the complete set of factor combinations [7] [8] [9]

[10] [11]

Definition 3: Four equations Q1, Q2, Q3, Q4 and 12 equations qij For each factor combination Ci, there are Q1:ax2 + by2 = cz2 Q2:by2 + cz2 = dw2 Q3:ax2 + dw2 = 2cz2 Q4:dw2 − ax2 = 2by2 q11:S2≡bc(mod |a|) q12: S2≡ac(mod |b|) q13: S2≡−ab(mod |c|)[7] [8] [9] [10] [11] q21:S2≡cd(mod |b|) q22: S2≡bc(mod |c|) q23: S2≡−bd(mod |d|) q31:S2≡2cd(mod |a|) q32: S2≡2ac(mod |d|) q33: S2≡−ac(mod |2c|) q41:S2≡2ab(mod |d|) q42: S2≡2bd(mod |a|) q43: S2≡ad(mod |2b|) These 12 qij, for every combination (a'b'c'd) of Mi, are used to determine whether Q1, Q2, Q3 and Q4 of the corresponding factor combination have solutions

Definition 4: The type upper bound estimator function of the solution Our definition of Z(A) serves two purposes: First Identify which combinations may have solutions [7] [8] [9] [10] [11]； Second Ten new algorithms are designed for these feasible combinations by using the new sufficient and necessary condition theorems and several identitie.

4. Our theorem for congruent number and necessary and sufficient conditions

퐚퐱ퟐ + 퐛퐲ퟐ=c퐳ퟐ (1) 퐛퐲ퟐ + 퐜퐳ퟐ=d퐰ퟐ (2) If and only if there is a positive integer solution A퐦ퟐ=st(퐬ퟐ − 퐭ퟐ) (3) There s>t,(s,t)=1,s,t one odd and one even (4) 퐱ퟐ||(s-t), 퐲ퟐ||t, 퐳ퟐ||s, 퐰ퟐ||(s+t),m=xyzw, a=(s-t)/퐱ퟐ,b=t/퐲ퟐ,c=s/퐳ퟐ,d=(s+t)/ 퐰ퟐ (5) Then A=abcd is congruent number. Symbols' | | 'said the biggest square factor. [Prove] by(4), s-t, t, s,s+t four positive integers are reciprocal prime By (5), s=cz2,t=by2,s-t=ax2,s+t=dw2 a,d It has to be odd, b,c at most have one even numbe. if equation (1) holds, this (s-t)+t=s=cz2 ,and equation (2) holds,this is t+s=s+t=dw2 then st(s2 − t2)=(cz2)(by2)(s+t)(s-t)= (cz2)(by2)(dw2)(ax2)=abcd(xyzw)2=Am2 this is (3) Conversely, condition (3) holds,by condition (5) s=cz2,t=by2,s-t=ax2,s+t=dw2,then ax2 + by2=(s-t)+t=s=cz2 this is (1)holds, by2 + cz2=(s)+(t)=s+t=dw2 this is (2) holds,this is (1)(2). Last (1)(2) ↔ (3), That is, if and only if the system has a solution (3),there A is congruent number. According to the different situations of a,b,c and d, we have found and designed the following ten algorithms, All of these algorithms are calculated by u,v, get The power of solving s,t, about u4, v4, the first time,we have connected the solution of congruent number to a system of quaternion quadratic equations, and achieved good results. A lot of congruences that you couldn't calculate before,now you can,and that's a big step forward.

When (a,b,c,d)=(1,A,1,1), Our theorem is Ferma's sufficient and necessary condition theorem. x2 + Ay2=z2 (1) Ay2 + z2=w2 (2) From (2) we get (3), and from (1) we get (4). z2 + Ay2=w2 (3) z2 − Ay2=x2 (4) Equations (3) and (4) are the equations of the original Fermat,s sufficient and necessary condition theorem in Section 1.

5. Ten new algorithms for congruent number

5.1 Algorithm 1 : N111 algorithm According to Theorem 2.3, we know that there are two types of solutions to primes p≡7 (mod 8), which are of type 1p11 and p111 according to Z(A) Where 1p11 is Fermat type,the number of solutions is larger, so let's start with type p111 This time ,the equations are as follows: px2 + y2=z2 (1) y2 + z2=w2 (2) By (1),z>y, and by (2) general solution is y=min(u2 − v2,2uv), z=max(u2 − v2,2uv),w=u2 + v2 (3) substitute the general solution (3) for the equation (1), As long as (1) is true in the range of positive integers px2=z2 − y2. For prime 127,we obtained u=145249 v=60248 x=97427521；y=17467450497；z=17501923504；w=24727093505 s=z2=306317326339867638016, t=y2=305111826865145547009 For prime 191, u=27469,v=11580,x=10173863;y=620449561;z=636182040；w=888642361 s= z2 =404727588018561600, t=y2=384957657745092721 If you use the original double cycle, in a finite lifetime, it's impossible to get to their solution. This method is also valid for composite numbers, as long as the Z(A) function is used and it is certain that it has A solution of type N111, for example 287=7*41

5.2 Algorithm 2 : 11N1 algorithm According to Theorem 2.3, we know that there are two types of solutions to primes p≡5 (mod 8), which are of type ,by Z(A) we know The types are 11p1 and 1p11.is Fermat type, The number of solutions is larger, So let's start with type 11p1: This time ,the equations are as follows: x2 + y2=pz2 (1) y2 + pz2=w2 (2) By (1)(2), w2 − x2=(y2 + pz2) − (pz2 − y2)=2y2, this x2+2y2=w2 (3) The general solution of equation (3) isx=|u2 − 2v2|, y=2uv, w=u2 + 2v2 (4) The general solution (4) The substitution equation pz2=z2 + y2, As long as this expression is true in the range of positive integers. For prime 157,We obtained u=356441, v=571761 x=526771095761；y=407598125202；z=53156661805；w=780871468723 s= pz2 =443624018997429899709925；t=y2 =166136231668185267540804 We obtained 173,u=177521 v=137059 x=6056633521；y=48661701478；z=3728226965；w=69084044403 s= pz2=2404644000341688241925 t=y2 =2367961190733987384484 We obtained 269, u=1185271,v=34891, X=1402432579679；y=82710580922；z=85656402325；w=1407302107203 s= pz2 =1973658180741549394113125 , t=y2=6841040196454710370084 The numbers of these solutions are so large that it is impossible to solve them using primitive methods. For composite numbers, just use the Z(A) function to be sure that it has A solution of type 11N1, This approach is also valid, for example 265=5*53, We obtainedu=21,v=4.

5.3 Algorithm 3 : 112N algorithm So let's take A=822, and we know from Z(822) that there are two types of solutions to 822 1.1.2.411 和 1.A.1.1, 1A11 .is Fermat type, The value of the solution is larger, so start from the type 1.1.2.411, At this time The equations are as follows: x2 + y2=2z2 (1) y2 + 2z2=411w2 (2) By (1), general solution is x = |u2 − v2 − 2uv|, y=|u2-v2+2uv|, z=u2+v2 (3) The substitution equation (2), 411w2=y2 + 2z2, As long as this expression is true in the range of positive integers For composite numbers 822,We obtained u=418, v=61 x=120007；y=221999；z=178445；w=16579 s= 2z2 =63685236050；t=y2 =49283556001 Even with this small solution, we won't be able to figure it out until 2020, and using this algorithm, it turns out that u and v are so small. We can also easily find the solution of 2*131,2*307 with this type of structure.

5.4 Algorithm 4: N121 Algorithm Here take A=302=2*151 as an example, according to Z(302), we know 302, There are two types of solutions: ,1.2.1.151 and 1.302.1.1, Here 1A11 .is Fermat type, The value of the solution is larger, so start from the type 1.2.1.151 , At this time The equations are as follows: 151x2 + y2=2z2 (1) y2 + 2z2=w2 (2) By (2), the general solution is y = |u2 − 2v2|, z=2uv, w=u2+2v2 (3) The substitution equation (2),get 151x2=2z2 − y2, As long as this expression is true in the range of positive integers. For 302, we find, u=60401 v=17945 X=49715543；y=3004234751；z=2167791890；w=4292326851 S= 2z2 =9398643356699544200；t=y2 =9025426439116032001 This solution is so large that it was impossible to solve it using primitive methods, The amount of computation. 1038,It will take 1010 years. We can also easily find the solutions of 2*103 or 2*167 and other congruent number with this type of structure.

5.5 Algorithm 5: 1k1N Algorithm Here take A=805=5*161 as an example According to Z(805), we know that 805 has two types of solutions:1.5.1.161and 1.805.1.1, Her 1A11 is Fermat type, The value of the solution is larger, so start from the type 1.5.1.161 , At this time The equations are as follows: x2 + 5y2=z2 (1) 5 y2 + z2=161w2 (2) By (1), the general solution is x = |5u2 − v2|, y=2uv, z=5u2+v2 (3) The substitution equation (2),get 161w2=5y2 + z2, As long as this expression is true in the range of positive integers. For 805,we get u=1, v=2, x=1；y=4；z=9；w=1, s= z2 = 81；t=5y2 = 80 The u v of this solution is much smaller than s t. ST/UV =3200, indicating that this method is 3200 times faster than the original method. If there are multiple factorizations of k, then there are multiple expressions for x,y, and z, and these different expressions may be evaluated at different speeds, but they all have the same result. Such as 15*3743689,There are the following situations: x = |au2 − bv2|, y=2uv, z=au2+bv2 g=gcd (y,z), The final result is that (s, t) are the same thing)

Table 1: many combinations of a ,b with different parameters u and v but the solutions (s,t) are equivalent a b u v x y z w g s t 15 1 11 48 489 1056 4119 3 3 16966161/9 16727040/9 5 3 11 16 163 352 1373 1 1 1885129 1858160 3 5 16 11 163 352 1373 1 1 1885129 1858160 1 15 48 15 489 1056 4119 3 3 16966161/9 16727040/9

5.6 Algorithm 6: k11N Algorithm The content of this algorithm is relatively rich. All congruent number with type structure K11N can be used with this method. For example, 303 has type 3.1.1.101 and 265 has type 5.1.1.53

All prime numbers for p≡7(mod 8) are as 7、23、31、47、71、79、103、127、167 Some composite numbers for N≡7(mod 8) are as 287=41*7 also has a 287.1.1.1 structure Where 1A11 Is Fermat type and the value of the solution is relatively large, Try to avoid it,start with other types,for 303,from type 3.1.1.101start, set k = k1 ∗ k2 k x2 + y2=z2 (1) y2 + z2=Nw2 (2) 2 2 2 2 By (1), the general solution is x = |k1u − k2v | , y=2uv, z=k1u +k2v (3) The substitution equation (2),get Nw2=y2 + z2, As long as this expression is true in the range of positive integers. For 303,we obtained u=109, v=607, a=3 b=1 g=2 X=132326,y=332806,z=404092,w=52090 S=z2 =163290344464/g2=40822586116 ,t= y2=110759833636/g2=27689958409 The u v of this solution is much smaller than s t, st/uv=1016. If there are multiple factorizations of k, then x,y,z There are multiple expressions, they might be evaluated at different speeds, but they give the same result. For 15*3743689, there is the following situation:x = |au2 − bv2|, y=2uv, z=au2+bv2 (s,t are equivalent in the following table, g=gcd(y,z)

Table 2: Different combinations of a,b but the solutions (s,t) are equivalent a b u v x y z w g s t 15 1 11 48 489 1056 4119 3 3 16966161/9 16727040/9 5 3 11 16 163 352 1373 1 1 1885129 1858160 3 5 16 11 163 352 1373 1 1 1885129 1858160 1 15 48 15 489 1056 4119 3 3 16966161/9 16727040/9 In particular, when N=1, primes can be computed as well as the algorithm N111 , here set k=p,N=1, but the formula is very different from the algorithm. k x2 + y2=z2 (1) y2 + z2=Nw2 (2) 2 2 2 2 By (1), the general solution is x=2uv,y=k1u − k2v , z=k1u − k2v , here k=k1 ∗ k2 , k1 = p, k2=1(或k1=1, k2=p) The substitution equation (2),we get w2= y2 + z2=2(p2u4+v4), So just by simple derivation,u and v are both odd, and you're going to have to do another 3/4 less computation. Here are a few examples: N u v x y z w Check formula w2 = 2(p2u4+v4) In integers 7 1 1 2 6 8 10 w2 = 2(7214+14)=100=102 23 1 17 34 266 312 410 w2 = 2(23214+174)=4102 31 1 7 14 18 80 82 w2 = 2(31214+74)=822 127 521 187001 x=194855042, y=34934900994, z=35003847008, w=49454187010 w2=2(12725214+1870014)=494541870102 It's interesting to compare algorithms 1 and 6,.(s,t) are the same,and the calculation amount is quite different. Algorithms 1,g=1 u=145249 v=60248 s= 306317326339867638016 t= 305111826865145547009 algorithms 6 g=2 u=521 v=187001 s=1225269305359470552064/4 t=1220447307460582188036/4 amount of calculation L1=145249*60248/2, L6=521*187001/4, L1/L2 ≈180 So for a congruent number, there's actually a choice of compute methods.

5.7 algorithms 7: N1k1 algorithms The situation of algorithm 7 is also rich, where N and k have various combinations, such as The composite number 395 has a solution of type 79.1.5.1 The composite number 22895= 4579*5 also has a solution of type 4579.1.5.1 In particular,N=1,k=p, can be solved p≡5(mod 8) ,or a composite number which has solution of type N.1.1.1, for 157, The available algorithm7 and the formula is different from the algorithm2, and the calculation is different, but the result is equivalent. For 157 , we use algorithm 2, get u145249,v=60428 by 157, algorithm7, methods the following: Nx2 + y2=kz2 (here N=1) (1) y2 + 157z2=w2 (2) By (2), We know that the general solution is y=pu2 − v2,z=2uv,w=pu2 + v2, substitution equation(1), we obtain Nx2=kz2 − y2, after searching, get u=87005,v=610961 g=2 x=1053542191522,y=815196250404,z=106313323610,w=1561742937446 s=1774496075989719598839700/4 t=664544926672741070163216/4 It is different from algorithm 2, u=356411,v= 571761, here, but x,y,z,w,s,t are equivalent by algorithms 2，u=356411,v= 571761, here u=87005,v=610961,but x,y,z,w,s,t to be equivalent. x=526771095761；y=407598125202；z=53156661805；w=780871468723 s= 443624018997429899709925 t=166136231668185267540804 exactly the same as The calculation of 87005*610960/2 than the algorithm 2 is about the same

5.8 algorithm 8 Nk11 algorithm Algorithm 8 is a variation of algorithm 7, the difference is that the 1k1 after N is replaced by k11. Therefore, the applicable structure is quite different from that of the congruent number . The corresponding quaternary quadratic equations are as follows:

Nx2 + ky2=z2 (1) ky2 + z2=w2 (2) By (2) get y =2uv, z=pu2 − v2,w=pu2 + v2, Any combination of N,k, suitable for congruent number of type N.K.1.1, may be prechecked by Z(A). For composite number,such as 1205=241*5,has a solution of type 241.5.1.1,which is obtained by calculation. u=1、 v=6、 x=1、 y=12、 z=31、 w=41、 s=961、 t=720 For composite number, such as 2614639*5 has a solution of type 2614639.1.1.1, which is obtained by calculation u=1、v=57、g=2、x=2、y=114、z=3244、w=3254、s=10523536/4、t=64980/4, 7123118430 times faster than the original algorithm. When N=1 and k is a positive integer without a square factor, it is equivalent to the following algorithm, 1N11.

5.9 algorithm 9: 1kN1 algorithm x2 + ky2=Nz2 (1) ky2 + Nz2=w2 (2) This algorithm uses equation Q4: w2-x2=2ky2 (3) here Ly2=2k 2 2 2 2 For equation (3), general solution is: x=L1u − L2v , y=2uv, w=L1u + L2v ,here L=L1 ∗ L2=2k Substitution equation (1),get Nz2 = x2 + ky2,Just check whether this expression is true in the range of positive integers. For composite number, such as 471=3*157,it’s type is 1.3.157.1 (Look at the intermediate result using the function Z(471).)

L1 ∗ L2 There are four combinations, and the calculation results are as follows: Table 3: 471 has different u and v values for different parameters

L1 L2 u v g x y z w s t 6 1 9 109 1 11395 1962 949 23843 141394357 3849444 3 2 18 109 2 22790 3924 1898 24086 565577428 46193328 2 3 27 109 3 34185 5886 2487 24491 1272549213 103934988 1 6 54 109 6 68390 11772 5694 26678 1272549213 415739952 When K=1, it is the same as algorithm 2, and can also be used for the calculation of prime 157.

5.10 algorithm 10 1N11 algorithm This algorithm is extremely rich in content, in fact it is the Fermat-Guanxungui sufficient and necessary condition theorem, but our content is much more varied. 5.10.0 Primitive Fermat – Guanxungui method z2 + Ny2=z2 (1) z2 − Ny2=w2 (2) Equations (1) and (2) are multiplied by each other get z4-(Ny2)2 = (zw)2, Let Z=z,X=zw,Y=Ny2, get X2 + Y2 = Z4, As we all know,the general solution of this equation is: X=|u4 + v4 −6u2v2| Y=4uv(u2 − v2) Z=u2 + v2 or X=4uv(u2 − v2) Y=|u4 + v4 −6u2v2| Z=u2 + v2 This Y= Ny2 = u4 + v4 − 6u2v2 , or 4uv u2 − v2 , We call it as solution for g_1,g_2, 2 2 In other words, Ny = g1or Ny = g_2, it is necessary and sufficient for N to be congruent number in the range of positive integers. Taking g_2 as the discriminant condition,the solution of 205 is calculated as follows: u=45,v=4 x=1649,y=84 z=2041 w=2369 S=z2=4165681 t=205y2=1446840

5.10.1 Zty method(1) : (1N1)1 x2 + Ny2=z2 (1) Ny2 + z2=w2 (2) 2 2 2 2 Starting from equation (1),its general solution is x=N1u − N2v ,y=2uv,z==N1u + N2v , here N==N1 ∗ N2 Substitution to equation (2) w2 = Ny2 + z2 ,If this expression is true in the range of positive integers, then N is congruent number,and s=z2,t= Ny2 Since N has a variety of combinations, it has a variety of formats. The solution u,v and g may be different,but the result s and t are equivalent.the example is as follows: Any congruent numberce must have a Fermat type 1N11. Take 205 for example. u, v,g are different, but s and t are the same.

Table 4:70 has different u and v values for different parameters 2 L1 L2 u v g x y z w s t(s,t must divided by g ) 205 1 3 14 1 1649 84 2041 2369 4165681 1446480 41 5 14 123 41 67609 3444 83681 97129 7002509761 2431532880 5 41 14 15 5 8245 420 10205 11845 104142025 36162000 1 205 14 3 1 1649 84 2041 2369 4165681 1446480 5.10.2 Zty method(1) 1(N11) x2 + Ny2=z2 (1) Ny2 + z2=w2 (2) Starting from equation (2),its general solution is 2 2 2 2 y=2uv z=N1u − N2v ,w==N1u + N2v , here N==N1 ∗ N2 Substitution to equation (1),x2 = z2 − Ny2 ,If this expression is true in the range of positive integers,then N is congruent number,and s=z2,t= Ny2. In 205 = 41 * 5, for example, Table 5: 205 has different u and v values for different parameters

L1 L2 u v g x y z w s t 205 1 2 105 5 8245 420 10205 11845 104142025 36162000 41 5 2 21 1 1649 84 2041 2369 4165681 1446480 5 41 21 2 1 1649 84 2041 2369 4165681 1446480 1 205 105 2 5 8245 420 10205 11845 104142025 36162000 5.10.3 Zty method(3) 1(N)11 x2 + Ny2=z2 (1) Ny2 + z2=w2 (2) 2 2 2 The third method uses Q4, w − x = 2Ny , let L=2N=L1 ∗ L2 2 2 2 2 The general solution is x=L1u − L2v , y=2uv, w=L1u + L2v Substitution to equation w2 − x2 ==(Ny2 + z2)-(z2 − Ny2)= 2Ny2, 2Ny2=w2 − x2 is a test expression. If this expression is true in the range of positive integers, then N is a congruent number.

Table 6: 205 has different u,v values for different parameters,but (s,t) are equivalent

L1 L2 u v g x y z w s t 410 1 6 287 41 67609 3444 83681 164774 7002509761 11861136 205 2 7 30 5 8245 420 10205 1849 104142025 176400 82 5 7 12 2 3298 168 4082 337 16662724 28244 41 10 7 6 1 1649 84 2041 121 4165681 7056 10 41 6 7 1 1649 84 2041 121 4165681 7056 5 82 12 7 2 3298 168 4082 337 16662724 28244 2 205 30 7 5 8245 420 10205 998 104142025 176400 1 410 60 7 10 16490 840 20410 3698 416568100 705600 5.10.4 Zty method(4) 1N(1)1 x2 + Ny2=z2 (1) Ny2 + z2=w2 (2) By Q3,this is x2 + w2=2z2 The general solution is x=|u2 − v2-2uv| w=|u2 − v2 +2uv| z=u2 + v2 Ny2 = z2 − x2 is a test expression. If this expression is true in the range of positive integers, then N is a congruent number.result as follow: u=45,v=4 x=1649,y=84 z=2041 w=2369 S=z2=4165681 t=205y2=1446840

Note 1:If the smallest solution is s,t,then the solution of Fermat type is S = (s2 + t2)2 ,T=4st((s2 − t2), it's terrible. Note 2: The new method adds 17 more methods to the original Fermat test, which had only two test formulas. If N has more factors, more tests can be added.from 5.10.1-5.10.4, The number of methods is related to the number of factors of N. This algorithm is nice, but it's easy not to use, Because the solution for Fermat type 1A11 is so large, when certain congruent number really do not have available non-fermat types, That is, you have to use it when there are no relevant types for algorithm 1-9. Only for algorithm 10, there are 2k+2+3 test formulas, where k is the number of factors of N, and +3 contains 3 methods of 5.10.0 and 5.10.4. 5.11 General algorithm: AST algorithm The algorithm is suitable for all congruent number,but the problem is that the calculation is too slow. We used this algorithm to collect 2145651 solutions for the congruent number, It's not complete, but it's useful. The algorithm is simply described as follows: for s=2 to M0:b=iif(mod(s,2)=0,1,2) && Guarantee one odd one even,M0 is upper bound of s for t=b to N0 step 2 && Guaranteed S and T reciprocal prime,N0 is upper bound of t if gcd(s,t)>1 loop endi if s*t*(s*s-t*t)/A Is A perfect square,then the solution of A is s,t endf endf note:The actual program is different from the one described here, using the bucket algorithm, each pair of s,t,because the bucket cannot be infinite,it limits the search scope of s and t and the upper bound of the maximum, for example M0=N0=107.

6 conclusion This paper is a comprehensive summary of the algorithm of congruent number,Carry forward and develop the congruent number calculated for the spirit of Zagier, the founder of the mountain. Our ten methods, all of which are the latest results of congruent number calculation It was first published by us, In the context of the current mathematical tools, which have been done to the extreme, it is very difficult to come up with new algorithms. Despite the discovery of ten new algorithms, the calculation of congruent number is only a small step in the long march, at present, there are 263,367(n≡7(mod 8) and 277,293,373,389(n≡5(mod 8) Without finding its solution, we expect a new breakthrough in mathematical methods.

7. Appendix Appendix 1: Generate the combinatorial complete set of factors Example: generate for 6 {(3211),(1611),(1231),(1213),(3121),(1321),(1161),(1123} Appendix 2: The original bucket search method

Set the width of the bucket Tk=107, search scope M0=3∗ 107 Parameter K,L,u0,v0 custom, double loop variable m,n 1) For m=2 to M0:b=iif(mod(m,2)=0,1,2) For n=b to M0-1 step 2:if gcd(m,n)>1 then skip 2) Remove the maximum squared factor of st(s+t)(s-t), the remaining integer A, and A≦Tk and If the position of the bucket is empty, record A into the bucket and placed in the holding sign. 3) Endf:Endf Were collected 2145651 solutions, They are solutions to 1954758 congruent number, most of which yield only one solution,

Appendix 3: Improved bucket search Set the width of the bucket Tk,Parameter K,L,u0,v0 custom 1) Double loop, variable m,n, generated s00 = K ∗ m2,t00=L∗ n2, if gcd(s00,t00)>1 then skip; s0=max(s00,t00) t0=min(s00,t00) : x=s0+t0:y=s0-t0 2) if s0 and t0 is all odd x=x/2:y=y/2 endi 3) let x,y remove maximum squared factor, remaining u0, v0 4) then congruent number is A=u0*v0*K*L;if A

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