数论小猜想

蔡家雄

【质数立方是三质数立方的完美立方数的质数P】

设 P^3=A^3+B^3+C^3，其中 P, A, B, C 都是质数，

且 A^3=a1^3+a2^3+a3^3，B^3=b1^3+b2^3+b3^3，C^3=c1^3+c2^3+c3^3 均为正整数解，

8000以内 P, A, B, C 均为素数的组合，由 Treenewbee 程序计算，

709 = [193, [25, 68, 190]] + [461, [5, 86, 460]] + [631, [120, 207, 622]]

2767 = [103, [12, 31, 102]] + [2179, [108, 235, 2178]] + [2213, [1238, 1373, 1852]]

蔡家雄

【质数立方是至少两组三质数立方的完美立方数的质数P】

设 P^3=A^3+B^3+C^3 = D^3+E^3+F^3，其中 P, A, B, C, D, E, F 都是质数，

且 A^3=a1^3+a2^3+a3^3，B^3=b1^3+b2^3+b3^3，C^3=c1^3+c2^3+c3^3 均为正整数解，

及 D^3=d1^3+d2^3+d3^3，E^3=e1^3+e2^3+e3^3，F^3= f1^3+f2^3+ f3^3 均为正整数解，

求 质数 P= ？  A= ?    B= ?    C= ?    D= ?    E= ?    F= ? （ 至少两组解！！！）

蔡家雄

【质数立方是至少两组三质数立方的完美立方数的质数P】

我相信一定有解！！！但我感觉满足此条件的最小质数P，大于十万，，，，，，

蔡家雄

锥形数：C(n) =n^2*(n+1)/2 ,

求证：若 n>=2，则 [n^2*(n+1)/2]^3=a^3+b^3+c^3 均有正整数解。


四维拟形数：C(n) =n*(n+1)^2*(n+2)/12 ,

求证：若 n>=2，则 [n*(n+1)^2*(n+2)/12]^3=a^3+b^3+c^3 均有正整数解。

设 U=n,  V=n*(n+1)/2,  W=U^2 - 2*U*V/3+2*V^2/3,

则 W(n+2)=C(n)+C(n+2) =n*(n+1)^2*(n+2)/12+(n+2)*(n+3)^2*(n+4)/12 .

Treenewbee

蔡家雄 发表于 2023-1-27 22:05
【质数立方是至少两组三质数立方的完美立方数的质数P】

设 P^3=A^3+B^3+C^3 = D^3+E^3+F^3，其中 P, A,  ...

33199^3=2833^3+19081^3+30941^3=15187^3+24197^3+26647^3
49069^3=661^3+37441^3+40343^3=22307^3+37243^3+38119^3

Treenewbee

只考虑第一层质数解，3000以内至少五组：709, 1033, 2767, 2791, 2917
OEIS搜到了该序列：
A114923                Primes p such that there exist three primes q, r and s with p^3=q^3+r^3+s^3.

        709, 1033, 2767, 2791, 2917, 3727, 3769, 5647, 5657, 5737, 7039, 7321, 8089, 8291, 8387, 9433, 9473, 9851, 12073, 12343, 13417, 14083, 14561, 14723, 14831, 14969, 15313, 18127, 19841, 25033, 28081, 28477, 29153, 29179, 32771, 33161, 33199, 33377, 34337, 36713

蔡家雄

设 p, p+2, p+6, p+8 是四生素数，

所谓 完美四生素数立方数，要很大，才可能有解，

p^3=a^3+b^3+c^3

(p+2)^3=a^3+b^3+c^3，a, b, c 可以是 合数，

(p+6)^3=a^3+b^3+c^3，a, b, c 可以是 合数，

(p+8)^3=a^3+b^3+c^3，a, b, c 可以是 合数，

需要四个式子同时有解，才叫 完美四生素数立方数，，，，

Treenewbee

10000以内有6组：
[191, [39, 146, 156]], [193, [25, 68, 190]], [197, [53, 58, 194]], [199, [6, 127, 180]]
[1871, [693, 809, 1785]], [1873, [6, 900, 1801]], [1877, [699, 743, 1803]], [1879, [300, 1087, 1746]]
[3251, [1028, 2355, 2724]], [3253, [525, 2128, 2910]], [3257, [297, 1926, 3014]], [3259, [72, 1315, 3186]]
[3461, [744, 941, 3426]], [3463, [135, 2316, 3076]], [3467, [248, 275, 3466]], [3469, [1723, 2367, 2859]]
[5651, 5653, 5657, 5659]
[9431, 9433, 9437, 9439]

Treenewbee

蔡家雄 发表于 2023-1-31 15:58
求解或反驳，
若 40k+31,  40k+33,  40k+37,  40k+39 是四生素数，

只有大约1/5的k满足条件

40k+31,  40k+33,  40k+37,  40k+39 是四生素数的k:
k={{191,1871,9431,18911,22271,101111,157271,187631,247991,294311,295871,299471,301991,346391,397751,412031,467471,500231,585911,597671,633791,666431,715151,734471,739391,768191,795791,833711,837071,857951,881471,889871,907391,930071,959471,978071,1027751,1043591,1093991,1100831,1121831,1146791,1156031,1157831,1163711,1210871,1228391,1233431,1257071,1285511,1298111,1322591,1381271,1451831,1468631,1524071,1584431,1588751,1678751,1681871,1748471,1798631,1834031,1837391,1857671,1904471,1954151,1959311,2007911,2075831,2079191,2083511,2159231,2176631,2202791,2223671,2266631,2290031,2303591,2362271,2418671,2546231,2552111,2576591,2594951,2822711,2927591,2952791,2989031,2990831}

1/(40k+33) 具有完全循环节长d＝ 40k＋32的k:

k={191,1871,157271,294311,397751,412031,500231,666431,715151,739391,768191,795791,833711,857951,1093991,1121831,1163711,1210871,1233431,1524071,1584431,1837391,1857671,1954151,2079191,2418671,2822711,2927591,2990831}

Treenewbee

40k+31={191,1871,157271,294311,397751,412031,500231,666431,715151,739391,768191,795791,833711,857951,1093991,1121831,1163711,1210871,1233431,1524071,1584431,1837391,1857671,1954151,2079191,2418671,2822711,2927591,2990831}