(a^2-1)(b^2-1)=c^3

王守恩

跑一跑题。

\(|1^3-1^2|=0\)
\(|2^3-3^2|=1\)
\(|3^3-5^2|=2\)
\(|1^3-2^2|=3\)
\(|2^3-2^2|=4\)
\(|?^3-?^2|=5\)
\(|?^3-?^2|=6\)
\(|2^3-1^2|=7\)
\(|1^3-3^2|=8\)
\(|3^3-6^2|=9\)
\(|?^3-?^2|=10\)
\(|3^3-4^2|=11\)
\(|13^3-47^2|=12\)
\(|17^3-70^2|=13\)
\(|?^3-?^2|=14\)
\(|1^3-4^2|=15\)
\(|?^3-?^2|=16\)
\(|2^3-5^2|=17\)
\(|3^3-3^2|=18\)
\(|5^3-12^2|=19\)
\(|6^3-14^2|=20\)
\(|?^3-?^2|=21\)
\(|3^3-7^2|=22\)
\(|3^3-2^2|=23\)
\(|1^3-5^2|=24\)
\(|5^3-10^2|=25\)
\(|3^3-1^2|=26\)
\(|?^3-?^2|=27\)
\(|2^3-6^2|=28\)
\(|?^3-?^2|=29\)
\(|19^3-83^2|=30\)

{{1, 1, 0}, {2, 3, 1}, {3, 5, 2}, {1, 2, 3}, {2, 2, 4}, {2, 1, 7}, {1, 3, 8}, {3, 6, 9}, {3, 4, 11}, {13, 47, 12}, {17, 70, 13}, {1, 4, 15},
{2, 5, 17}, {3, 3, 18}, {5, 12, 19}, {6, 14, 20}, {3, 7, 22}, {3, 2, 23}, {1, 5, 24}, {5, 10, 25}, {3, 1, 26}, {2, 6, 28}, {19, 83, 30}}
SortBy[DeleteDuplicatesBy[SortBy[Flatten[Table[With[{d = Abs[a^3 - b^2]}, If[d <= 30, {a, b, d}, Nothing]], {a, n = 99}, {b, n}], 1], Last], Last], Last]

{0, 1, 2, 3, 4, 7, 8, 9, 11, 12, 13, 15, 17, 18, 19, 20, 22, 23, 24, 25, 26, 28, 30}
Sort@DeleteDuplicates[Reap[Do[a2 = a^2; bo = Max[1, Ceiling[If[a2 > 30, (a2 - 30)^(1/3), 0]]]; bH = Floor[(a2 + 30)^(1/3)]; Do[d = Abs[a2 - b^3]; If[d <= 30, Sow[d]], {b, bo, bH}], {a, 999999999}]][[2, 1]]]

猜想 5, 6, 10, 14, 16, 21, 27, 29, ...... ——不会有解。

王守恩

\((2^2-1)(2^2-1)(5^2-1)=6^3\)
\((3^2-1)(6^2-1)(99^2-1)=140^3\)
\((4^2-1)(4^2-1)(11^2-1)=30^3\)
\((5^2-1)(2^2-1)(5^2-1)=12^3\)
\((6^2-1)(3^2-1)(99^2-1)=140^3\)
\((7^2-1)(3^2-1)(17^2-1)=48^3\)
\((8^2-1)(3^2-1)(97^2-1)=168^3\)
\((9^2-1)(9^2-1)(13719^2-1)=10640^3\)
\((10^2-1)(45^2-1)(1057^2-1)=6072^3\)
\((11^2-1)(4^2-1)(11^2-1)=60^3\)
\((12^2-1)(99^2-1)(846^2-1)=10010^3\)
\((13^2-1)(2^2-1)(97^2-1)=168^3\)
\((14^2-1)(14^2-1)(1249^2-1)=3900^3\)
\((15^2-1)(9^2-1)(99^2-1)=560^3\)
\((16^2-1)(97^2-1)(2024^2-1)=21420^3\)
\((17^2-1)(3^2-1)(7^2-1)=48^3\)
\((18^2-1)(305^2-1)(647^2-1)=23256^3\)
\((19^2-1)(2^2-1)(26^2-1)=90^3\)
\((20^2-1)(2449^2-1)(2889^2-1)=271320^3\)
\((21^2-1)(21^2-1)(109^2-1)=1320^3\)
\((22^2-1)(22^2-1)(323^2-1)=2898^3\)
\((23^2-1)(23^2-1)(65^2-1)=1056^3\)
\((24^2-1)(24^2-1)(143749^2-1)=189750^3\)
\((25^2-1)(3^2-1)(24335^2-1)=14352^3\)
\((26^2-1)(2^2-1)(19^2-1)=90^3\)
\((27^2-1)(27^2-1)(9827^2-1)=37128^3\)
\((28^2-1)(?^2-1)(?^2-1)=?^3\)
\((29^2-1)(15^2-1)(449^2-1)=3360^3\)
\((30^2-1)(?^2-1)(?^2-1)=?^3\)

{{2, 2, 5, 6}, {3, 6, 99, 140}, {4, 4, 11, 30}, {5, 2, 5, 12}, {6, 3, 99, 140}, {7, 3, 17, 48}, {8, 3, 97, 168}, {9, 9, 13719, 10640}, {10, 45, 1057, 6072}, {11, 4, 11, 60}, {12, 99, 846, 10010},
{13, 2, 97, 168}, {14, 14, 1249, 3900}, {15, 9, 99, 560}, {16, 97, 2024, 21420}, {17, 3, 7, 48}, {18, 305, 647, 23256}, {19, 2, 26, 90}, {20, 2449, 2889, 271320}, {21, 21, 109, 1320},
{22, 22, 323, 2898}, {23, 23, 65, 1056}, {24, 24, 143749, 189750}, {25, 3, 24335, 14352}, {26, 2, 19, 90}, {27, 27, 9827, 37128}, {28, 0, 0, 0}, {29, 15, 449, 3360}, {30, 0, 0, 0}}

1. cubefree[x_] := Module[{f = FactorInteger[x], d = 1, c = 1}, Do[{p, e} = f[[i]]; d *= p^Mod[e, 3]; c *= p^Quotient[e, 3];, {i, Length[f]}];{d, c}];
   
2. n = 999999;(*可自行调大，比如 9999*)Print["预计算 a^2-1 的无立方因子..."]; data = Table[Module[{d, c}, {d, c} = cubefree[a^2 - 1]; {a, d, c}], {a, 2, n}];
   
3. grouped = GroupBy[data, #[[2]] &];(*D\[Rule]{{a,d,c},...}*)Print["开始搜索 k..."]; results = Table[Module[{Dk, Ck, targetD, sol = {k, 0, 0, 0}}, {Dk, Ck} = cubefree[k^2 - 1];
   
4. Do[{a, Da, Ca} = entry; (*需要的 D_b=cubefree((Da*Dk)^2)*)targetD = cubefree[(Da*Dk)^2][[1]]; If[KeyExistsQ[grouped, targetD],(*在 targetD 组中找第一个 b>a*) Do[{b, Db, Cb} = entry2;
   
5. If[b > a, prod = (k^2 - 1)*(a^2 - 1)*(b^2 - 1); v = Round[prod^(1/3)]; If[v^3 == prod, sol = {k, a, b, v};
   
6. Break[]]], {entry2, grouped[targetD]}]; If[sol[[2]] != 0, Break[]];], {entry, data}]; sol], {k, 2, 30}];
   
7. results

复制代码

王守恩

小结。\((a^2-1)(b^2-1)=c^3\)  只有这12组解。

组合(1)。\((3^2-1)(3^2-1)=4^3\)

组合(2)。\((7^2-1)(17^2-1)=24^3\)

组合(3)。\((6^2-1)(99^2-1)=70^3\)

组合(4)。\((8^2-1)(97^2-1)=84^3\)

组合(5)。\((97^2-1)(251^2-1)=840^3\)

组合(6)。\((199^2-1)(241^2-1)=1320^3\)

组合(7)。\((151^2-1)(721^2-1)=2280^3\)

组合(8)。\((129^2-1)(1351^2-1)=3120^3\)

组合(9)。\((25^2-1)(24335^2-1)=7176^3\)

组合(10)。\((1351^2-1)(13311^2-1)=68640^3\)

组合(11)。\((10405^2-1)(3786751^2-1)=11579040^3\) ——谢谢网友 northwolves ！

组合(12)。\((708158977^2-1)(3730418047^2-1)=1910991897024^3\)——谢谢网友 nindilg ！


组合(1)。a = 3; b = 3; c = 4;
a^2-1 = {{2,3}}
b^2-1 = {{2,3}}
c^3    = {{2,6}}

组合(2)。a = 7; b = 17; c = 24;
a^2-1 = {{2,4},{3,1}}
b^2-1 = {{2,5},{3,2}}
c^3    = {{2,9},{3,3}}

组合(3)。a = 6; b = 99; c = 70;
a^2-1 = {{5,1},{7,1}}
b^2-1 = {{2,3},{5,2},{7,2}}
c^3    = {{2,3},{5,3},{7,3}}

组合(4)。a = 8; b = 97; c = 84;
a^2-1 = {{3,2},{7,1}}
b^2-1 = {{2,6},{3,1},{7,2}}
c^3    = {{2,6},{3,3},{7,3}}

组合(5)。a = 97; b = 251; c = 840;
a^2-1 = {{2,6},{3,1},{7,2}}
b^2-1 = {{2,3},{3,2},{5,3},{7,1}}
c^3    = {{2,9},{3,3},{5,3},{7,3}}

组合(6)。a = 199; b = 241; c = 1320;
a^2-1 = {{2,4},{3,2},{5,2},{11,1}}
b^2-1 = {{2,5},{3,1},{5,1},{11,2}}
c^3    = {{2,9},{3,3},{5,3},{11,3}}

组合(7)。a = 151; b = 721; c = 2280;
a^2-1 = {{2,4},{3,1},{5,2},{19,1}}
b^2-1 = {{2,5},{3,2},{5,1},{19,2}}
c^3    = {{2,9},{3,3},{5,3},{19,3}}

组合(8)。a = 129; b = 1351; c = 3120;
a^2-1 = {{2,8},{5,1},{13,1}}
b^2-1 = {{2,4},{3,3},{5,2},{13,2}}
c^3    = {{2,12},{3,3},{5,3},{13,3}}

组合(9)。a = 25; b = 24335; c = 7176;
a^2-1 = {{2,4},{3,1},{13,1}}
b^2-1 = {{2,5},{3,2},{13,2},{23,3}}
c^3    = {{2,9},{3,3},{13,3},{23,3}}

组合(10)。a = 1351; b = 13311; c = 68640;
a^2-1 = {{2,4},{3,3},{5,2},{13,2}}
b^2-1 = {{2,11},{5,1},{11,3},{13,1}}
c^3    = {{2,15},{3,3},{5,3},{11,3},{13,3}}

组合(11)。a = 10405; b = 3786751; c = 11579040;——谢谢网友 northwolves ！
a^2-1 = {{2,3},{3,2},{11,2},{17,2},{43,1}}
b^2-1 = {{2,12},{3,4},{5,3},{11,1},{17,1},{43,2}}
c^3    = {{2,15},{3,6},{5,3},{11,3},{17,3},{43,3}}

组合(12)。a = 708158977; b = 3730418047; c = 1910991897024;——谢谢网友 nindilg ！
a^2-1 = {{2,10},{3,1},{7,2},{31,2},{97,2},{607,2}}
b^2-1 = {{2,8},{3,2},{7,1},{19,3},{31,1},{41,3},{97,1},{607,1}}
c^3    = {{2,18},{3,3},{7,3},{19,3},{31,3},{41,3},{97,3},{607,3}}


{4, 24, 70, 84, 840, 1320, 2280, 3120, 7176, 68640, 11579040, 1910991897024}——OEIS没有。

Flatten[Table[c = Power[(a^2 - 1) (b^2 - 1), (3)^-1]; If[IntegerQ[c], {a, b, c}, Nothing], {a, 2, 9999}, {b, a, 9999}], 1]

风花飘飘

王守恩 发表于 2026-9-18 18:46
跑一跑题。

\(|1^3-1^2|=0\)

反着来：
把底数统一改成负值即可，革命队伍整齐划一
1.
2.
3.
4.
5.=|(-1)^3-(-2)^2|=|-1-4|
6.
7.
8.
9.
……
n.

风花飘飘

AI都学会谦虚了么？

王守恩

风花飘飘 发表于 2026-9-21 21:21
AI都学会谦虚了么？

是的。应该允许有 "0" 与 "负数"。OEIS 也是这样想的。

OEIS——A110223——不是立方数和平方数的绝对差的数字。

6, 14, 21, 29, 32, 34, 42, 46, 51, 58, 59, 62, 66, 69, 70, 75, 77, 78, 84, 85, 86, 88, 90, 93, 96, 102, 103, 110, 111, 114, 115, 123, 130, 133, 137, 140,——这串数(无解的数)会越来越多——会超过有解的数。
Complement[Range[99], Union[Flatten[Table[Select[Table[Abs[a^3 - b^2], {b, 0, 999}], # < 100 &], {a, -4, 99}]]]]

{{0, 0, 0}, {-1, 0, 1}, {-1, 1, 2}, {1, 2, 3}, {0, 2, 4}, {-1, 2, 5}, {2, 1, 7}, {-2, 0, 8}, {-2, 1, 9}, {-1, 3, 10}, {3, 4, 11}, {-2, 2, 12}, {17, 70, 13}, {1, 4, 15}, {0, 4, 16}, {-2, 3, 17}, {3, 3, 18}, {5, 12, 19},
{6, 14, 20}, {3, 7, 22}, {3, 2, 23}, {-2, 4, 24}, {0, 5, 25}, {-1, 5, 26}, {-3, 0, 27}, {-3, 1, 28}, {19, 83, 30}, {-3, 2, 31}, {-2, 5, 33}, {1, 6, 35}, {-3, 3, 36}, {-1, 6, 37}, {11, 37, 38}, {4, 5, 39}, {6, 16, 40},
{2, 7, 41}, {-3, 4, 43}, {-2, 6, 44}, {21, 96, 45}, {6, 13, 47}, {1, 7, 48}, {0, 7, 49}, {-1, 7, 50}, {-3, 5, 52}, {9, 26, 53}, {3, 9, 54}, {4, 3, 55}, {2, 8, 56}, {-2, 7, 57}, {4, 2, 60}, {5, 8, 61}, {-3, 6, 63},
{-4, 0, 64}, {-4, 1, 65}, {23, 110, 67}, {-4, 2, 68}, {5, 14, 71}, {-2, 8, 72}, {-4, 3, 73}, {99, 985, 74}, {-3, 7, 76}, {20, 89, 79}, {-4, 4, 80}, {0, 9, 81}, {-1, 9, 82}, {27, 140, 83}, {7, 16, 87}, {-4, 5, 89},
{-3, 8, 91}, {2, 10, 92}, {3, 11, 94}, {6, 11, 95}, {18, 77, 97}, {7, 21, 98}, {1, 10, 99}}, {6, 14, 21, 29, 32, 34, 42, 46, 51, 58, 59, 62, 66, 69, 70, 75, 77, 78, 84, 85, 86, 88, 90, 93, 96}
SortBy[DeleteDuplicatesBy[SortBy[Flatten[Table[With[{d = Abs[a^3 - b^2]}, If[d <= 99, {a, b, d}, Nothing]], {a, -4, 99}, {b, 0, 999}], 1], Last], Last], Last]

风花飘飘

王守恩 发表于 2026-9-22 06:21
是的。应该允许有 "0" 与 "负数"。OEIS 也是这样想的。

OEIS——A110223——不是立方数和平方数的绝对 ...

import math

def is_square(m):
    if m < 0: return False
    r = int(math.isqrt(m))
    return r*r == m

def representable(n, Amax=2000):
    # 找 |a^3 - b^2| = n  => b^2 = a^3 ± n
    for a in range(-Amax, Amax+1):
        a3 = a*a*a
        for sign in (1, -1):
            val = a3 + sign*n   # b^2 = a^3 + n 或 a^3 - n
            if is_square(val):
                b = int(math.isqrt(val))
                # 验证
                if abs(a3 - b*b) == n:
                    return True, (a, b, abs(a3-b*b))
    return False, None

# 生成前若干“不可表”数
gaps = []
n = 1
while len(gaps) < 30:
    ok, _ = representable(n, Amax=500)
    if not ok:
        gaps.append(n)
    n += 1

print(gaps)

风花飘飘

这个帖子讨论完了，应该沉淀。