不定方程的整数解

王守恩

主帖好难！！！还是做点容易的——前面2串在OEIS有。后面的就没有了。

Table[Length@Solve[{((a + k)/a) ((b + k)/b) == (n + k)/n, (n + k + 1) n ≥ a ≥ b > 0}, {b, a}, Integers], {k, 9}, {n, 49}]

{1, 2, 3, 3, 4, 4, 4, 6, 6, 4, 6, 6, 4, 8, 10, 5, 6, 6, 6, 12, 8, 4, 8, 12, 6, 8, 12, 6, 8, 8, 6, 12, 8, 8, 18, 9, 4, 8, 16, 8, 8, 8, 6, 18, 12, 4, 10, 15, 9},
{1, 2, 2, 4, 2, 5, 3, 5, 3, 8, 2, 8, 4, 6, 4, 9, 2, 12, 4, 8, 4, 10, 3, 10, 6, 8, 4, 16, 2, 14, 4, 7, 8, 12, 4, 12, 4, 10, 4, 20, 2, 16, 6, 8, 6, 12, 3, 18, 6},
{2, 2, 3, 3, 4, 4, 4, 4, 6, 4, 4, 9, 5, 4, 8, 5, 6, 8, 4, 6, 12, 6, 4, 10, 9, 4, 10, 6, 6, 12, 4, 12, 12, 4, 8, 12, 8, 4, 12, 8, 6, 16, 4, 6, 20, 6, 6, 15, 9},
{1, 3, 2, 3, 3, 6, 2, 6, 3, 6, 4, 7, 2, 9, 4, 7, 4, 9, 2, 12, 6, 6, 4, 12, 3, 12, 4, 8, 4, 12, 4, 12, 4, 6, 8, 18, 2, 12, 4, 12, 6, 12, 2, 14, 9, 9, 4, 14, 3},
{2, 2, 4, 5, 3, 4, 6, 4, 6, 6, 5, 6, 6, 4, 9, 10, 4, 6, 8, 6, 8, 8, 6, 8, 8, 4, 12, 12, 4, 12, 9, 6, 8, 8, 12, 9, 8, 4, 12, 18, 4, 8, 10, 9, 12, 8, 6, 10, 12},
{1, 3, 2, 4, 2, 6, 2, 5, 4, 6, 2, 8, 2, 8, 6, 6, 2, 10, 3, 8, 5, 8, 2, 15, 3, 7, 5, 8, 4, 16, 2, 7, 6, 10, 4, 16, 2, 8, 8, 10, 2, 18, 3, 12, 8, 8, 2, 15, 6},
{2, 3, 4, 3, 6, 4, 3, 8, 8, 4, 6, 6, 6, 6, 8, 5, 8, 9, 4, 12, 9, 4, 8, 8, 9, 8, 8, 9, 9, 8, 4, 12, 16, 4, 12, 9, 6, 12, 8, 8, 10, 8, 6, 12, 18, 4, 8, 20, 8},
{2, 3, 2, 5, 2, 6, 4, 4, 3, 9, 2, 10, 4, 6, 4, 8, 3, 9, 4, 10, 4, 12, 2, 9, 6, 6, 8, 15, 2, 12, 4, 9, 4, 12, 4, 15, 6, 6, 4, 16, 3, 18, 4, 10, 6, 12, 4, 16, 6},
{2, 2, 5, 3, 4, 6, 5, 4, 5, 4, 6, 9, 4, 4, 12, 8, 4, 6, 6, 6, 12, 4, 6, 12, 6, 8, 9, 6, 4, 12, 8, 6, 12, 4, 12, 15, 4, 4, 15, 12, 6, 12, 6, 6, 12, 8, 8, 15, 6}}

Table[Length@Solve[{((a + k)/a) ((b + k)/b) == (n + k)/n, (n + k + 1) n ≥ a > b > 0}, {b, a}, Integers], {k, 9}, {n, 49}]

{1, 2, 3, 3, 4, 4, 4, 6, 6, 4, 6, 6, 4, 8, 10, 5, 6, 6, 6, 12, 8, 4, 8, 12, 6, 8, 12, 6, 8, 8, 6, 12, 8, 8, 18, 9, 4, 8, 16, 8, 8, 8, 6, 18, 12, 4, 10, 15, 9},
{1, 2, 2, 4, 2, 5, 3, 5, 3, 8, 2, 8, 4, 6, 4, 9, 2, 12, 4, 8, 4, 10, 3, 10, 6, 8, 4, 16, 2, 14, 4, 7, 8, 12, 4, 12, 4, 10, 4, 20, 2, 16, 6, 8, 6, 12, 3, 18, 6},
{1, 2, 3, 3, 4, 4, 4, 4, 6, 4, 4, 9, 5, 4, 8, 5, 6, 8, 4, 6, 12, 6, 4, 10, 9, 4, 10, 6, 6, 12, 4, 12, 12, 4, 8, 12, 8, 4, 12, 8, 6, 16, 4, 6, 20, 6, 6, 15, 9},
{1, 3, 2, 3, 3, 6, 2, 6, 3, 6, 4, 7, 2, 9, 4, 7, 4, 9, 2, 12, 6, 6, 4, 12, 3, 12, 4, 8, 4, 12, 4, 12, 4, 6, 8, 18, 2, 12, 4, 12, 6, 12, 2, 14, 9, 9, 4, 14, 3},
{2, 2, 4, 4, 3, 4, 6, 4, 6, 6, 5, 6, 6, 4, 9, 10, 4, 6, 8, 6, 8, 8, 6, 8, 8, 4, 12, 12, 4, 12, 9, 6, 8, 8, 12, 9, 8, 4, 12, 18, 4, 8, 10, 9, 12, 8, 6, 10, 12},
{1, 2, 2, 4, 2, 6, 2, 5, 4, 6, 2, 8, 2, 8, 6, 6, 2, 10, 3, 8, 5, 8, 2, 15, 3, 7, 5, 8, 4, 16, 2, 7, 6, 10, 4, 16, 2, 8, 8, 10, 2, 18, 3, 12, 8, 8, 2, 15, 6},
{2, 3, 4, 3, 6, 4, 3, 8, 7, 4, 6, 6, 6, 6, 8, 5, 8, 9, 4, 12, 9, 4, 8, 8, 9, 8, 8, 9, 9, 8, 4, 12, 16, 4, 12, 9, 6, 12, 8, 8, 10, 8, 6, 12, 18, 4, 8, 20, 8},
{1, 3, 2, 5, 2, 6, 4, 4, 3, 9, 2, 10, 4, 6, 4, 8, 3, 9, 4, 10, 4, 12, 2, 9, 6, 6, 8, 15, 2, 12, 4, 9, 4, 12, 4, 15, 6, 6, 4, 16, 3, 18, 4, 10, 6, 12, 4, 16, 6},
{2, 2, 4, 3, 4, 6, 5, 4, 5, 4, 6, 9, 4, 4, 12, 7, 4, 6, 6, 6, 12, 4, 6, 12, 6, 8, 9, 6, 4, 12, 8, 6, 12, 4, 12, 15, 4, 4, 15, 12, 6, 12, 6, 6, 12, 8, 8, 15, 6}}

王守恩

这类题——解应该是有限的吧？

Table[FindInstance[{((a + 1)/a) ((b + 1)/b) ((c + 1)/c) == (n + 1)/n, ((a + 1)/a) ((b + 1)/b) == (x + 1)/x, ((a + 1)/a) ((c + 1)/c) == (y + 1)/y, ((b + 1)/b) ((c + 1)/c) == (z + 1)/z, c > b > a > n}, {a, b, c, x, y, z}, Integers, 1], {n, 21}]

{{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {{a -> 20, b -> 84, c -> 119, x -> 16, y -> 17, z -> 49}}, {}, {}, {}, {}, {}, {{a -> 35, b -> 90, c -> 104, x -> 25, y -> 26, z -> 48}}, {}}

王守恩

$(1+\frac5a)(1+\frac5b) =1+\frac5x,(1+\frac5a)(1+\frac5c)=1+\frac5y,(1+\frac5a)(1+\frac5d)=1+\frac5z,(1+\frac5b)(1+\frac5c)=1+\frac5w,(1+\frac5b)(1+\frac5d)=1+\frac5s,(1+\frac5c)(1+\frac5d)=1+\frac5t,$

$(1+\frac5a)(1+\frac5b)(1+\frac5c)(1+\frac5d)=1+\frac5n,\ \ a<b<c<d,\ \  a,b,c,d,n,x,y,z,w,s,t都是正整数。5可以统一换。$——有解吗？

王守恩

来1组解？谢谢！！！

$(1+\frac1a)(1+\frac1b) =1+\frac1x,(1+\frac1a)(1+\frac1c)=1+\frac1y,(1+\frac1b)(1+\frac1c)=1+\frac1z,(1+\frac1a)(1+\frac1b) (1+\frac1c)=1+\frac{1}{2024}$

譬如。

$(1+\frac{1}{119})(1+\frac{1}{560}) =1+\frac{1}{98},(1+\frac{1}{119})(1+\frac{1}{594})=1+\frac{1}{99},(1+\frac{1}{560})(1+\frac{1}{594})=1+\frac{1}{288},(1+\frac{1}{119})(1+\frac{1}{560}) (1+\frac{1}{594})=1+\frac{1}{84}$

northwolves

王守恩 发表于 2025-5-3 11:00
来1组解？谢谢！！！

$(1+\frac1a)(1+\frac1b) =1+\frac1x,(1+\frac1a)(1+\frac1c)=1+\frac1y,(1+\frac1b) ...

{20,84,119}        14
{35,90,104}        20
{54,440,539}        44
{119,560,594}        84
{104,1260,1455}        90
{170,2736,3059}        152
{440,539,1539}        209
{279,1890,1952}        216
{390,935,1104}        220
{560,594,935}        220
{252,5060,5543}        230
{350,8424,9099}        324
{464,13020,13919}        434
{560,3135,5984}        440
{539,4752,4850}        440
{740,1664,7695}        480
{696,2870,4059}        492
{594,19040,20195}        560
{1456,2015,2820}        650
{740,26676,28119}        702
{1332,3440,3999}        774
{923,10010,10152}        780
{1785,2736,3059}        798
{1599,2664,4550}        819
{902,36120,37883}        860
{1325,4080,9009}        900
{1935,3267,4256}        945

northwolves

2024似乎无解