不定方程的整数解
已知 $(1+\frac4a)(1+\frac4b)(1+\frac4c)(1+\frac4d)=\frac{120}{114} ,0<a<=b<=c<=d $有623554组正整数解，下面的不定方程的整数解有多少组呢？
$(1+\frac5a)(1+\frac5b)(1+\frac5c)(1+\frac5d)=\frac{120}{119} ,0<a<=b<=c<=d $
从简单开始。\(\frac{n+1}{n}=(\frac{a+1}{a})(\frac{b+1}{b})\)
\(a(1)=1,\frac{2}{1}=(\frac{3}{2})(\frac{4}{3})\)
\(a(2)=2,\frac{3}{2}=(\frac{4}{3})(\frac{9}{8}=(\frac{5}{4})(\frac{6}{5})\)
\(a(3)=3,\frac{4}{3}=(\frac{5}{4})(\frac{16}{15})=(\frac{6}{5})(\frac{10}{9})=(\frac{7}{6})(\frac{8}{7})\)
\(a(4)=3,\frac{5}{4}=(\frac{6}{5})(\frac{25}{24})=(\frac{7}{6})(\frac{15}{14})=(\frac{9}{8})(\frac{10}{9})\)
\(a(5)=4,\frac{6}{5}=(\frac{7}{6})(\frac{36}{35})=(\frac{8}{7})(\frac{21}{20})=(\frac{9}{8})(\frac{16}{15})=(\frac{11}{10})(\frac{12}{11})\)
\(a(6)=4,\frac{7}{6}=(\frac{8}{7})(\frac{49}{48})=(\frac{9}{8})(\frac{28}{27})=(\frac{10}{9})(\frac{21}{20})=(\frac{13}{12})(\frac{14}{13})\)
\(a(7)=4,\frac{8}{7}=(\frac{9}{8})(\frac{64}{63})=(\frac{10}{9})(\frac{36}{35})=(\frac{12}{11})(\frac{22}{21})=(\frac{15}{14})(\frac{16}{15})\)
\(a(8)=6,\frac{9}{8}=(\frac{10}{9})(\frac{81}{80})=(\frac{11}{10})(\frac{45}{44})=(\frac{12}{11})(\frac{33}{32})=(\frac{13}{12})(\frac{27}{26})=(\frac{25}{14})(\frac{21}{20})=(\frac{17}{16})(\frac{18}{17})\)
\(a(9)=6,\frac{10}{9}=(\frac{11}{10})(\frac{100}{99})=(\frac{12}{11})(\frac{55}{54})=(\frac{13}{12})(\frac{40}{39})=(\frac{15}{14})(\frac{28}{27})=(\frac{16}{15})(\frac{25}{24})=(\frac{19}{18})(\frac{20}{19})\)
得到一串数。{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, 12, 12, 6, 8, 16, 16} \(\frac{n+1}{n}=(\frac{a+1}{a})(\frac{b+1}{b})(\frac{c+1}{c})\)
\(a(1)=0,\frac{2}{1}\)
\(a(2)=1,\frac{3}{2}=(\frac{7}{6})(\frac{8}{7})(\frac{9}{8})\)
\(a(3)=4,\frac{4}{3}=(\frac{7}{6})(\frac{15}{14})(\frac{16}{15})=(\frac{8}{7})(\frac{13}{12})(\frac{14}{13})=(\frac{9}{8})(\frac{10}{9})(\frac{16}{15})=(\frac{10}{9})(\frac{11}{10})(\frac{12}{11})\)
\(a(4)=7,\frac{5}{4}=(\frac{8}{7})(\frac{21}{20})(\frac{25}{24})=(\frac{9}{8})(\frac{16}{15})(\frac{25}{24})=(\frac{9}{8})(\frac{19}{18})(\frac{20}{19})=(\frac{10}{9})(\frac{15}{14})(\frac{21}{20})=(\frac{10}{9})(\frac{17}{16})(\frac{18}{17})=(\frac{11}{10})(\frac{12}{11})(\frac{25}{24})=(\frac{13}{12})(\frac{14}{13})(\frac{15}{14})\)
\(a(5)=13,\frac{6}{5}=(\frac{9}{8})(\frac{28}{27})(\frac{36}{35})=(\frac{9}{8})(\frac{31}{30})(\frac{32}{31})=(\frac{10}{9})(\frac{21}{20})(\frac{36}{35})=(\frac{10}{9})(\frac{26}{25})(\frac{27}{26})=(\frac{11}{10})(\frac{18}{17})(\frac{34}{33})=(\frac{11}{10})(\frac{23}{22})(\frac{24}{23})\)
\(=(\frac{12}{11})(\frac{16}{15})(\frac{33}{32})=(\frac{12}{11})(\frac{21}{20})(\frac{22}{21})=(\frac{13}{12})(\frac{14}{13})(\frac{36}{35})=(\frac{13}{12})(\frac{16}{15})(\frac{27}{26})=(\frac{14}{13})(\frac{15}{14})(\frac{26}{25})=(\frac{15}{14})(\frac{16}{15})(\frac{21}{20})=(\frac{16}{15})(\frac{17}{16})(\frac{18}{17})\)
\(a(6)=16,\frac{7}{6}=(\frac{10}{9})(\frac{36}{35})(\frac{49}{48})=(\frac{10}{9})(\frac{41}{40})(\frac{42}{41})=(\frac{11}{10})(\frac{28}{27})(\frac{45}{44})=(\frac{11}{10})(\frac{34}{33})(\frac{35}{34})=(\frac{12}{11})(\frac{22}{21})(\frac{49}{48})=(\frac{12}{11})(\frac{28}{27})(\frac{33}{32})=(\frac{13}{12})(\frac{21}{20})(\frac{40}{39})\)
\(=(\frac{13}{12})(\frac{27}{26})(\frac{28}{27})=(\frac{14}{13})(\frac{19}{18})(\frac{39}{38})=(\frac{14}{13})(\frac{25}{24})(\frac{26}{25})=(\frac{15}{14})(\frac{16}{15})(\frac{49}{48})=(\frac{15}{14})(\frac{21}{20})(\frac{28}{27})=(\frac{16}{15})(\frac{17}{16})(\frac{35}{34})=(\frac{16}{15})(\frac{21}{20})(\frac{125}{24})=(\frac{17}{16})(\frac{18}{17})(\frac{28}{27})=(\frac{19}{18})(\frac{20}{19})(\frac{21}{20})\)
\(a(7)=23,\frac{8}{7}=(\frac{11}{10})(\frac{45}{44})(\frac{64}{63})=(\frac{11}{10})(\frac{50}{49})(\frac{56}{55})=(\frac{12}{11})(\frac{33}{32})(\frac{64}{63})=(\frac{12}{11})(\frac{36}{35})(\frac{55}{54})=(\frac{12}{11})(\frac{43}{42})(\frac{44}{43})=(\frac{13}{12})(\frac{27}{26})(\frac{64}{63})=(\frac{13}{12})(\frac{36}{35})(\frac{40}{39})=(\frac{14}{13})(\frac{26}{25})(\frac{50}{49})=(\frac{15}{14})(\frac{21}{20})(\frac{64}{63})=(\frac{15}{14})(\frac{22}{21})(\frac{56}{55})=(\frac{15}{14})(\frac{24}{23})(\frac{46}{45})\)
\(=(\frac{15}{14})(\frac{26}{25})(\frac{40}{39})=(\frac{15}{14})(\frac{28}{27})(\frac{36}{35})=(\frac{15}{14})(\frac{31}{30})(\frac{32}{31})=(\frac{16}{15})(\frac{20}{19})(\frac{57}{56})=(\frac{16}{15})(\frac{21}{20})(\frac{50}{49})=(\frac{16}{15})(\frac{22}{21})(\frac{45}{44})=(\frac{16}{15})(\frac{25}{24})(\frac{36}{35})=(\frac{16}{15})(\frac{29}{28})(\frac{30}{29})=(\frac{17}{16})(\frac{18}{17})(\frac{64}{63})=(\frac{18}{17})(\frac{22}{21})(\frac{34}{33})=(\frac{19}{18})(\frac{20}{19})(\frac{36}{35})=(\frac{22}{21})(\frac{23}{22})(\frac{24}{23})\)
\(a(8)=32,\frac{9}{8}=\)
\(a(9)=41,\frac{10}{9}=\)
得到一串数。{0, 1, 4, 7, 13, 16, 23, 32, 41, 38, 44, 47, 53, 89, 97, 60, 76, 75, 91, 160, 134, 78, 104, 182, 139}——OEIS没有这串数。 $a,b,c,d$只能是119的因子吧, 这样的话,就只能取1,7,17了 比如a=1007，b=3220时有10组解：
{{1007,3220,3225,15895},{1007,3220,3281,14663},{1007,3220,3311,14093},{1007,3220,3440,12155},{1007,3220,3655,10065},{1007,3220,4023,8041},{1007,3220,4235,7310},{1007,3220,4859,5984},{1007,3220,5049,5719},{1007,3220,5083,5676}} a=1007时，共有7100组解 a Solutions
--------------------
1001 4620
1002 1620
1003 21112
1004 792
1005 5411
1006 917
1007 7100
1008 2457
1009 1527
1010 4468
$a \in $ (1+5/a)^4 > 120/119 > (1+5/a)
然后给定a可以估算b的范围，最后c,d最好通过因子分解来求解 我用了两天计算出$(1+\frac5a)(1+\frac5b)(1+\frac5c)(1+\frac5d)=\frac{130}{126}, 0 < a<=b<=c<=d $ 有3220235组解。 其它计算结果：
$(1+\frac5a)(1+\frac5b)(1+\frac5c)(1+\frac5d)=\frac{140}{133}, 0 < a<=b<=c<=d$有586573组解。
$(1+\frac5a)(1+\frac5b)(1+\frac5c)(1+\frac5d)=\frac{150}{140}, 0 < a<=b<=c<=d$有446895组解。
$(1+\frac5a)(1+\frac5b)(1+\frac5c)(1+\frac5d)=\frac{160}{147}, 0 < a<=b<=c<=d$有413302组解。
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