x(x+a)=2(x+b)(x+c)有两个正整数解
如何选择正整数a、b、c，使得x(x+a)=2(x+b)(x+c)有两个正整数解？ 首先求根公式里面的(b+c-a)(b+c-a)-bc要为完全平方数，而且a>b+c+[(b+c-a)^2-bc]^(1/2)或[(b+c-a)^2-bc]^(1/2)+a>b+c 先看答案(好像没有规律)。
Table, {x, 1, 4}, {k, 1, 6}]
{x=1,k=1, 6}{c -> 1, b -> 1, a -> 7}}, {}, {{c -> 1, b -> 2,a -> 11}}, {}, {{c -> 1, b -> 3, a -> 15}}, {}},
{x=2,k=1, 6}{c -> 1, b -> 3, a -> 13}}, {{c -> 1, b -> 4, a -> 16},{c -> 2, b -> 2, a -> 14}}, {{c -> 1, b -> 5, a -> 19}}, {{c -> 1, b -> 6,a -> 22},{c->2,b->3,a->18}},{{c->1,b->7,a->25}},{{c->1,b->8,a->28},{c->2,b->4,a->22}},
{x=3,k=1, 6}{c -> 1, b -> 6, a -> 21}, {c -> 2, b -> 3, a -> 17}}, {}, {{c -> 1, b -> 9, a -> 29}, {c -> 3, b -> 3, a -> 21}}, {}, {{c -> 1, b -> 12, a -> 37}, {c -> 2, b -> 6,a -> 27}, {c -> 3, b -> 4, a -> 25}}, {}},
{x=4,k=1, 6}{c -> 1, b ->10, a ->31}, {c -> 2, b -> 5, a -> 23}}, {{c -> 1, b -> 12, a -> 36}, {c -> 2, b -> 6, a -> 26}, {c -> 3, b -> 4,a -> 24}}, {{c -> 1, b -> 14, a -> 41}, {c -> 2, b -> 7, a -> 29}},
{{c -> 1, b -> 16, a -> 46}, {c -> 2, b -> 8, a -> 32}, {c -> 4, b -> 4, a -> 28}}, {{c -> 1, b -> 18, a -> 51}, {c -> 2, b -> 9, a -> 35}, {c -> 3, b -> 6, a -> 31}}, {{c -> 1, b -> 20, a -> 56}, {c->2,b->10,a->38},{c->4,b->5,a->32}} Select] x-2(x+k[])(x+k[])==0,x,PositiveIntegers]},{k,Subsets}],#[]!={}&]
{{{10,2,1},{2,2}},{{11,2,1},{1,4}},{{13,3,1},{2,3}},{{15,3,1},{1,6}},{{16,4,1},{2,4}},{{19,4,1},{1,8}},{{19,5,1},{2,5}},{{23,5,1},{1,10}},{{21,6,1},{3,4}},{{22,6,1},{2,6}},{{27,6,1},{1,12}},{{25,7,1},{2,7}},{{26,8,1},{4,4}},{{28,8,1},{2,8}},{{29,9,1},{3,6}},{{17,3,2},{3,4}},{{18,3,2},{2,6}},{{23,3,2},{1,12}},{{20,4,2},{4,4}},{{22,4,2},{2,8}},{{29,4,2},{1,16}},{{23,5,2},{4,5}},{{26,5,2},{2,10}},{{26,6,2},{4,6}},{{27,6,2},{3,8}},{{30,6,2},{2,12}},{{29,7,2},{4,7}},{{24,4,3},{4,6}},{{25,4,3},{3,8}},{{28,4,3},{2,12}},{{27,5,3},{5,6}},{{29,5,3},{3,10}},{{30,6,3},{6,6}}} 若b=c,以下情况有解：
{{{7,1},{1,2}},{{14,2},{2,4}},{{17,2},{1,8}},{{21,3},{3,6}},{{23,3},{2,9}},{{31,3},{1,18}},{{28,4},{4,8}},{{34,4},{2,16}},{{49,4},{1,32}},{{35,5},{5,10}},{{47,5},{2,25}},{{71,5},{1,50}},{{41,6},{8,9}},{{42,6},{6,12}},{{46,6},{4,18}},{{51,6},{3,24}},{{62,6},{2,36}},{{97,6},{1,72}},{{49,7},{7,14}},{{79,7},{2,49}},{{56,8},{8,16}},{{68,8},{4,32}},{{98,8},{2,64}},{{63,9},{9,18}},{{69,9},{6,27}},{{93,9},{3,54}},{{70,10},{10,20}},{{73,10},{8,25}},{{85,10},{5,40}},{{94,10},{4,50}},{{77,11},{11,22}},{{82,12},{16,18}},{{84,12},{12,24}},{{89,12},{9,32}},{{92,12},{8,36}},{{91,13},{13,26}},{{98,14},{14,28}}}
{{7,1},{14,2},{17,2},{21,3},{23,3},{31,3},{28,4},{34,4},{49,4},{35,5},{47,5},{71,5},{41,6},{42,6},{46,6},{51,6},{62,6},{97,6},{49,7},{79,7},{56,8},{68,8},{98,8},{63,9},{69,9},{93,9},{70,10},{73,10},{85,10},{94,10},{77,11},{82,12},{84,12},{89,12},{92,12},{91,13},{98,14}} 易知$a=7n,b=c=n$时方程有两个正整数解${n,2n}$ 易知$a=3n+4,b=1,c=n$时方程有两个正整数解${2,n}$ 也就是两根之积2bc,两根之和a-2(b+c).
所以对于任意选择的正整数b,c选择2bc任意一个正因子x1,记x2=2bc/x1,取a=x1+x2+2(b+c)即可
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