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# [求助] a^2+ab+b^2=c^2的整数解

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a^2+ab+b^2=c^2的所有整数解

楼主| 发表于 2014-7-1 17:44:44 | 显示全部楼层
 比如 7,8,13是一组解答

 一般都是考虑对于哪些c有解

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楼主| 发表于 2014-7-1 18:16:26 | 显示全部楼层
 我给出mathematica的代码,能够找到一部分解答. 暴力求解 (*求整数解答*) Do[out={ToRules@Reduce[a^2+a*b+b^2==c^2&&a>0&&a这类符号,只保留结果*)      Map[Print,outabc](*通过映射打印出所有解答*)   ],    {n,0,200} ] 复制代码 求解结果. {3,5,7} {7,8,13} {6,10,14} {5,16,19} {9,15,21} {14,16,26} {12,20,28} {11,24,31} {15,25,35} {7,33,37} {10,32,38} {21,24,39} {18,30,42} {13,35,43} {16,39,49} {21,35,49} {28,32,52} {24,40,56} {15,48,57} {9,56,61} {22,48,62} {27,45,63} {35,40,65} {32,45,67} {30,50,70} {17,63,73} {14,66,74} {20,64,76} {33,55,77} {42,48,78} {40,51,79} {36,60,84} {26,70,86} {11,85,91} {19,80,91} {39,65,91} {49,56,91} {33,72,93} {25,80,95} {55,57,97} {32,78,98} {42,70,98} {40,77,103} {56,64,104} {45,75,105} {24,95,109} {21,99,111} {48,80,112} {30,96,114} {63,72,117} {51,85,119} {18,112,122} {44,96,124} {54,90,126} {13,120,127} {39,105,129} {70,80,130} {23,120,133} {35,112,133} {57,95,133} {65,88,133} {64,90,134} {69,91,139} {60,100,140} {77,88,143} {34,126,146} {48,117,147} {63,105,147} {28,132,148} {56,115,151} {40,128,152} {66,110,154} {55,120,155} {84,96,156} {25,143,157} {80,102,158} {69,115,161} {75,112,163} {72,120,168} {15,161,169} {91,104,169} {45,144,171} {52,140,172} {75,125,175} {104,105,181} {22,170,182} {38,160,182} {78,130,182} {98,112,182} {27,168,183} {35,165,185} {66,144,186} {81,135,189} {50,160,190} {32,175,193} {110,114,194} {105,120,195} {64,156,196} {84,140,196} {56,165,199}复制代码

楼主| 发表于 2014-7-1 18:20:26 | 显示全部楼层
 再加一个限制条件吧, 如果a=0或者b=0,这个解答是平凡的,没有意义的. 同时限制c>0 以及由于对称,限制a0 GCD[a,b]=1 这是几个求解条件

楼主| 发表于 2014-7-1 18:23:55 | 显示全部楼层
 cn8888 发表于 2014-7-1 18:20 再加一个限制条件吧, 如果a=0或者b=0,这个解答是平凡的,没有意义的. 同时限制c>0 在上面的条件限制下: (*求整数解答*) Do[out={ToRules@Reduce[a^2+a*b+b^2==c^2&&0这类符号,只保留结果*)      Map[Print,outabc](*通过映射打印出所有解答*)   ],    {n,0,200} ] 复制代码 求解结果如下: {3,5,7} {7,8,13} {5,16,19} {11,24,31} {7,33,37} {13,35,43} {16,39,49} {9,56,61} {32,45,67} {17,63,73} {40,51,79} {11,85,91} {19,80,91} {55,57,97} {40,77,103} {24,95,109} {13,120,127} {23,120,133} {65,88,133} {69,91,139} {56,115,151} {25,143,157} {75,112,163} {15,161,169} {104,105,181} {32,175,193} {56,165,199} 复制代码

 $$\D a^2+ab+b^2=c^2 \implies \left(\frac a c\right)^2+\left(\frac a c\right)\left(\frac b c\right)+\left(\frac b c\right)^2=1\implies m^2+mn+n^2=1$$ 令 $$n=km-1$$， 可得到 $$m^2+m(km-1)+(km-1)^2=1$$ $$(k^2+k+1)m^2-(2k+1)m=0$$ $$\D m=0;\ m=\frac{2k+1}{k^2+k+1}$$

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 对于$m=0$的情况，可得到一组通解$(0,k,k)$

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楼主| 发表于 2014-7-1 18:30:50 | 显示全部楼层
 cn8888 发表于 2014-7-1 18:23 在上面的条件限制下: 求解结果如下: 在上面的解答中,只有 {49, 91, 91, 133, 133, 169} 是合数 其中49与169是完全平方 91与133都是给出了两组解答

楼主| 发表于 2014-7-1 18:39:45 | 显示全部楼层
 在c的可能解答中 {199, 211, 217, 217, 223, 229, 241, 247, 247, 259, 259, 271, 277, \ 283, 301, 301, 307, 313, 331, 337, 343, 349, 361, 367, 373, 379, 397} 不是素数的 {217, 217, 247, 247, 259, 259, 301, 301, 343, 361} 要么是两组解答,要么是立方数或者平方数

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343=7^3,361=19^2  发表于 2014-7-1 19:00

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