找回密码
 欢迎注册
查看: 23762|回复: 2

[原创] \(a^2+b^2=c^c\)的整数解

[复制链接]
发表于 2014-7-1 19:44:44 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能,让你轻松玩转社区。

您需要 登录 才可以下载或查看,没有账号?欢迎注册

×
\[0 < a < b,c > 0\]
解答如下:
  1. {38,41,5}
  2. {3627003,17021162,13}
  3. {14978251196,24553864319,17}
  4. {110422359737857437,276811749100242716,25}
  5. {62654932136711087245,1601174728953762642038,29}
  6. {17653223669804406176810517719,101210170683788696336691161166,37}
  7. {660838610868797899598276970535204,945605605428452134699502122600795,41}
  8. {2838225339535590212632171720895586458614263073,4037489948847030354123278413390735412626034362,53}
  9. {92026335203365641228883569927785073219758320252972194,2833551078801118340582525994017176224987125885357998195,61}
  10. {18875148354976973140591021651659505535006602180542443896312,80944307640120876607176276669958114893060784557971487651841,65}
  11. {57126109178946065305412311372108933705914213036776348574713,60372674405198043668582460291247584088877551967558143668484,65}
  12. {50396194793822973702009500340813859234181306665908286244614111229768,89407095339857322318933460546771615780038748828047071595622728780003,73}
  13. {2598665847742448698837725382524318067791561204721325487322399378645545949318239238,9663704731635322472680884378920954654270215212690385335099545362573606642252021809,85}
  14. {5170456435303741627879005738368973703146134647337675750137591256890885847847734034,8567767129134639625749091074053976671552877258530108901800404176374377681662817313,85}
  15. {292549082370793816905920485683507600364345498566184017363777667021347308854194889889755,477006161396549254567288221363910786495605911702402615432675935333541420930803977693128,89}
  16. {615315211574957490824498594839203172855365521205325625783084250481112631694459689756470377660804,2198097616149661986424224025061103655172296200442738822651691575987582884691493412810060218373111,97}
复制代码
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2014-7-2 17:59:16 | 显示全部楼层
本帖最后由 282842712474 于 2014-7-2 20:33 编辑

$$|p+qi|^{2(p^2+q^2)}=|(p+qi)^{p^2+q^2}|^2=|c+di|^2=c^2+d^2$$
可以发现,你的c都是$p^2+q^2$的形式。

点评

你这符号用的,尽量不要跟题目混淆,以免别人看错。  发表于 2014-7-2 20:17
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
您需要登录后才可以回帖 登录 | 欢迎注册

本版积分规则

小黑屋|手机版|数学研发网 ( 苏ICP备07505100号 )

GMT+8, 2024-11-24 08:04 , Processed in 0.022396 second(s), 18 queries .

Powered by Discuz! X3.5

© 2001-2024 Discuz! Team.

快速回复 返回顶部 返回列表