One is 0; the other one is any square.

来源: tieddog 2014-02-05 09:43:52 [] [旧帖] [给我悄悄话] 本文已被阅读: 次 (489 bytes)
Assume x<=y.
If x^2+y=m^2, y=m^2-x^2.
If x is not zero, y^2<y^2+x.
We just need to show that y^2+x <(y+1)^2 to see that y^2+x cannot be a square.
The above inequality is equivalet to x<2y+1. (we already assumed x<=y).
So x =0. Done.





 


 
 





 

所有跟帖: 

AH, the second line is useless. -tieddog- 给 tieddog 发送悄悄话 (0 bytes) () 02/05/2014 postreply 09:45:33

不是吧 -怪哉- 给 怪哉 发送悄悄话 怪哉 的博客首页 (29 bytes) () 02/05/2014 postreply 10:38:35

The 2nd line in my solution is useless. -tieddog- 给 tieddog 发送悄悄话 (0 bytes) () 02/05/2014 postreply 10:45:57

true -怪哉- 给 怪哉 发送悄悄话 怪哉 的博客首页 (0 bytes) () 02/05/2014 postreply 10:56:58

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