※ 引述《adamchi (adamchi)》之銘言：
: 1.a,b,c是整數 , 且(a,b)=1
: 試證: 若ax + by = c 有整數解 則(a,b)是c的因數
: (也要證回去: 若(a,b)是c的因數 ,則ax + by = c 有整數解 )
: 2. abc不等於0,
: a + b + c 不等於0
: 若 1/a +1/b +1/c = 1/(a+b+c)
: 試證: 1/a^2011 + 1/b^2011 +1/c^2011 = 1/(a+b+c)^2011
2.
we have (bc+ca+ab)(a+b+c) = abc
consider x^3 + px^2 +qx +r = 0
with x = a, b, c
x^3 +px^2 +qx +pq = 0 {note that p,q =! 0}
= (x^3 +qx) + p(x^2 +q ) = (x^2 +q)( x+p )
since x = -p = a +b +c = a or b or c
then b+c =0 or c+a =0 or a+b =0
enough to solve this problem.
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