1.
proof ax+by=c exist Z solution, then (a,b)=k|c
if (a,b)=k then a=ka', b=kb', (a', b')=1
ax+by = ka'x +kb'y =k(a'x+b'y)=c
if (x,y) exist Z solution, and (a',b')=1
obvious (a'x,b'y)=1 (if (x,y)=1) then k|c proved.
if (x,y) is not equal to 1 , this condition can be condenced to the above.
xxx
proof (a,b)=k|c, then ax+by=c exist Z solution.
if (a,b)=k|c, then we have c=kc', and a=ka', b=kb', (a',b')=1,
then ax+by=c => ka'x+kb'y=kc' => a'x+b'y=c' , cause (a',b')=1|c,
then (x,y) exist Z solution.
is that right? ^^"
※ 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
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