由 skywalker 於 星期日 十二月 24, 2006 8:46 am
#ed_op#P#ed_cl##ed_op#FONT face=Verdana#ed_cl#令Ai=4#ed_op#SUP#ed_cl#i#ed_op#/SUP#ed_cl#+3#ed_op#SUP#ed_cl#i#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#A(i+4)=256*4#ed_op#SUP#ed_cl#i#ed_op#/SUP#ed_cl#+81*3#ed_op#SUP#ed_cl#i#ed_op#/SUP#ed_cl#≡6Ai(mod25)#ed_op#BR#ed_cl#A1,A3,A4≠0(mod25),A2≡0(mod25),故僅A(2),A(6),..A(4i+2)≡0(mod25)#ed_op#BR#ed_cl#令2i+1=m,故可令(3#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#)^2+(4#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#)^2=(5#ed_op#SUP#ed_cl#p#ed_op#/SUP#ed_cl#)^2#ed_op#/FONT#ed_cl##ed_op#/P#ed_cl##ed_op#P#ed_cl##ed_op#FONT face=Verdana#ed_cl#以下借galaxy作法:#ed_op#BR#ed_cl#3#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#=hh-kk,4#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#=2hk,h,k一奇一偶,(h,k)=1#ed_op#BR#ed_cl#由4#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#=2hk,得h=4#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#/2,k=1#ed_op#BR#ed_cl#當m>1時,3#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#=hh-kk=(4#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#/2)^2-1#ed_op#BR#ed_cl#=4#ed_op#SUP#ed_cl#2m-1#ed_op#/SUP#ed_cl#-1>3#ed_op#SUP#ed_cl#m#ed_op#/SUP#ed_cl#,故m=1為唯一上解,即n=2為唯一解#ed_op#/FONT#ed_cl##ed_op#/P#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#以上是轉貼自昌爸yani的證法#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#