YLL討論網

#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#

#ed_op#FONT face=Verdana#ed_cl#(i)n為正整數#ed_op#BR#ed_cl#5^n的因數必為5^k型，若(3^n+4^n)｜5^n#ed_op#BR#ed_cl#則存在正整數m，使得3^n+4^n=5^m#ed_op#BR#ed_cl#方程式兩邊對3同餘，得0+1=2^m，所以m為偶數#ed_op#BR#ed_cl#方程式兩邊對4同餘，得(-1)^n+0=1，所以n為偶數#ed_op#BR#ed_cl#假設m=2a，n=2b，則有(3^b)^2+(4^b)^2=(5^a)^2#ed_op#BR#ed_cl#則3^b，4^b，5^a是一組畢氏數#ed_op#BR#ed_cl#必存在一奇一偶的h,k，且(h,k)=1#ed_op#BR#ed_cl#使得3^b=h^2-k^2...(1)，4^b=2hk...(2)#ed_op#BR#ed_cl#由(2)得2^(2b-1)=hk#ed_op#BR#ed_cl#因為h,k為一奇一偶且(h,k)=1，所以k=1，h=2^(2b-1)#ed_op#BR#ed_cl#代入(1)中得3^b=4^(2b-1)-1#ed_op#BR#ed_cl#=> 3^b+1=4^(2b-1)#ed_op#BR#ed_cl#若b≧2，上式右邊為8的倍數，上式左邊用8除餘數為2或4#ed_op#BR#ed_cl#所以只能b=1，即n=2#ed_op#BR#ed_cl##ed_op#BR#ed_cl#(ii)若n為負整數#ed_op#BR#ed_cl#設n=-m，m為正整數#ed_op#BR#ed_cl#(5^n)/(3^n+4^n)#ed_op#BR#ed_cl#=(1/5)^m/[(1/3)^m+(1/4)^m]#ed_op#BR#ed_cl#=12^m/(15^m+20^m)為一正真分數，不為整數#ed_op#/FONT#ed_cl##ed_op#BR#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#以上是轉貼自昌爸的#ed_op#STRONG#ed_cl##ed_op#FONT face=Verdana#ed_cl#galaxy的證法#ed_op#/FONT#ed_cl##ed_op#/STRONG#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

在此只討論n為整數的情況...

擴分得(分母 分子同*5^n)


(3^2+4^2)^n當然被5^整除.
若(3^2+4^2)^n被(3^n+4^n)整除
則3^2+4^2必須被(3^n+4^n)整除...
即n只能=2.

#ed_op#DIV#ed_cl#n為實數,已知5#ed_op#SUP#ed_cl#n#ed_op#/SUP#ed_cl#/(3#ed_op#SUP#ed_cl#n#ed_op#/SUP#ed_cl#+4#ed_op#SUP#ed_cl#n#ed_op#/SUP#ed_cl#)是整數,試求n值#ed_op#/DIV#ed_cl#