n為實數,已知5n/(3n+4n)是整數,試求n值
令Ai=4i+3i
A(i+4)=256*4i+81*3i≡6Ai(mod25)
A1,A3,A4≠0(mod25),A2≡0(mod25),故僅A(2),A(6),..A(4i+2)≡0(mod25)
令2i+1=m,故可令(3m)^2+(4m)^2=(5p)^2
以下借galaxy作法:
3m=hh-kk,4m=2hk,h,k一奇一偶,(h,k)=1
由4m=2hk,得h=4m/2,k=1
當m>1時,3m=hh-kk=(4m/2)^2-1
=42m-1-1>3m,故m=1為唯一上解,即n=2為唯一解