形如2m(m+1)的Fibonacci数仅有0和144

Only Fibonacci numbers with form 2m(m + 1) are 0 and 144

应用雅可比符号恒等式及Fibonacci数关于模m的剩余类的周期性,通过求解序列{2F_(n)+1}中的平方数,证明可表为4倍于三角形数即形如2m(m+1)的Fibonacci数仅有F_(0)=0和F_(12)=144。

By using the Jacobi symbol identity and the periodicity of F_(n)(mod m),it is proved that the only Fibonacci numbers with the form 2m(m+1)are F_(0)=0 and F_(12)=144 through solving the squares in{2F_(n)+1}.