Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas: $$ f_n =\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5} $$ $$ We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad Note\space that \qquad{ D= f_{n+20}- f_n=\frac{\left(\phi \right)^{n+20} - \left(- \phi \right)^{-(n+20)}}{\sqrt5}-\frac{\left(\phi \right)^n - \left(- \phi \right)^{-n}}{\sqrt5}=\left( \frac{\left(\phi \right)^{10} - \left(- \phi \right)^{-10}}{\sqrt5} \right) \left({\left(\phi \right)^{n+10} - \left(- \phi \right)^{-(n+10)}}\right)=(f_{10})\left({\left(\phi \right)^{n+10} - \left(- \phi \right)^{-(n+10)}}\right)} $$ $$ but\qquad 5\mid (f_{10}=55) \Longrightarrow 5\mid D\Longrightarrow{f_n≡f_{n+20}(mod5)} $$

Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas: $$ f_n =\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5} $$ $$ We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad Note\space that \qquad{ D= f_{n+20}- f_n=\frac{\left(\phi \right)^{n+20} - \left(1- \phi \right)^{-(n+20)}}{\sqrt5}-\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5}=\left( \frac{\left(\phi \right)^{10} - \left(1- \phi \right)^{-10}}{\sqrt5} \right) \left({\left(\phi \right)^{n+10} - \left(1- \phi \right)^{-(n+10)}}\right)=(f_{10})\left({\left(\phi \right)^{n+10} - \left(1- \phi \right)^{-(n+10)}}\right)} $$ $$ but\qquad 5\mid (f_{10}=55) \Longrightarrow 5\mid D\Longrightarrow{f_n≡f_{n+20}(mod5)} $$

Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas: $$ f_n =\frac{\left(\phi \right)^n - \left(- \phi \right)^{-n}}{\sqrt5} $$ $$ We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad Note\space that \qquad{ D= f_{n+20}- f_n=\frac{\left(\phi \right)^{n+20} - \left(- \phi \right)^{-(n+20)}}{\sqrt5}-\frac{\left(\phi \right)^n - \left(- \phi \right)^{-n}}{\sqrt5}=\left( \frac{\left(\phi \right)^{10} - \left(- \phi \right)^{-10}}{\sqrt5} \right) \left({\left(\phi \right)^{n+10} - \left(- \phi \right)^{-(n+10)}}\right)=(f_{10})\left({\left(\phi \right)^{n+10} - \left(- \phi \right)^{-(n+10)}}\right)} $$ $$ but\qquad 5\mid (f_{10}=55) \Longrightarrow 5\mid D\Longrightarrow{f_n≡f_{n+20}(mod5)} $$

Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas: $$ f_n =\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5} $$ $$ We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad Note\space that \qquad{ D= f_{n+20}- f_n=\frac{\left(\phi \right)^{n+20} - \left(- \phi \right)^{-(n+20)}}{\sqrt5}-\frac{\left(\phi \right)^n - \left(- \phi \right)^{-n}}{\sqrt5}=\left( \frac{\left(\phi \right)^{10} - \left(- \phi \right)^{-10}}{\sqrt5} \right) \left({\left(\phi \right)^{n+10} - \left(- \phi \right)^{-(n+10)}}\right)=(f_{10})\left({\left(\phi \right)^{n+10} - \left(- \phi \right)^{-(n+10)}}\right)} $$ $$ but\qquad 5\mid (f_{10}=55) \Longrightarrow 5\mid D\Longrightarrow{f_n≡f_{n+20}(mod5)} $$