prime or composite?

\hspace{-16}$Here $\mathbf{2^{58}+1=4.(2)^{56}+1}$\\\\ $\mathbf{=4.(2^{14})^{4}+1=(2.2^{28}+1-2.2^{14}).(2.2^{28}+1+2.2^{14})}$\\\\ $\mathbf{=(2^{29}+1-2^{15}).(2^{29}+1+2^{15})}$ \\\\Using Sophie Jermain Identity\\\\ $\mathbf{4x^4+y^4=(2x^2+y^2-2xy).(2x^2+y^2+2xy)}$\\\\ Now Here $\mathbf{(2^{29}+1+2^{15})=5.K\;,}$ Where $\mathbf{K\in\mathbb{Z^{+}}}$\\\\ bcz Last Digit of $\mathbf{(2^{29}+1+2^{15})=}$\; Last Digit of $\mathbf{2^9+1+2^5=5}$\\\\ So $\mathbf{\frac{2^{58}+1}{5}=\frac{5.K.(2^{29}+1-2^{15})}{5}=K.(2^{29}+1-2^{15})}$\\\\ So It has two factor \\\\ So $\mathbf{\frac{2^{58}+1}{5}}$ is a Composite no.

Nice Solution man111.

Thanks. but the proof is really good..

Composite