Factoring : \(2^{58}+1=\left(2^{29}+2^{15}+1\right)\cdot\left(2^{29}-2^{15}+1\right)\)

Because : \(\left\{{}\begin{matrix}2^{29}+2^{15}+1=536903681\\2^{29}-2^{15}+1=536838145=5\cdot107367629\end{matrix}\right.\)

So \(2^{58}+1=\left(2^{29}+2^{15}+1\right)\cdot\left(2^{29}-2^{15}+1\right)=536903681\cdot107367629\cdot5\)

\(\rightarrow\dfrac{2^{58}+1}{5}=\dfrac{536903681\cdot107367629\cdot5}{5}=536903681\cdot107367629\)

So the number \(\dfrac{2^{58}+1}{5}\) is composite.

I wish ForOver hadn't written 2²⁹ + 2¹⁵ + 1 and 2²⁹ - 2¹⁵ + 1 as 2 numbers. I'll do it again :

\(2^{58}+1=\left(2^{29}\right)^2+1^2+2.2^{29}.1-2^{30}=\left(2^{29}+1\right)^2-\left(2^{15}\right)^2\)

\(=\left(2^{29}-2^{15}+1\right)\left(2^{29}+2^{15}+1\right)\)

\(2^{29}\equiv\left(2^4\right)^7.2\equiv6.2\equiv2\left(mod10\right)\)

\(2^{15}\equiv\left(2^4\right)^3.2^3\equiv6.8\equiv8\left(mod10\right)\)

\(\Rightarrow2^{29}-2^{15}+1\equiv2-8+1\equiv12-8+1\equiv5\left(mod10\right)\)

\(\Rightarrow2^{29}-2^{15}+1⋮5\)

So, \(\dfrac{2^{58}+1}{5}\) is the product of 2 natural numbers \(2^{29}+2^{15}+1;\dfrac{2^{29}-2^{15}+1}{5}\). Hence, it's a composite number