Well this problem has been confusing on how I should solve it for a long time, please give answer and explanation. thanks you

Here's another way of doing this for those who don't have a handy facility for factorising large numbers:

what is the largest value of \(n\) for which \(n^2\) is a factor of \(6^{101}+10^{101}\)?

Formula:

For an odd positive integer \(n\)
\(a^n+b^n=\left(a+b \right)\left(a^{n-1}-a^{n-2}b+\cdots -ab^{n-2}+b^{n-1}\right) \)

Source "What is the formula for \(a^n + b^n\)?" see: https://www.quora.com/What-is-the-formula-for-a-n-b-n

\(\begin{array}{|rcll|} \hline \mathbf{a^n+b^n} &=& \mathbf{\left(a+b \right)\left(a^{n-1}-a^{n-2}b+\cdots -ab^{n-2}+b^{n-1}\right)} \\ && \boxed{n=101,\ a=6,\ b=10} \\\\ 6^{101}+10^{101} &=& (6+10)\left(6^{101-1}-6^{101-2}*10+\cdots -6*10^{101-2}+10^{101-1}\right) \\\\ 6^{101}+10^{101} &=& 16\left(6^{100}-6^{99}*10+\cdots -6*10^{99}+10^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^4\left(6^{100}-6^{99}*10+6^{98}*10^{2}-6^{97}*10^{3}+ \\\\ \cdots -6^{3}*10^{97}+6^{2}*10^{98}-6*10^{99}+10^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^4\left(2^{100}3^{100}-2^{99}3^{99}*2*5+2^{98}3^{98}*2^{2}*5^{2}-2^{97}3^{97}*2^{3}*5^{3}+ \\\\ \cdots -2^{3}3^{3}*2^{97}5^{97}+2^{2}3^{2}*2^{98}5^{98}-2*3*2^{99}5^{99}+2^{100}5^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^4\left(2^{100}3^{100}-2^{100}3^{99}*5+2^{100}3^{98}*5^{2}-2^{100}3^{97}*5^{3}+ \\\\ \cdots -2^{100}3^{3}*5^{97}+2^{100}3^{2}*5^{98}-2^{100}*3*5^{99}+2^{100}5^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^42^{100}\left(3^{100}-3^{99}*5+3^{98}*5^{2}-3^{97}*5^{3}+ \\\\ \cdots -3^{3}*5^{97}+3^{2}*5^{98}-3*5^{99}+5^{100}\right) \\\\ 6^{101}+10^{101} &=& \mathbf{{\color{red}2^{104}}}\left(3^{100}-3^{99}*5+3^{98}*5^{2}-3^{97}*5^{3}+ \\\\ \cdots -3^{3}*5^{97}+3^{2}*5^{98}-3*5^{99}+5^{100}\right) \\ \hline n^2 &=& \mathbf{{\color{red}2^{104}}} \\ n &=& 2^{\frac{104}{2}} \\ \mathbf{n} &=& \mathbf{2^{52}} \\ \hline \end{array}\)

without latex: