2016
Problem - 3119
Find the largest of three prime divisors of $13^4+16^5-172^2$.
By Sophie Germain identity (see # 3863). \begin{align*} &13^4+16^5-172^2\\ =\quad & 13^4 + (2^5)^4 - 172^2\\ =\quad & (13^2 + 32 \end{align*} $13^4+16^5-172^2 = 2^{20} - 2^{10}+1 =\frac{2^{30}+1}{2^{10}+1}$ $2^{30} + 1 = 4\times (128)^4 + 1 =(2 \times 128^2 + 1 + 2\times 128)\times(2\times 128^2 + 1 - 2\times 128 )$