I'm having a go at BMO 2006/7 Q1 which states: "Find four prime numbers less than 100 which are factors of $3^{32}-2^{32}$."
My working is as follows (basically just follows difference of two squares loads of times):
$$3^{32}-2^{32}$$ $$=(3^{16}+2^{16})(3^{16}-2^{16})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{8}-2^{8})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{4}-2^{4})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)(3-2)$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)$$
Now it is simple to get $3$ of the primes here: $$3+2=5$$ $$3^2+2^2=13$$ $$3^4+2^4=97$$
Now I was having some trouble with finding the fourth prime. It's clear that the fourth prime must either be a prime factor of $3^8+2^8$ or $3^{16}+2^{16}$. I started off with $3^8+2^8$:
$$3^8+2^8$$ $$=3^{4^2}+256$$ $$=81^2+256$$ $$=6561+256$$ $$=6817$$
Here is where I am stuck because google tells me that $6817=17*401$ and so the fourth prime is $17$. But this is a non-calculator paper so is there a way of working out this answer without working out $\frac{6817}{3},\frac{6817}{7},\frac{6817}{11}$ etc until one of them is an integer solution?
Finally, before anyone says this is a duplicate, there is a similar question here (prime factors of $3^{32}-2^{32}$) that I didn't know existed until writing this question. However, none of those answers give a way of finding $17$ without using any computational methods. So are they just expecting you to essentially do trial and error throughout all of the primes under $100$ until one works?