Find the largest of the three prime divisors of the number $13^4 + 16^5 - 172^2$ I was able to factor out only the prime 13,thus $13^4 + 16^5 - 172^2=13\cdot 80581$
What should be done to solve it? (Maybe some clever factorization, modulo, or anything else?)

Since $13^4−172^2 = (13^2-172)(13^2+172) = -3 \times 341 = -1023 = 1-2^{10}$ we see that
$$13^4+16^5−172^2 = 2^{20}-2^{10}+1.$$
Next, since
$$\frac{1+x^{30}}{1+x^{10}} = x^{20}-x^{10}+1,$$
then, using the Sophie Germain identity on both numerator and denominator, we have that
$$2^{20}-2^{10}+1 = \frac{1+4 \times (2^7)^4}{1+4 \times (2^2)^4} = 
\frac{1-2^8+2^{15}}{1+2^3+2^5} \times \frac{1+2^8+2^{15}}{1-2^3+2^5}=
\frac{32513}{41} \times \frac{33025}{25}=793 \times 1321.$$
It's easy to see that $793 = 13 \times 61$ and, as the solution is a product of three primes, we are done. So the largest prime divisor is 1321.
