subtracting exponents with different bases and finding all the prime factors of it

Question

Here is the question:

Find all prime factors of $3^{18}-2^{18}$

No Calculators allowed, and it is a competition problem, fastest answers are the best.

How can I approach this problem? If it was just $3^{18}$, it would just be 3 and likewise for 2. What do I do?

Answer 1

Note that $3^{18}-2^{18} = (3^{9}-2^{9})(3^{9} + 2^{9})$ by Difference of Squares. Using Difference and Sum of Cubes:

$$(3^{9}-2^{9})(3^{9} + 2^{9})$$

$$(3^{3}-2^{3})(3^{6} + 3^{3}2^{3} + 2^{6})(3^{3}+2^{3})(3^{6}-3^{3}2^{3}+2^{6})$$

Now the factors are a bit more manageable:

$$19\cdot 1009\cdot 35\cdot 577$$

Both of the larger numbers are prime (which can be verified by a bit of trial division), so we have:

$$3^{18}-2^{18} = \boxed{5\cdot 7\cdot 19\cdot 577\cdot 1009}$$

Answer 2

You are supposed to see it as a difference of squares and cubes to start. $3^{18}-2^{18}=(3^9-2^9)(3^9+2^9)$ and you can factor each of these as a sum or difference of cubes. That gets the numbers down to primes you may know, or you may need to do some trial division.