prime factors of $3^{32}-2^{32}$ The question asks to find 4 prime factors of $3^{32}-2^{32}$ under $100$.
My take:I factorized it and the obvious ones are $5, 97$ and $13$.I cannot find the last one ,however.I was wondering if we could prove the number is even but that has not worked out well.So what is the other factor and how do we find it?
 A: If you are able to factor the number (which I did with wolframalpha), you will get $$5\cdot 13\cdot 17\cdot 97\cdot 401\cdot 3041\cdot 14177$$
which immediately answers your question.
However most likely the intent of the person asking was for you to use indirect means to determine small factors.  For example, to tell whether $5$ is a factor, we calculate the expression modulo 5: $$3^{32}-2^{32}\equiv 3^{32}-(-3)^{32}\equiv 3^{32}-(-1)^{32}3^{32}\equiv 0$$
Additional example as requested, modulo 13:
$$3^4= 81\equiv 3, 3^{16}=(3^4)^4\equiv 3, 3^{32}=(3^{16})^2\equiv 9$$
$$2^4=16\equiv 3, 2^{16}=(2^4)^4\equiv 3^4\equiv 3, 2^{32}=(2^{16})^2\equiv 9$$
A: If you wanted to find the other factor via your method, you would have to say:
$$
3^{32}-2^{32}=(3^8-2^8)(3^8+2^8)(3^{16}+2^{16})
$$
and then note that $(3^8+2^8)=6817=401\cdot17$
A: Since the algebraic factors of $3^{32}-2^{32}$ are all sums of two (coprime) squares, we only need to look for primes that are $1$ mod $4$.  This suggests checking $17$ once $5$ and $13$ have been found, and we quickly see that Fermat's theorem shows $3^{32} \equiv 2^{32} \equiv 1 \pmod{17}$.
Edit: By "algebraic factors" I mean those arising from the factorization of $x^{32} - y^{32}$ over $\mathbb Q[x,y]$:
$$3^{32} - 2^{32} = (3^{16}+2^{16})(3^8+2^8)(3^4+2^4)(3^2+2^2)(3+2)(3-2).$$
A: Actually breakimg that by a^2-b^2 is just useful for only the 3 answers you have got, but im sure you did not check out the nos. with the bigger powers. If you had checked 3^8 +2^8, which is 6817 , it is divisible by 17 (and 401). And by the way anything odd minus even is always odd. So this no. only has odd factors (because all of them are in form 3^x+-2^y. Here i mean plus minus.)
