[Math] 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?

Best Answer

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:

$$3^4= 81\equiv 3, 3^{16}=(3^4)^4\equiv 3, 3^{32}=(3^{16})^2\equiv 9$$ $$2^4=16\equiv 3, 2^{16}=(2^4)^4\equiv 3^4\equiv 3, 2^{32}=(2^{16})^2\equiv 9$$

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