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Prime factor question

Find 4 prime factors of 332−232 3^{32} - 2^{32} which are below 100.

I've managed to show 5 is one (and 1 obviously) but I'm a bit stuck as to where to go next.

I've tried to simplify the expression by putting it in terms of 2 or 3 but to no avail. To show 5 is a factor, i showed 3^32 ends in 1 and 2^32 ends in 6, so the value of the expression ends in 5 whcih is a multiple of 5.

I dont really know where to go next.
Reply 1
Avatar for SimonM
SimonM
18
1 isn't prime.

Recall the 'difference of two squares' identity.
Reply 2
Avatar for My Alt
My Alt
12
You could try x=332−232=(316+216)(316−216)=(316+216)(38+28)(38−28)...x=3^{32} - 2^{32} = (3^{16} + 2^{16})(3^{16} - 2^{16}) = (3^{16} + 2^16)(3^8 + 2^8)(3^8 - 2^8)...

To find smaller and smaller prime factors. I think this is what it wants you to do, as 32 is a power of two (meaning you can continnually find the difference of two squares.)
Reply 3
Avatar for kabbers
kabbers
4
note 332−232=(316−216)(316+216)3^{32} - 2^{32} = (3^{16}-2^{16})(3^{16}+2^{16})

you can continuously factorise in this way until you find factors below 100.
this can be done by finding the least n for which 3n+2n>1003^{n}+2^{n} > 100 and ignoring appropriate factors.

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