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Source: British Math Olympiad, 2006.

This factors nicely as \(3^{32}-2^{32} = \left(3^{16}+2^{16}\right)\left(3^{16}-2^{16}\right)\), and we can continue factoring in this way to get \[3^{32}-2^{32} = \left(3^{16}+2^{16}\right)\left(3^8+2^8\right)\left(3^4+2^4\right)\left(3^2+2^2\right)\left(3^2-2^2\right).\]The final three terms are \(5, 13, 97\), so we have three of the four required primes. For another prime divisor, consider \(3^{16}-2^{16}\). By Fermat's Little Theorem \(a^{16}-1\equiv 0 \bmod 17\) for all \(a\) with \((a,17)=1\), and so it follows that \(3^{16}-2^{16}\equiv 0 \bmod 17\), and we therefore have \(17\) as a fourth such prime divisor.

Alternatively, note \( \left(\dfrac{3}{17}\right)=-1, \left(\dfrac{2}{17}\right)=1\), hence by Euler's Criterion \(3^8\equiv -1 \bmod 17\) and \(2^8\equiv 1 \bmod 17\), giving \(3^8+2^8\equiv 0\bmod 17\).

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