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manhnguyen0164

Áp dụng liên tiếp hằng đẳng thức $a^2-b^2=(a-b)(a+b)$.

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)=...=2^{32}-1$
 
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