Question

(2 +1)(2 в кв +1)(2в 4 +1)(2 в 16+1)(2 в 32+1)=?
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Answer 1

$(2+1)*(2^2+1)*(2^4+1)*(2^8+1)(2^{16}+1}*(2^{32}+1))=\\(2-1)(2+1)*(2^2+1)*(2^4+1)*(2^8+1)(2^{16}+1}*(2^{32}+1))=\\(2^2-1)*(2^2+1)*(2^4+1)*(2^8+1)(2^{16}+1}*(2^{32}+1))=\\(2^4-1)*(2^4+1)*(2^8+1)(2^{16}+1}*(2^{32}+1))=\\(2^8-1)*(2^8+1)(2^{16}+1}*(2^{32}+1))=\\(2^{16}-1)*(2^{16}+1}*(2^{32}+1))=\\(2^{32}-1)*(2^{32}+1))=\\2^{64}-1$