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Number theory Difficulty 6.4 National olympiad Prove it

1-151 1 Prove that 21992−12^{1992}-1 can be expressed as the product of six integers, each greater than 22482^{248}.

Solution

[Solution] 1992=8×2491992=8 \times 249 \quad and 249×4=332×3249 \times 4=332 \times 3
21992−1=(2249)8−1={(2249)4−1}{(2249)4+1}=(2249−1)(2249+1){(2249)2+1}{(2332)3+1}=(2249−1)(2249+1){(2249)2+2×2249+1−2250}(2332+1)(2664−2332+1)=(2249−1)(2249+1)(2249+1−2125)(2249+1+2125)(2332+1)(2664−2332+1).\begin{aligned} 2^{1992}-1= & \left(2^{249}\right)^{8}-1 \\ = & \left\{\left(2^{249}\right)^{4}-1\right\}\left\{\left(2^{249}\right)^{4}+1\right\} \\ = & \left(2^{249}-1\right)\left(2^{249}+1\right)\left\{\left(2^{249}\right)^{2}+1\right\}\left\{\left(2^{332}\right)^{3}+1\right\} \\ = & \left(2^{249}-1\right)\left(2^{249}+1\right)\left\{\left(2^{249}\right)^{2}+2 \times 2^{249}+1-\right. \\ & \left.2^{250}\right\}\left(2^{332}+1\right)\left(2^{664}-2^{332}+1\right) \\ = & \left(2^{249}-1\right)\left(2^{249}+1\right)\left(2^{249}+1-2^{125}\right)\left(2^{249}+1+\right. \\ & \left.2^{125}\right)\left(2^{332}+1\right)\left(2^{664}-2^{332}+1\right) . \end{aligned}

It is clear that each of the six factors is greater than 22482^{248}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.