017. Can Be Factorized Further

Show that $2^{1992} - 1$ can be expressed as the product of six integers greater than $2^{248}$.

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Since $1992 = 8 \times 249$, $$\begin{aligned} 2^{1992} - 1 &= (2^{249})^8 - 1 = \left( (2^{249})^4 - 1 \right)\left( (2^{249})^4 + 1 \right) \\\\ &= \left( (2^{249})^2 - 1 \right)\left( (2^{249})^2 + 1 \right)\left( (2^{249})^4 + 1 \right) \\\\ &= \left( 2^{249} - 1 \right)\left( 2^{249} + 1 \right)\left( (2^{249})^2 + 1 \right)\left( (2^{249})^4 + 1 \right) \\\\ &= \left( 2^{249} - 1 \right)\left( 2^{249} + 1 \right)\left( (2^{249})^2 + 1 \right)\left( (2^{332})^3 + 1 \right) \\\\ &= \left( 2^{249} - 1 \right)\left( 2^{249} + 1 \right)\left( (2^{249})^2 + 1 \right)\left( 2^{332} + 1 \right)\left( (2^{332})^2 - 2^{332} + 1 \right) \end{aligned}$$

It seems that the above is the final factorized, but it can be factorized further. $t^2 + 1$ can be factorized if some $a$ can be found such that $t^2 + 1 = (t + 1 + a)(t + 1 - a)$. $$\begin{aligned} t^2 + 1 = (t + 1 + a)(t + 1 - a) = (t + 1)^2 - a^2 \implies a^2 = 2t \end{aligned}$$

Let $t = 2^{249}$. Then, $$\begin{aligned} a^2 = 2t = 2^{250} \implies a = 2^{125} \end{aligned}$$

This means $$\begin{aligned} (2^{249})^2 + 1 = (2^{249} + 1 + 2^{125})(2^{249} + 1 - 2^{125}) \end{aligned}$$

Therefore, $$\begin{aligned} 2^{1992} - 1 = \left( 2^{249} - 1 \right)\left( 2^{249} + 1 \right)\left( 2^{249} + 2^{125} + 1 \right)\left( 2^{249}- 2^{125} + 1 \right)\left( 2^{332} + 1 \right)\left( 2^{664} - 2^{332} + 1 \right) \end{aligned}$$

Note that the six integers are greater than $2^{248}$.