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Question Number 190286 by Abdullahrussell last updated on 31/Mar/23

Answered by talminator2856792 last updated on 31/Mar/23

$$b^2-1=1001a$$$$(b-1)(b+1)=1001a$$$$\frac{(b-1)(b+1)}{1001}=a$$$$\frac{(b-1)(b+1)}{7\times 11\times 13}=a$$$$\mathrm{each}\mathrm{of}7,11\mathrm{and}13\mathrm{must}\mathrm{divide}\mathrm{either}b-1\mathrm{or}b+1$$$$\mathrm{now}\mathrm{the}\mathrm{necessary}\mathrm{combinations}\mathrm{of}\mathrm{multiplying}7,11,13\mathrm{must}\mathrm{be}\mathrm{checked}.$$$$\mathrm{we}\mathrm{have}\mathrm{six}\mathrm{cases}:$$$$$$$$\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{cases}:$$$$b+1\mathrm{is}\mathrm{divisible}\mathrm{by}7\times 11\mathrm{and}b-1\mathrm{divisible}\mathrm{by}13$$$$\mathrm{this}\mathrm{could}\mathrm{be}\mathrm{expressed}\mathrm{as}$$$$77x-2\equiv 0\left(\mathrm{mod}13\right)$$$$\Rightarrow 77x\equiv 2\left(\mathrm{mod}13\right)\mathrm{for}\mathrm{some}\mathrm{non}\mathrm{negative}x.$$$$77\equiv 12\left(\mathrm{mod}13\right)$$$$x=11$$$$b+1=847$$$$b=846$$$$a=715$$$$$$$$\mathrm{after}\mathrm{applying}\mathrm{this}\mathrm{approach}\mathrm{to}\mathrm{all}\mathrm{other}\mathrm{cases}$$$$\mathrm{the}\mathrm{following}\mathrm{case}\mathrm{yields}\mathrm{the}\mathrm{value}\mathrm{for}a\mathrm{that}\mathrm{satisfies}\mathrm{the}\mathrm{conditions}.$$$$$$$$b+1\mathrm{is}\mathrm{divisible}\mathrm{by}11\times 13\mathrm{and}b-1\mathrm{divisible}\mathrm{by}7$$$$143x-2\equiv 0\left(\mathrm{mod}7\right)$$$$\Rightarrow 143x\equiv 2\left(\mathrm{mod}7\right)$$$$143\equiv 3\left(\mathrm{mod}7\right)$$$$x=3$$$$b+1=429$$$$b=428$$$$$$$$a=183$$

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