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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 (((b − 1)(b + 1))/(1001)) = a (((b − 1)(b + 1))/(7 × 11 × 13 )) = a each of 7, 11 and 13 must divide either b − 1 or b + 1 now the necessary combinations of multiplying 7, 11, 13 must be checked. we have six cases: one of the cases: b + 1 is divisible by 7 × 11 and b − 1 divisible by 13 this could be expressed as 77x −2 ≡ 0 (mod 13) ⇒ 77x ≡ 2 (mod 13) for some non negative x. 77 ≡ 12 (mod 13) x = 11 b + 1 = 847 b = 846 a = 715 after applying this approach to all other cases the following case yields the value for a that satisfies the conditions. b + 1 is divisible by 11 × 13 and b − 1 divisible by 7 143x − 2 ≡ 0 (mod 7) ⇒ 143x ≡ 2 (mod 7) 143 ≡ 3 (mod 7) x = 3 b + 1 = 429 b = 428 a = 183

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

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