a-2-1-b-2-b-2-c-2-b-4-ab-c-solve-for-a-b-c-

Question Number 55887 by behi83417@gmail.com last updated on 05/Mar/19

a^{2} + 1 = b^{2}

b^{2} + c^{2} = b^{4}

a b = c

solve for : a, b, c .

Answered by MJS last updated on 05/Mar/19

b^{2} = a^{2} + 1 \Rightarrow {(a^{2} + 1)}^{2} - (a^{2} + 1) - c^{2} = 0

a^{4} + a^{2} - c^{2} = 0

c = a b \Rightarrow a^{4} + a^{2} - a^{2} (a^{2} + 1) = 0 true for a \in R

\Rightarrow a \in R \land b = \pm \sqrt{a^{2} + 1} \land c = \pm a \sqrt{a^{2} + 1}

Answered by JDamian last updated on 06/Mar/19

As the 2 nd . and the 3 rd . expressions

combined are the first; the only one to solve is :

a^{2} + 1 = b^{2} \Rightarrow b = \pm \sqrt{a^{2} + 1} \Rightarrow c = \pm a \sqrt{a^{2} + 1}

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