what are the solutions of (√(3x^2 +1))=n where n∈N


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Question Number 73308 by wo1lxjwjdb last updated on 10/Nov/19

$w h a t a r e t h e s o l u t i o n s$ $o f \sqrt{3 x^{2} + 1} = n w h e r e n \in N$
Commented by MJS last updated on 10/Nov/19

$n_{k} = \frac{1}{2} ({(2 - \sqrt{3})}^{k} + {(2 + \sqrt{3})}^{k}) = \cosh (k \ln (2 - \sqrt{3}))$ $x_{k} = - \frac{\sqrt{3}}{3} \sinh (k \ln (2 - \sqrt{3}))$ $k, n_{k}, x_{k} \in N$
Commented by MJS last updated on 10/Nov/19

$you mean n \in N$
Commented by kaivan.ahmadi last updated on 10/Nov/19

$n \in N$ $3 x^{2} + 1 = n^{2} \Rightarrow 3 x^{2} = n^{2} - 1 \Rightarrow x^{2} = \frac{n^{2} - 1}{3} \Rightarrow$ $x = \sqrt{\frac{n^{2} - 1}{3}}$
Commented by wo1lxjwjdb last updated on 10/Nov/19

$w h e n i s i t a w h o l e n u m b e r ?$
Commented by MJS last updated on 10/Nov/19
$n∈N x∈? if x∈R x=±(√((n^2 −1)/3)); n>0 [for x∈C we also get n=0; x=((√3)/3)i] if x∈N same equation, solutions are: x∈{0, 1, 4, 15, 56, 209, 780, ...} n∈{1, 2, 7, 26, 97, 362, 1351, ...}$
$n \in N$ $x \in ?$ $if x \in R$ $x = \pm \sqrt{\frac{n^{2} - 1}{3}}; n > 0$ $[for x \in C we also get n = 0; x = \frac{\sqrt{3}}{3} i]$ $if x \in N$ $same equation, solutions are :$ $x \in {0, 1, 4, 15, 56, 209, 780, . . .}$ $n \in {1, 2, 7, 26, 97, 362, 1351, . . .}$
Commented by kaivan.ahmadi last updated on 10/Nov/19

$M r M J S a n s w e r e d t h i s q u o e s t i o n .$
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