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Question Number 112058 by mohssinee last updated on 05/Sep/20

	Answered by MJS_new last updated on 06/Sep/20
	$⇒ x=−((19±(√(4a^2 −407)))/2) 4a^2 −407=b^2 with b∈Z 4a^2 =b^2 +407 let 4a^2 =(b+n)^2 with n∈Z ⇒ (b+n)^2 −b^2 =407 ⇒ b=((407−n^2 )/(2n)) ∣((407−n^2 )/(2n))∣≥1 ⇒ −19≤n≤19 ⇒ n=±11∧b=±13 ⇒ a=±12∧x= { ((−3)),((−16)) :} n=±1∧b=±203 ⇒ a=±102∧x= { ((−111)),((92)) :} no other solutions$
	$\Rightarrow x = - \frac{19 \pm \sqrt{4 a^{2} - 407}}{2}$ $4 a^{2} - 407 = b^{2} with b \in Z$ $4 a^{2} = b^{2} + 407$ $let 4 a^{2} = {(b + n)}^{2} with n \in Z$ $\Rightarrow {(b + n)}^{2} - b^{2} = 407$ $\Rightarrow b = \frac{407 - n^{2}}{2 n}$ $∣ \frac{407 - n^{2}}{2 n} ∣⩾ 1 \Rightarrow - 19 ⩽ n ⩽ 19$ $\Rightarrow$ $n = \pm 11 \land b = \pm 13 \Rightarrow a = \pm 12 \land x = {\begin{cases} - 3 \\ - 16 \end{cases}$ $n = \pm 1 \land b = \pm 203 \Rightarrow a = \pm 102 \land x = {\begin{cases} - 111 \\ 92 \end{cases}$ $no other solutions$
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