if positive integer x satisfies x^2 −4x+56 ≡14 (mod 17) , what is the minimum value of x.


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Question Number 110358 by bobhans last updated on 28/Aug/20

$i f p o s i t i v e i n t e g e r x s a t i s f i e s x^{2} - 4 x + 56 \equiv 14 (m o d 17)$ $, w h a t i s t h e m i n i m u m v a l u e o f x .$
	Answered by john santu last updated on 28/Aug/20
	$⇔x^2 −4x+4 + 52 = 14 (mod 17) ⇔ (x−2)^2 = −38 (mod 17 ) ⇔ (x−2)^2 = 13 (mod 17) now we need to see whether 13 is square modulo 17. ⇒ 13 = 13+ 3.17 = 64 (mod 17) then ⇔ (x−2)^2 =64 (mod 17) ⇔ (x−2)^2 = 8^2 (mod 17) { ((x−2=8 (mod 17))),((x−2=−8 (mod 17))) :} { ((x=10 (mod 17))),((x=11 (mod 17))) :} minimum value of x is 10$
	$\Leftrightarrow x^{2} - 4 x + 4 + 52 = 14 (m o d 17)$ $\Leftrightarrow {(x - 2)}^{2} = - 38 (m o d 17)$ $\Leftrightarrow {(x - 2)}^{2} = 13 (m o d 17)$ $n o w w e n e e d t o s e e w h e t h e r 13 i s$ $s q u a r e m o d u l o 17 .$ $\Rightarrow 13 = 13 + 3.17 = 64 (m o d 17)$ $t h e n \Leftrightarrow {(x - 2)}^{2} = 64 (m o d 17)$ $\Leftrightarrow {(x - 2)}^{2} = 8^{2} (m o d 17)$ ${\begin{cases} x - 2 = 8 (m o d 17) \\ x - 2 = - 8 (m o d 17) \end{cases}$ ${\begin{cases} x = 10 (m o d 17) \\ x = 11 (m o d 17) \end{cases}$ $m i n i m u m v a l u e o f x i s 10$
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