Be p a prime number , arbitrary. Solve on positive integers (x;y;z) { ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :}


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 160711 by HongKing last updated on 05/Dec/21
$Be p a prime number , arbitrary. Solve on positive integers (x;y;z) { ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :}$
$Be p a prime number, arbitrary .$ $Solve on positive integers (x; y; z)$ ${\begin{cases} xy + z^{2} = 3 p + 4 \\ x + {yz}^{2} = p + 4 \end{cases}$
	Answered by mr W last updated on 06/Dec/21
	$(i)−(ii): x(y−1)+z^2 (1−y)=2p (x−z^2 )(y−1)=2p= { ((1×2p)),((2×p)) :} y−1=1, x−z^2 =2p y=2, x=z^2 +2p z^2 +2p+2z^2 =p+4 3z^2 =4−p ⇒not possible, since LHS=odd ≥3, RHS=even≤2 y−1=2p, x−z^2 =1 y=2p+1, x=z^2 +1 z^2 +1+(2p+1)z^2 =p+4 2(p+1)z^2 =p+3 (p+1)(2z^2 −1)=2 ⇒not possible, since LHS≥3 y−1=2, x−z^2 =p y=3, x=z^2 +p z^2 +p+3z^2 =p+4 z^2 =1 ⇒z=1, x=1+p ✓ y−1=p, x−z^2 =2 y=p+1, x=z^2 +2 z^2 +2+(p+1)z^2 =p+4 z^2 =1 ⇒z=1, x=3, y=p+1 ✓ summary of solution: x=3, y=p+1, z=1 x=p+1, y=3, z=1$
	$(i) - (i i) :$ $x (y - 1) + z^{2} (1 - y) = 2 p$ $(x - z^{2}) (y - 1) = 2 p = {\begin{cases} 1 \times 2 p \\ 2 \times p \end{cases}$ $y - 1 = 1, x - z^{2} = 2 p$ $y = 2, x = z^{2} + 2 p$ $z^{2} + 2 p + 2 z^{2} = p + 4$ $3 z^{2} = 4 - p \Rightarrow n o t p o s s i b l e,$ $s i n c e L H S = o d d ⩾ 3, R H S = e v e n ⩽ 2$ $y - 1 = 2 p, x - z^{2} = 1$ $y = 2 p + 1, x = z^{2} + 1$ $z^{2} + 1 + (2 p + 1) z^{2} = p + 4$ $2 (p + 1) z^{2} = p + 3$ $(p + 1) (2 z^{2} - 1) = 2 \Rightarrow n o t p o s s i b l e,$ $s i n c e L H S ⩾ 3$ $y - 1 = 2, x - z^{2} = p$ $y = 3, x = z^{2} + p$ $z^{2} + p + 3 z^{2} = p + 4$ $z^{2} = 1$ $\Rightarrow z = 1, x = 1 + p ✓$ $y - 1 = p, x - z^{2} = 2$ $y = p + 1, x = z^{2} + 2$ $z^{2} + 2 + (p + 1) z^{2} = p + 4$ $z^{2} = 1$ $\Rightarrow z = 1, x = 3, y = p + 1 ✓$ $s u m m a r y o f s o l u t i o n :$ $x = 3, y = p + 1, z = 1$ $x = p + 1, y = 3, z = 1$
	Commented by HongKing last updated on 05/Dec/21

	$very nice thank you dear Sir$
	Answered by Rasheed.Sindhi last updated on 06/Dec/21
	${ ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :} { ((x=((3p + 4−z^2 )/y))),((x= p + 4−yz^2 )) :} ⇒3p + 4−z^2 =py + 4y−y^2 z^2 y^2 z^2 −z^2 +3p−py+4−4y=0 z^2 (y−1)(y+1)−4(y−1)=p(y−3) determinant ((((y−1)( z^2 (y+1)−4 )=p(y−3)))) •∴ y−1=p ∧ z^2 (y+1)−4=y−3 y=p+1 ∧ z^2 (p+1+1)−4=p+1−3 z^2 =((p+2)/(p+2))=1⇒z=1 x=p+4−(p+1)(1)=3 (x,y,z)=(3,p+1,1) • (y−1)∣(y−3)⇒((y−3)/(y−1))∈{0,1,2,...} y−1=±1 or y−3=0 y=2,0,3 y=2 ⇒ z^2 (2+1)−4=p(2−3) z^2 =((−p+4)/3)⇒z=(√((4−p)/3)) ⇒p=1∉P y≠0 ∵ x=((3p + 4−z^2 )/y) y=3 ⇒ (3−1)( z^2 (3+1)−4 )=p(3−3) z^2 −1=0⇒z=1 x= p + 4−yz^2 =p+4−(3)(1)^2 x=p+1 (x,y,z)=(p+1,3,1)$
	${\begin{cases} xy + z^{2} = 3 p + 4 \\ x + {yz}^{2} = p + 4 \end{cases}$ ${\begin{cases} x = \frac{3 p + 4 - z^{2}}{y} \\ x = p + 4 - {yz}^{2} \end{cases} \Rightarrow 3 p + 4 - z^{2} = py + 4 y - y^{2} z^{2}$ $y^{2} z^{2} - z^{2} + 3 p - py + 4 - 4 y = 0$ $z^{2} (y - 1) (y + 1) - 4 (y - 1) = p (y - 3)$ $\begin{matrix} (y - 1) (z^{2} (y + 1) - 4) = p (y - 3) \end{matrix}$ $∙ ∴ y - 1 = p \land z^{2} (y + 1) - 4 = y - 3$ $\underset{―}{y = p + 1} \land z^{2} (p + 1 + 1) - 4 = p + 1 - 3$ $z^{2} = \frac{p + 2}{p + 2} = 1 \Rightarrow z = 1$ $x = p + 4 - (p + 1) (1) = 3$ $(x, y, z) = (3, p + 1, 1)$ $∙ (y - 1) ∣ (y - 3) \Rightarrow \frac{y - 3}{y - 1} \in {0, 1, 2, . . .}$ $y - 1 = \pm 1 or y - 3 = 0$ $y = 2, 0, 3$ $y = 2 \Rightarrow z^{2} (2 + 1) - 4 = p (2 - 3)$ $z^{2} = \frac{- p + 4}{3} \Rightarrow z = \sqrt{\frac{4 - p}{3}} \Rightarrow p = 1 \notin P$ $y \neq 0 ∵ x = \frac{3 p + 4 - z^{2}}{y}$ $\underset{―}{y = 3} \Rightarrow (3 - 1) (z^{2} (3 + 1) - 4) = p (3 - 3)$ $z^{2} - 1 = 0 \Rightarrow z = 1$ $x = p + 4 - {yz}^{2} = p + 4 - (3) {(1)}^{2}$ $x = p + 1$ $(x, y, z) = (p + 1, 3, 1)$
	Commented by mr W last updated on 06/Dec/21

	$j u s t f o r d i s c u s s i o n s i r :$ $i n$ $\begin{matrix} (y - 1) (z^{2} (y + 1) - 4) = p (y - 3) \end{matrix}$ $w e a r e n o t s u r e i f y - 3 i s p r i m e . i f$ $i t i s n o t p r i m e, i t c a n c o n t a i n m o r e$ $f a c t o r s, t h e n t h e r e a r e m u c h m o r e$ $p o s s i b i l i t i e s .$
	Commented by Rasheed.Sindhi last updated on 06/Dec/21

	$Y e s s i r I r e a l i z e d i t . N o w I^{'} m g o i n g$ $t o c h a n g e m y a n s w e r i n t h i s l i g h t .$ $T h a n k s t o g u i d e m e s i r! P l e a s e s e e$ $m y a n s w e r o n c e m o r e .$
	Commented by mr W last updated on 08/Dec/21

	$v e r y g o o d s i r!$
	Commented by Rasheed.Sindhi last updated on 09/Dec/21

	$G r a t e f u l s i r!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum