Q#168480 reposted. n^2 +n+109=x^2 x∈Z, n(∈Z^+ )=?


Question and Answers Forum

All Questions Topic List
Number Theory Questions
Previous in All Question Next in All Question
Previous in Number Theory Next in Number Theory
Question Number 168823 by Rasheed.Sindhi last updated on 18/Apr/22

$You can't use 'macro parameter character #' in math mode$ $n^{2} + n + 109 = x^{2}$ $x \in Z, n (\in Z^{+}) = ?$
	Answered by Rasheed.Sindhi last updated on 18/Apr/22
	$n^2 +n+109=x^2 x∈Z, n(∈Z^+ )=? There′s no harm to assume x as non-negative because x^2 is anyway non-negative. n^2 +n+109−x^2 =0 n=((−1±(√(1−4(109−x^2 ))))/2) n=((−1±(√(4x^2 −435)))/2) ∵ n>0 ∴ n=((−1+(√(4x^2 −435)))/2) ∵n∈Z^+ ∴4x^2 −435 { ((>1)),(( perfect square)) :} x^2 >((436)/4)⇒x^2 ≥109⇒x≥11 determinant ((x,((√(4x^2 −435)) ),(n=((−1+(√(4x^2 −435)))/2))),((11),7,3),((23),(41),(20)),((37),(71),(35)),((109),(217),(108)),((...),(...),(...)))$
	$n^{2} + n + 109 = x^{2}$ $x \in Z, n (\in Z^{+}) = ?$ ${There}^{'} s no harm to assume x as$ $non - negative because x^{2} is anyway$ $non - negative .$ $n^{2} + n + 109 - x^{2} = 0$ $n = \frac{- 1 \pm \sqrt{1 - 4 (109 - x^{2})}}{2}$ $n = \frac{- 1 \pm \sqrt{4 x^{2} - 435}}{2}$ $∵ n > 0$ $∴ n = \frac{- 1 + \sqrt{4 x^{2} - 435}}{2}$ $∵ n \in Z^{+}$ $∴ 4 x^{2} - 435 {\begin{cases} > 1 \\ perfect square \end{cases}$ $x^{2} > \frac{436}{4} \Rightarrow x^{2} ⩾ 109 \Rightarrow x ⩾ 11$ $\begin{array}{cccccc} x & \sqrt{{4 x}^{2} - 435} & n = \frac{- 1 + \sqrt{4 x^{2} - 435}}{2} \\ 11 & 7 & 3 \\ 23 & 41 & 20 \\ 37 & 71 & 35 \\ 109 & 217 & 108 \\ . . . & . . . & . . . \end{array}$
	Answered by mindispower last updated on 18/Apr/22
	$⇔4n^2 +4n+436=(2x)^2 ⇔(2n+1)^2 +435=(2x)^2 ⇔(2x−2n−1)(2x+2n+1)=435 Solve for x∈Z_+ 2x+2n+1>2x−2n−1,x>n 435=5.3.29 ⇔ { ((2x+2n+1=435)),((2x−2n−1=1)) :} x=109,n=108 { ((2x+2n+1=87)),((2x−2n−1=5)) :} x=23 n=20 { ((2x+2n+1=145)),((2x−2n−1=3)) :} ⇒x=37,n=35 { ((2x+2n+1=29)),((2x−2n−1=15)) :} x=11,n=3 S=(x,n)∈{(11,3);(23,20);(37,35);(109,108);(−11,3), (−23,20);(−37,35);(−109,108)}$
	$\Leftrightarrow 4 n^{2} + 4 n + 436 = {(2 x)}^{2}$ $\Leftrightarrow {(2 n + 1)}^{2} + 435 = {(2 x)}^{2}$ $\Leftrightarrow (2 x - 2 n - 1) (2 x + 2 n + 1) = 435$ $S o l v e f o r x \in Z_{+}$ $2 x + 2 n + 1 > 2 x - 2 n - 1, x > n$ $435 = 5.3.29$ $\Leftrightarrow {\begin{cases} 2 x + 2 n + 1 = 435 \\ 2 x - 2 n - 1 = 1 \end{cases}$ $x = 109, n = 108$ ${\begin{cases} 2 x + 2 n + 1 = 87 \\ 2 x - 2 n - 1 = 5 \end{cases}$ $x = 23$ $n = 20$ ${\begin{cases} 2 x + 2 n + 1 = 145 \\ 2 x - 2 n - 1 = 3 \end{cases}$ $\Rightarrow x = 37, n = 35$ ${\begin{cases} 2 x + 2 n + 1 = 29 \\ 2 x - 2 n - 1 = 15 \end{cases}$ $x = 11, n = 3$ $S = (x, n) \in {(11, 3); (23, 20); (37, 35); (109, 108); (- 11, 3),$ $(- 23, 20); (- 37, 35); (- 109, 108)}$
	Commented by Rasheed.Sindhi last updated on 18/Apr/22

	$\lor \cap i \subset\in solution!$
	Commented by Rasheed.Sindhi last updated on 18/Apr/22

	$T han k s sir!$
	Commented by mindispower last updated on 19/Apr/22

	$w i t h e P l e a s u r H a v e a n i c e D a y$
	Commented by Rasheed.Sindhi last updated on 19/Apr/22
	,��
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum