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Question Number 141221 by Engr_Jidda last updated on 16/May/21

Commented by mr W last updated on 16/May/21

$n o s u c h k e x i s t s .$
Commented by Engr_Jidda last updated on 16/May/21

$h o w s i r ? ?$
	Answered by MJS_new last updated on 16/May/21
	$k^2 =x^2 +22 k=x+n∧n≥1 x^2 +2nx+n^2 =x^2 +22 2nx+n^2 =22 x=((22−n^2 )/(2n)) (1) x>0 ⇒ n<5 we only have to try n∈{1, 2, 3, 4} ⇒ ⇒ no integer x exists (2) if you don′t want to try: x∈N ⇒ 2∣(22−n^2 ) ⇒ n=2m x=((22−4m^2 )/(4m))=((11−2m^2 )/(2m)) x∈N ⇒ 2∣(11−2m^2 ) impossible$
	$k^{2} = x^{2} + 22$ $k = x + n \land n ⩾ 1$ $x^{2} + 2 n x + n^{2} = x^{2} + 22$ $2 n x + n^{2} = 22$ $x = \frac{22 - n^{2}}{2 n}$ $(1) x > 0 \Rightarrow n < 5 we only have to try n \in {1, 2, 3, 4} \Rightarrow$ $\Rightarrow no integer x exists$ $(2) if you {don}^{'} t want to try :$ $x \in N \Rightarrow 2 ∣ (22 - n^{2}) \Rightarrow n = 2 m$ $x = \frac{22 - 4 m^{2}}{4 m} = \frac{11 - 2 m^{2}}{2 m}$ $x \in N \Rightarrow 2 ∣ (11 - 2 m^{2}) impossible$
	Commented by Engr_Jidda last updated on 20/May/21

	$t h a n k y o u s i r$
	Answered by Rasheed.Sindhi last updated on 17/May/21
	$k^2 =x^2 +22 k^2 −x^2 =22 (k−x)(k+x)=1×22 ,2×11 { ((k−x=1 ∧ k+x=22)),((k−x=22 ∧ k+x=1)) :}⇒k=11.5∉N { ((k−x=2 ∧ k+x=11)),((k−x=11 ∧ k+x=2)) :}⇒k=6.5∉N No possible k$
	$k^{2} = x^{2} + 22$ $k^{2} - x^{2} = 22$ $(k - x) (k + x) = 1 \times 22, 2 \times 11$ ${\begin{cases} k - x = 1 \land k + x = 22 \\ k - x = 22 \land k + x = 1 \end{cases} \Rightarrow k = 11.5 \notin N$ ${\begin{cases} k - x = 2 \land k + x = 11 \\ k - x = 11 \land k + x = 2 \end{cases} \Rightarrow k = 6.5 \notin N$ $N o p o s s i b l e k$
	Commented by Engr_Jidda last updated on 20/May/21

	$t h a n k y o u s i r$
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