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Question Number 196637 by sonukgindia last updated on 28/Aug/23

	Answered by deleteduser1 last updated on 28/Aug/23
	$a=(√(12−(√m)));b=(√(12+(√m))),a^2 +b^2 =24 ab=(√(144−m));(a+b)^2 =24+2(√(144−m)) ⇒a+b=(√(24+2(√(144−m))))=n ⇒p^2 =144−m≥0⇒n=(√(24+2p))⇒24+2p=x^2 p∈{1,2,..,n,..11}⇒p=6⇒n=6⇒m=108 ⇒m+n=114$
	$a = \sqrt{12 - \sqrt{m}}; b = \sqrt{12 + \sqrt{m}}, a^{2} + b^{2} = 24$ $a b = \sqrt{144 - m}; {(a + b)}^{2} = 24 + 2 \sqrt{144 - m}$ $\Rightarrow a + b = \sqrt{24 + 2 \sqrt{144 - m}} = n$ $\Rightarrow p^{2} = 144 - m ⩾ 0 \Rightarrow n = \sqrt{24 + 2 p} \Rightarrow 24 + 2 p = x^{2}$ $p \in {1, 2, . ., n, . . 11} \Rightarrow p = 6 \Rightarrow n = 6 \Rightarrow m = 108$ $\Rightarrow m + n = 114$
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