Let k2=n2+19n+92 On multiplying both sides by 4 , we get 4n2+76n+368=4k2 ⇒(2n)2+2×19×2n+361+7=4k2 ⇒(2n+19)2=(4k2−7)...(i) Let (2n+19)=m Then, 4k2−7=m2 4k2−m2=7 (2k−m)(2k+m)=7 [∵a2−b2=(a+b)(a−b)] Factor of 7 are 7 and 1 So, 2k+m=7 2k−m=1 By solving this, we get &k=2 Now, putting k=2 in Eq. (i), we get (2n+19)2=(4(2)2−7)=16−7 ⇒(2n+19)2=9 ⇒2n+19=±3 ⇒2n+19=+3⇒n=−8 ⇒2n+19=−3⇒n=−11 Sum of integer value of n=−8+(−11) =−19