Find all positive integers n such that n^2 +7n+6 is perfect square.


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Question Number 216783 by ArshadS last updated on 20/Feb/25

$F i n d a l l p o s i t i v e i n t e g e r s n s u c h t h a t$ $n^{2} + 7 n + 6 i s p e r f e c t s q u a r e .$
	Answered by mehdee7396 last updated on 20/Feb/25

	$n^{2} + 7 n + 6 = k^{2}$ $n^{2} + 7 n + 6 - k^{2} = 0$ $\Rightarrow n = \frac{- 7 + \sqrt{4 k^{2} + 25}}{2}$ $4 k^{2} + 25 = m^{2} \Rightarrow (m + 2 k) (m - 2 k) = 25$ $1) m + 2 k = 25 & m - 2 k = 1$ $\Rightarrow m = 13 & k = 6 \Rightarrow n = 3 ✓$ $2) m + 2 k = 5 & m - 2 k = 5$ $\Rightarrow m = 5 & k = 0 \Rightarrow n = - 1 \times$
	Commented by ArshadS last updated on 21/Feb/25

	$T hanks sir!$
	Answered by Hanuda354 last updated on 21/Feb/25

	$n^{2} + 7 n + 6 = p^{2}$ $n^{2} + 7 n + 6 - p^{2} = 0, n > 0$ $n = \frac{- 7 + \sqrt{49 - 4 (6 - p^{2})}}{2} = \frac{- 7 + \sqrt{4 p^{2} + 25}}{2}$ $Let q^{2} = 4 p^{2} + 25 \Leftrightarrow (q + 2 p) (q - 2 p) = 25$ $Then we have n = \frac{- 7 + q}{2}$ $where q = 2 n + 7, n \in N$ $Since n > 0, giving us 3 possibilities :$ $(1) q + 2 p = 25 & q - 2 p = 1 \Rightarrow q = 13$ $(2) q + 2 p = 5 & q - 2 p = 5 \Rightarrow q = 5$ $(3) q + 2 p = 1 & q - 2 p = 25 \Rightarrow q = 13$ $For q = 13 \Rightarrow n = 3 (a c c e p t e d)$ $For q = 5 \Rightarrow n = - 1 (r e j e c t e d)$
	Commented by ArshadS last updated on 21/Feb/25

	$T h a n k s s i r!$
	Answered by Rasheed.Sindhi last updated on 21/Feb/25
	$let n^2 +7n+6=m^2 n^2 +7n=m^2 −6 n^2 +7n+((7/2))^2 =m^2 −6+((7/2))^2 (n+(7/2))^2 =m^2 +((−24+49)/4) Multiplying by 4 to get rid of denominator (2n+7)^2 =4m^2 +25 (2n+7)^2 −4m^2 =25 (2n−2m+7)(2n+2m+7)=25 25=1×25 or 5×5 possibilities: { (((2n−2m+7)=1 &(2n+2m+7)=25...(1))),(((2n−2m+7)=25 &(2n+2m+7)=1...(2))),(((2n−2m+7)=5 &(2n+2m+7)=5...(3))) :} { (((1),(2)⇒ 4n+14=26⇒n=3 ✓)),(((3)⇒ 4n+14=10⇒n=−1∉Z^+ ×)) :}$
	$l e t n^{2} + 7 n + 6 = m^{2}$ $n^{2} + 7 n = m^{2} - 6$ $n^{2} + 7 n + {(\frac{7}{2})}^{2} = m^{2} - 6 + {(\frac{7}{2})}^{2}$ ${(n + \frac{7}{2})}^{2} = m^{2} + \frac{- 24 + 49}{4}$ $M u l t i p l y i n g b y 4 t o g e t r i d o f d e n o m i n a t o r$ ${(2 n + 7)}^{2} = 4 m^{2} + 25$ ${(2 n + 7)}^{2} - 4 m^{2} = 25$ $(2 n - 2 m + 7) (2 n + 2 m + 7) = 25$ $25 = 1 \times 25 o r 5 \times 5$ $p o s s i b i l i t i e s :$ ${\begin{cases} (2 n - 2 m + 7) = 1 & (2 n + 2 m + 7) = 25 . . . (1) \\ (2 n - 2 m + 7) = 25 & (2 n + 2 m + 7) = 1 . . . (2) \\ (2 n - 2 m + 7) = 5 & (2 n + 2 m + 7) = 5 . . . (3) \end{cases}$ ${\begin{cases} (1), (2) \Rightarrow 4 n + 14 = 26 \Rightarrow n = 3 ✓ \\ (3) \Rightarrow 4 n + 14 = 10 \Rightarrow n = - 1 \notin Z^{+} \times \end{cases}$
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