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

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

$${Find}\:{all}\:{positive}\:{integers}\:{n}\:{such}\:{that} \\ $$$${n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}\:{is}\:{perfect}\:{square}. \\ $$

Answered by mehdee7396 last updated on 20/Feb/25

n^2 +7n+6=k^2 n^2 +7n+6−k^2 =0 ⇒n=((−7+(√(4k^2 +25)))/2) 4k^2 +25=m^2 ⇒(m+2k)(m−2k)=25 1)m+2k=25 & m−2k=1 ⇒m=13 & k=6⇒n=3 ✓ 2)m+2k=5 & m−2k=5 ⇒m=5 & k=0⇒n=−1 ×

$${n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}={k}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}−{k}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{n}=\frac{−\mathrm{7}+\sqrt{\mathrm{4}{k}^{\mathrm{2}} +\mathrm{25}}}{\mathrm{2}} \\ $$$$\mathrm{4}{k}^{\mathrm{2}} +\mathrm{25}={m}^{\mathrm{2}} \Rightarrow\left({m}+\mathrm{2}{k}\right)\left({m}−\mathrm{2}{k}\right)=\mathrm{25} \\ $$$$\left.\mathrm{1}\right){m}+\mathrm{2}{k}=\mathrm{25}\:\&\:\:{m}−\mathrm{2}{k}=\mathrm{1} \\ $$$$\Rightarrow{m}=\mathrm{13}\:\&\:{k}=\mathrm{6}\Rightarrow{n}=\mathrm{3}\:\:\checkmark \\ $$$$\left.\mathrm{2}\right){m}+\mathrm{2}{k}=\mathrm{5}\:\:\&\:\:{m}−\mathrm{2}{k}=\mathrm{5} \\ $$$$\Rightarrow{m}=\mathrm{5}\:\:\&\:\:{k}=\mathrm{0}\Rightarrow{n}=−\mathrm{1}\:× \\ $$$$ \\ $$

Commented by ArshadS last updated on 21/Feb/25

Thanks sir!

$$\mathbb{T}\mathrm{hanks}\:\mathrm{sir}! \\ $$

Answered by Hanuda354 last updated on 21/Feb/25

n^2 +7n+6 = p^2 n^2 +7n+6−p^2 = 0 , n > 0 n = ((−7+(√(49−4(6−p^2 ))))/2) = ((−7+(√(4p^2 +25)))/2) Let q^2 = 4p^2 + 25 ⇔ (q+2p)(q−2p) = 25 Then we have n = ((−7+q)/2) where q = 2n+7 , n ∈ N Since n > 0 , giving us 3 possibilities: (1) q+2p = 25 & q−2p = 1 ⇒ q = 13 (2) q+2p = 5 & q−2p = 5 ⇒ q = 5 (3) q+2p = 1 & q−2p = 25 ⇒ q = 13 For q = 13 ⇒ n = 3 (accepted) For q = 5 ⇒ n = −1 (rejected)

$${n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}\:=\:{p}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}−{p}^{\mathrm{2}} \:=\:\mathrm{0}\:,\:{n}\:>\:\mathrm{0} \\ $$$$ \\ $$$${n}\:=\:\frac{−\mathrm{7}+\sqrt{\mathrm{49}−\mathrm{4}\left(\mathrm{6}−{p}^{\mathrm{2}} \right)}}{\mathrm{2}}\:=\:\frac{−\mathrm{7}+\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{25}}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Let}\:\:{q}^{\mathrm{2}} \:=\:\mathrm{4}{p}^{\mathrm{2}} \:+\:\mathrm{25}\:\Leftrightarrow\:\left({q}+\mathrm{2}{p}\right)\left({q}−\mathrm{2}{p}\right)\:=\:\mathrm{25} \\ $$$$\mathrm{Then}\:\mathrm{we}\:\mathrm{have}\:\:{n}\:=\:\frac{−\mathrm{7}+{q}}{\mathrm{2}} \\ $$$$\mathrm{where}\:\:{q}\:=\:\mathrm{2}{n}+\mathrm{7}\:,\:\:{n}\:\in\:\mathrm{N} \\ $$$$ \\ $$$$\mathrm{Since}\:{n}\:>\:\mathrm{0}\:,\:\mathrm{giving}\:\:\mathrm{us}\:\:\mathrm{3}\:\mathrm{possibilities}: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{q}+\mathrm{2}{p}\:=\:\mathrm{25}\:\&\:{q}−\mathrm{2}{p}\:=\:\mathrm{1}\:\Rightarrow\:{q}\:=\:\mathrm{13} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{q}+\mathrm{2}{p}\:=\:\mathrm{5}\:\&\:{q}−\mathrm{2}{p}\:=\:\mathrm{5}\:\Rightarrow\:{q}\:=\:\mathrm{5} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\:{q}+\mathrm{2}{p}\:=\:\mathrm{1}\:\&\:{q}−\mathrm{2}{p}\:=\:\mathrm{25}\:\Rightarrow\:{q}\:=\:\mathrm{13} \\ $$$$ \\ $$$$\mathrm{For}\:\:{q}\:=\:\mathrm{13}\:\:\:\Rightarrow\:{n}\:=\:\mathrm{3}\:\left({accepted}\right) \\ $$$$\mathrm{For}\:\:{q}\:=\:\mathrm{5}\:\:\:\:\:\:\Rightarrow\:{n}\:=\:−\mathrm{1}\:\left({rejected}\right) \\ $$

Commented by ArshadS last updated on 21/Feb/25

Thanks sir!

$$\mathcal{T}{hanks}\:{sir}! \\ $$

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^+ ×)) :}

$${let}\:{n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}={m}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{n}^{\mathrm{2}} +\mathrm{7}{n}={m}^{\mathrm{2}} −\mathrm{6} \\ $$$$\:\:\:\:\:\:\:{n}^{\mathrm{2}} +\mathrm{7}{n}+\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} −\mathrm{6}+\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\left({n}+\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} +\frac{−\mathrm{24}+\mathrm{49}}{\mathrm{4}} \\ $$$${Multiplying}\:{by}\:\mathrm{4}\:{to}\:{get}\:{rid}\:{of}\:{denominator} \\ $$$$\left(\mathrm{2}{n}+\mathrm{7}\right)^{\mathrm{2}} =\mathrm{4}{m}^{\mathrm{2}} +\mathrm{25} \\ $$$$\left(\mathrm{2}{n}+\mathrm{7}\right)^{\mathrm{2}} −\mathrm{4}{m}^{\mathrm{2}} =\mathrm{25} \\ $$$$\left(\mathrm{2}{n}−\mathrm{2}{m}+\mathrm{7}\right)\left(\mathrm{2}{n}+\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{25} \\ $$$$\mathrm{25}=\mathrm{1}×\mathrm{25}\:{or}\:\mathrm{5}×\mathrm{5} \\ $$$${possibilities}: \\ $$$$\begin{cases}{\left(\mathrm{2}{n}−\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{1}\:\&\left(\mathrm{2}{n}+\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{25}...\left(\mathrm{1}\right)}\\{\left(\mathrm{2}{n}−\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{25}\:\&\left(\mathrm{2}{n}+\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{1}...\left(\mathrm{2}\right)}\\{\left(\mathrm{2}{n}−\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{5}\:\&\left(\mathrm{2}{n}+\mathrm{2}{m}+\mathrm{7}\right)=\mathrm{5}...\left(\mathrm{3}\right)}\end{cases} \\ $$$$\begin{cases}{\left(\mathrm{1}\right),\left(\mathrm{2}\right)\Rightarrow\:\:\:\:\:\mathrm{4}{n}+\mathrm{14}=\mathrm{26}\Rightarrow{n}=\mathrm{3}\:\checkmark}\\{\left(\mathrm{3}\right)\Rightarrow\:\:\:\:\:\mathrm{4}{n}+\mathrm{14}=\mathrm{10}\Rightarrow{n}=−\mathrm{1}\notin\mathbb{Z}^{+} \:×}\end{cases} \\ $$

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