Menu Close

Let-S-n-n-2-20n-12-n-a-positive-integer-What-is-the-sum-of-all-possible-values-of-n-for-which-S-n-is-a-perfect-square-

Tinku Tara 0 Comments
Pin




Question Number 19332 by Tinkutara last updated on 09/Aug/17
Let S_n = n^2 + 20n + 12, n a positive integer. What is the sum of all possible values of n for which S_n is a perfect square?
$$\mathrm{Let}\:{S}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{20}{n}\:+\:\mathrm{12},\:{n}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{values}\:\mathrm{of}\:{n}\:\mathrm{for}\:\mathrm{which}\:{S}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect} \\ $$$$\mathrm{square}? \\ $$
Commented by mrW1 last updated on 09/Aug/17
Σn=3+13=16
$$\Sigma\mathrm{n}=\mathrm{3}+\mathrm{13}=\mathrm{16} \\ $$
Answered by mrW1 last updated on 09/Aug/17
n^2 + 20n + 12=m^2 n^2 + 20n + 12−m^2 =0 ⇒n=((−20±(√(20^2 −4(12−m^2 ))))/2)=−10±(√(88+m^2 )) 88+m^2 =k^2 k^2 −m^2 =88 (k−m)(k+m)=88=a×b 88=2^3 ×11 a×b=1×88=2×44=4×22=8×11 k−m=a k+m=b ⇒k=((a+b)/2) for k to be integer, a and b must be both odd or both even. ⇒k=((2+44)/2)=23 ⇒k=((4+22)/2)=13 ⇒n=−10±23=−33, 13 ⇒n=−10±13=−23, 3 sum of +ve n=3+13=16
$${n}^{\mathrm{2}} \:+\:\mathrm{20}{n}\:+\:\mathrm{12}=\mathrm{m}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} \:+\:\mathrm{20}{n}\:+\:\mathrm{12}−\mathrm{m}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{n}=\frac{−\mathrm{20}\pm\sqrt{\mathrm{20}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{12}−\mathrm{m}^{\mathrm{2}} \right)}}{\mathrm{2}}=−\mathrm{10}\pm\sqrt{\mathrm{88}+\mathrm{m}^{\mathrm{2}} } \\ $$$$\mathrm{88}+\mathrm{m}^{\mathrm{2}} =\mathrm{k}^{\mathrm{2}} \\ $$$$\mathrm{k}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} =\mathrm{88} \\ $$$$\left(\mathrm{k}−\mathrm{m}\right)\left(\mathrm{k}+\mathrm{m}\right)=\mathrm{88}=\mathrm{a}×\mathrm{b} \\ $$$$\mathrm{88}=\mathrm{2}^{\mathrm{3}} ×\mathrm{11} \\ $$$$\mathrm{a}×\mathrm{b}=\mathrm{1}×\mathrm{88}=\mathrm{2}×\mathrm{44}=\mathrm{4}×\mathrm{22}=\mathrm{8}×\mathrm{11} \\ $$$$\mathrm{k}−\mathrm{m}=\mathrm{a} \\ $$$$\mathrm{k}+\mathrm{m}=\mathrm{b} \\ $$$$\Rightarrow\mathrm{k}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{k}\:\mathrm{to}\:\mathrm{be}\:\mathrm{integer},\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{both}\:\mathrm{odd}\:\mathrm{or}\:\mathrm{both}\:\mathrm{even}. \\ $$$$\Rightarrow\mathrm{k}=\frac{\mathrm{2}+\mathrm{44}}{\mathrm{2}}=\mathrm{23} \\ $$$$\Rightarrow\mathrm{k}=\frac{\mathrm{4}+\mathrm{22}}{\mathrm{2}}=\mathrm{13} \\ $$$$\Rightarrow\mathrm{n}=−\mathrm{10}\pm\mathrm{23}=−\mathrm{33},\:\mathrm{13} \\ $$$$\Rightarrow\mathrm{n}=−\mathrm{10}\pm\mathrm{13}=−\mathrm{23},\:\mathrm{3} \\ $$$$\mathrm{sum}\:\mathrm{of}\:+\mathrm{ve}\:\mathrm{n}=\mathrm{3}+\mathrm{13}=\mathrm{16} \\ $$
Commented by Tinkutara last updated on 09/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *