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Question Number 19332 by Tinkutara last updated on 09/Aug/17

$$\mathrm{Let}{S}_n=n^2+20n+12,n\mathrm{a}\mathrm{positive}$$$$\mathrm{integer}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{possible}$$$$\mathrm{values}\mathrm{of}n\mathrm{for}\mathrm{which}{S}_n\mathrm{is}\mathrm{a}\mathrm{perfect}$$$$\mathrm{square}?$$

Commented by mrW1 last updated on 09/Aug/17

$$\mathrm{\Sigma }\mathrm{n}=3+13=16$$

Answered by mrW1 last updated on 09/Aug/17

$$n^2+20n+12={\mathrm{m}}^2$$$$n^2+20n+12-{\mathrm{m}}^2=0$$$$\Rightarrow \mathrm{n}=\frac{-20\pm \sqrt{{20}^2-4(12-{\mathrm{m}}^2)}}{2}=-10\pm \sqrt{88+{\mathrm{m}}^2}$$$$88+{\mathrm{m}}^2={\mathrm{k}}^2$$$${\mathrm{k}}^2-{\mathrm{m}}^2=88$$$$(\mathrm{k}-\mathrm{m})(\mathrm{k}+\mathrm{m})=88=\mathrm{a}\times \mathrm{b}$$$$88=2^3\times 11$$$$\mathrm{a}\times \mathrm{b}=1\times 88=2\times 44=4\times 22=8\times 11$$$$\mathrm{k}-\mathrm{m}=\mathrm{a}$$$$\mathrm{k}+\mathrm{m}=\mathrm{b}$$$$\Rightarrow \mathrm{k}=\frac{\mathrm{a}+\mathrm{b}}{2}$$$$\mathrm{for}\mathrm{k}\mathrm{to}\mathrm{be}\mathrm{integer},\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{must}\mathrm{be}$$$$\mathrm{both}\mathrm{odd}\mathrm{or}\mathrm{both}\mathrm{even}.$$$$\Rightarrow \mathrm{k}=\frac{2+44}{2}=23$$$$\Rightarrow \mathrm{k}=\frac{4+22}{2}=13$$$$\Rightarrow \mathrm{n}=-10\pm 23=-33,13$$$$\Rightarrow \mathrm{n}=-10\pm 13=-23,3$$$$\mathrm{sum}\mathrm{of}+\mathrm{ve}\mathrm{n}=3+13=16$$

Commented by Tinkutara last updated on 09/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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