Question #147460
Find the sum of squares of all numbers x such that both expressions x^2+5x and x+(1/x) are integer numbers.
1
Expert's answer
2020-12-01T01:58:32-0500

let x+1x=k,kinteger numberx2+1x=kx2kx+1=0x=k±k242let x2+5x=nx2+5xn=0x=5±254n2k±k242=5±254n2k±k24=5±254nwhich equations have 3 integer solutionsk=5,n=1k=2,n=4k=2,n=6x=1,x=1,x=5±212S=12+(1)2+(5+212)2+(5212)2==1+1+(25+1021+21+25+1021+214)=25answer:25let \ x+\dfrac{1}{x} = k , k- integer\ number\\ \dfrac{x^2+1}{x} = k\\ x^2-kx+1 = 0\\ x = \dfrac{k \pm \sqrt{k^2-4}}{2}\\ let \ x^2+5x = n\\ x^2+5x-n = 0\\ x = \dfrac{-5 \pm \sqrt{25-4n}}{2}\\ \dfrac{k \pm \sqrt{k^2-4}}{2} = \dfrac{-5 \pm \sqrt{25-4n}}{2}\\ k \pm \sqrt{k^2-4} = -5 \pm \sqrt{25-4n}\\ \text{which equations have 3 integer solutions} \\ k = -5 , n = 1\\ k = -2, n = 4\\ k = 2, n = -6\\ x = 1, x=-1 , x = \dfrac{-5 \pm \sqrt{21}}{2}\\ S= 1^2+(-1)^2 + (\dfrac{-5 + \sqrt{21}}{2})^2 +(\dfrac{-5 - \sqrt{21}}{2})^2 = \\ = 1+1 +(\dfrac{25 + -10\sqrt{21} +21 +25+10\sqrt{21}+21}{4}) = 25 \\answer: 25


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