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Ask iit jee aieee pet cbse icse state board experts Expert Question: Prove that the integer next greater than (sqrt 7+sqrt3)^n is divisble by 2^2n
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mroobm (0)

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Prove that the integer next greater than (sqrt 7+sqrt3)^n is divisble by 2^2n
    
Avinash_Bhat (655)

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For n  =  1, the next integer greater than (  ) is 5, which is not divisible 22  =  4.


Some mistake in the question???

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chinnu (83)

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for n=1 it s 20
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1249111521 (22)

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for n=1


it is 4


ques is absol  ,.correct


doceed ti u nac ?
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krishna.gopal (2600)

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Avinash is right. sqrt(7)+sqrt(3) is 4.3 something so next integer is 5 which is not dvisible by 4. Please check

Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi
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mroobm (0)

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In this question there is one hint given and that is:


first prove that (sqrt7+sqrt3)^n + (sqrt7-sqrt3)^n is an integer


then prove that he above integer is divisible by 2^2n


as,0<sqrt7-sqrt3<1


but i'm not able to understand that y sqrt7-sqrt3 since that is not the integer next greater than p(n)

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