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2bornot2b

  • 2 years ago

Prove that if an integer is a perfect square and a perfect cube, simultaneously, then it is either of the form 7k or 7k+1 (The problem belongs to the chapter, division algorithm)

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  1. 2bornot2b
    • 2 years ago
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    @Mani Jha @Zarkon @FoolForMath @JamesJ Please help!

  2. 2bornot2b
    • 2 years ago
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    @Ishaan94 can you help?

  3. FoolForMath
    • 2 years ago
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    Standard number theory problem. This should help : http://www.math.wisc.edu/~jensen/567/hwk1.pdf

  4. Mani_Jha
    • 2 years ago
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    If an integer is both a square and a cube, it can be of the form: \[(a ^{3})^{2}\] Now, since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath), we write \[a ^{3}=7k\] and get the no to be 49k^2, which is in the form of 7 times something \[49k ^{2}=7\times(7k ^{2})\] Now put \[a ^{3}=7k+-1\] Square it and you'll get a number in the form of (7times something +1) Do you want me to show the steps that I skipped?

  5. Mani_Jha
    • 2 years ago
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    Luis Rivera and I are the only people here who have usernames with a space. So, we can't be pinged :(

  6. 2bornot2b
    • 2 years ago
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    @FoolForMath can you provide me a similarn pdf containing problem and solution set for topics related to divisibility, prime numbers, and Greatest Common divisor?

  7. 2bornot2b
    • 2 years ago
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    I didn't get this line "If an integer is both a square and a cube, it can be of the form: (a^3)^2" written by @Mani Jha

  8. Mani_Jha
    • 2 years ago
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    Consider 64. It is the cube of 4 and square of 8. It can be written as: 8^2=(2^3)^2 I've assumed that this is valid for all numbers which are powers of 6(64=2^6) or of multiples of 6. \[a ^{6}=(a ^{3})^{2}\] Any other power like a^2 a^3 a^4 can't be both a square and cube of some integer. Clear?

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