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Thread: Find a whole number that is both perfect square and cube found between 2000 and 10000

  1. #1

    Find a whole number that is both perfect square and cube found between 2000 and 10000

    "Find a whole number that is both perfect square and cube found between 2000 and 10000?"

    I am kind of stumped on this question. If anyone can help me that would great.

  2. #2
    Elite Member
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    Quote Originally Posted by Sparticusretiarius View Post
    "Find a whole number that is both perfect square and cube found between 2000 and 10000?"

    I am kind of stumped on this question. If anyone can help me that would great.
    Brute Force!!

    How many integer square numbers are there between 2000 ↔ 10000 (452 ↔ 1002) - list those

    How many integer cubic numbers are there between 2000 ↔ 10000 (133 ↔ 213) - list those

    Hint: How many "square numbers" do you have within 13 ↔ 21?
    Last edited by Subhotosh Khan; 02-25-2018 at 08:14 PM.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Quote Originally Posted by Sparticusretiarius View Post
    "Find a whole number that is both perfect square and cube found between 2000 and 10000?"

    "Brute force" was stated, but it is not required.

    Because it is a perfect square, it is of the form [tex] \ N^2[/tex].

    Because it is also a perfect cube, it is also of the form [tex] \ M^3 [/tex].

    [tex] x^6 \ = \ (x^3)^2 \ = \ (x^2)^3 \ [/tex] satisfies both. (The exponent on x is the least common multiple of the other two exponents.)


    Then we have:

    [tex] 2,000 \ < x^6 \ < 10,000[/tex]

    [tex]\sqrt[6]{2,000} \ < x \ < \sqrt[6]{10,000}[/tex]

    [tex]3.5... \ < x \ < 4.6...[/tex]


    What can you conclude?
    Last edited by lookagain; 02-26-2018 at 11:06 AM.

  4. #4
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    It is obvious that

    [tex]a \in \mathbb Z \implies a^2,\ a^3, a^6 \in \mathbb Z \text { and } (a^2)^3 \equiv a^6 \equiv (a^3)^2.[/tex]

    It is not obvious, however, that

    [tex]b^2 = c^3 \in \mathbb Z \implies \sqrt{b} \in \mathbb Z.[/tex]

    It may be true, but it requires a proof.

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