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Math Help - no positive integers

  1. #1
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    no positive integers

    Show no positive integers m for which m^4 + 2m^3 + 2m^2 + 2m + 1 is a perfect square. Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?

    If so, find all such m.
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  2. #2
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    Quote Originally Posted by Fayyaz View Post
    Show no positive integers m for which m^4 + 2m^3 + 2m^2 + 2m + 1 is a perfect square. Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?

    If so, find all such m.
    Hi

    m^2 + 1)" alt="m^4 + 2m^3 + 2m^2 + 2m + 1 = m^4 + 2m^2 + 1 + 2m^3 + 2m
    = (m^2 + 1)^2 + 2m\m^2 + 1)" />

    m^2 + 2m + 1) = (m^2 + 1)\m+1)^2" alt="m^4 + 2m^3 + 2m^2 + 2m + 1 = (m^2 + 1)\m^2 + 2m + 1) = (m^2 + 1)\m+1)^2" />
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  3. #3
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    Hello, Fayyaz!

    running-gag has an excellent proof, but omitted the punchline.


    N \;=\;m^4 + 2m^3 + 2m^2 + 2m + 1 \;=\; m^4 + 2m^2  + 1 + 2m^3  + 2m

    . . =\; (m^2 + 1)^2 + 2m\,(m^2 + 1) \;=\; (m^2 + 1)\,(m^2 + 2m + 1) \;=\; (m^2 + 1)\,(m+1)^2

    If N is a square, then (m^2+1) must be a square.

    . . Hence, m^2 + 1 \:=\:a^2\quad\hdots for some integer a.

    Then: . a^2-m^2 \:=\:1 \quad\hdots two squares differ by 1.

    . . But the only two squares that differ by 1 are: .  a = \pm1,\text{ and }\, m = 0


    Since m must be a positive integer, no solution exists.

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  4. #4
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    thanks guys

    how should i proceed with the next bit:

    Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?
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  5. #5
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    Hello Fayyaz
    Quote Originally Posted by Fayyaz View Post
    thanks guys

    how should i proceed with the next bit:

    Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?
    When m = 3, m^4 +m^3 +m^2 +m+1=121 = 11^2. So there's certainly one value. Whether there are any more, I don't know.

    Grandad
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  6. #6
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    hi

    thanks, yeah i figured that out using excel spreadsheet, but anyway I can do an algebraic proof?
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