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Math Help - proof please...

  1. #1
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    Red face proof please...

    can anyone show me the proof that "the product of four positive integers in arithmetic progression cannot be the square of an integer".

    thanks!
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  2. #2
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    Hello, sheryl!

    I know a proof for a similar problem, but I've hit a wall.
    Let me show you the groundwork . . . maybe you can finish it.


    The product of four positive integers in arithmetic progression
    cannot be the square of an integer.

    Let the four numbers be: . a,\;a+d,\;a+2d,\;a+3d

    Suppose their product is a square.
    Then we have: . a(a+d)(a+2d)(a+3d)\;=\;k^2

    Regroup: . [a(a+3d)]\,[(a+d)(a+2d)] \;=\;k^2

    Multiply: . [a^2+3ad)\,(a^2+3ad + 2d^2)\;=\;k^2

    And we have: . [(a^2 + 3ad + d^2) - d^2]\,[(a^2+3ad + d^2) + d^2] \;= \;k^2

    Multiply: . (a^2+3ad + d^2)^2 - d^4 \;= \;k^2


    And this is supposed to lead to a contradiction . . . but I can't find it.

    Anyone? .Anyone?

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  3. #3
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    Hello, all!

    The problem I referred to is:
    "Prove that the product of four consecutive positive integers cannot be a square."


    Let the integers be: a,\;a+1,\;a+2,\;a+3

    Suppose their product is a square.
    Then we have: . a(a+1)(a+2)(a+3) \:=\:k^2

    Regrouping: . \left[a(a+3)\right]\,\left[(a+1)(a+2)\right] \:=\:k^2

    We have: . [a^2 + 3a]\,[a^2+3a +2] \:=\:k^2

    . . then: . \left[(a^2 + 3a + 1) - 1\right]\,\keft[(a^2 + 3a + 1) + 1\right] \:= \:k^2

    Hence: . (a^2 + 3a + 1)^2 - 1^2\:=\:k^2\quad\Rightarrow\quad (a^2 + 3a + 1)^2 - k^2\:=\:1


    We have: "The difference of two squares is 1."

    But this is true for (\pm1)^2 - 0^2 only.

    Then: . a^2 + 3a + 1 \:= \:\pm1\quad\Rightarrow\quad\left\{\begin{array}{cc  }a(a +3) \:= \:0 \\ (a+1)(a+2) \:= \:0\end{array}

    . . and a is not a positive integer.

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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, sheryl!

    I know a proof for a similar problem, but I've hit a wall.
    Let me show you the groundwork . . . maybe you can finish it.


    [size=3]
    Let the four numbers be: . a,\;a+d,\;a+2d,\;a+3d

    Suppose their product is a square.
    Then we have: . a(a+d)(a+2d)(a+3d)\;=\;k^2
    It is just enthusiasm you talked about/


    Let gcd(a,d)=c
    Let a=cm, d=cn
    Then your expression will be c^4 m(m+n)(m+2n)(m+3n)
    Now, since m and n are relatively prime, any two factors are relatively prime to each other(may have common factor of 2 or 3)
    Hence their product cannot be a cannot be a perfect square.

    Keep Smiling
    Malay
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