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Math Help - Proof, that product of 5 numbers cannot be square of a positive interger

  1. #1
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    Proof, that product of 5 numbers cannot be square of a positive interger

    "Prove that the product of five consecutive numbers cannot be the square of a positive integer."

    I have tried using examples, and it shows that this is true, but how do i prove this? Please show working/reasoning if possible.

    Thanks
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  2. #2
    Super Member Bacterius's Avatar
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    Hello,
    easy enough. Given five consecutive numbers a, b, c, d and e, for their product to be a square they need to combine all their prime factors with even powers. For instance, taking only two numbers for simplicity, for their product to be a square we need for instance 7 \times 7 = 7^2 or 7^4 \times 3^2 = (49 \times 3)^2, and 7 \times 3 = 7^1 \times 3^1 is not a square because every prime factor doesn't have an even power.

    Now assume some prime number greater than three divides a. It obviously doesn't divide a + 1 = b, a + 2 = c, a + 3 = d, a + 4 = e. Therefore, this prime factor is only present once in the product abcde, hence has an odd exponent (or power) ( 1) and thus abcde is not a square.

    Now just validate the statement with the two prime numbers smaller or equal to three, namely 2 and 3 (just take a = 2 and a = 3, since any multiple is suitable), to conclude the proof. Oh, and also validate for a = 1 and a = 0 ! And for negative numbers, you only have to check up to a = -4, because the product of five negative numbers is a negative number which cannot be a square !

    Does it make sense ?
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  3. #3
    Super Member Bacterius's Avatar
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    I just noticed a flaw in my proof. I'll correct it here : Note that if p^2, p being a prime number greater than three, divides a, then abcde has p with an even power. However, if a = p^2 with p prime and p > 3, then neither a + 1, a + 2, a + 3, a + 4 can be squares, applying the same reasoning and introducing the linear increase of the difference between two consecutive squares.

    Does ... does it make sense ?
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  4. #4
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    Hello, cedricc!

    Prove that the product of five consecutive integers
    cannot be the square of a positive integer.

    Let the five consecutive integers be: . n-2,\:n-1,\:n,\:n+1,\:n+2


    Their product is: . P \:=\:(n-2)(n-1)n(n+1)(n+2) \;=\;n(n^2-1)(n^2-4) \;=\;n(n^4-5n+4)


    Assume P is a square.

    Since P contains the factor n, it must contain the factor n^2.
    . . That is, n^4-5n^2+4 has a factor of n . . . and we can factor it out.

    Hence: . P \;=\;n^2\left(n^3-5n + \frac{4}{n}\right)


    Since P is an integer, the cubic expression must be an integer.

    . . Then \frac{4}{n} must be an integer . . . n is a factor of 4.

    Hence: . n \:=\:\pm1,\:\pm2,\:\pm4


    And we have: . \begin{array}{|c|c|} n & P \\ \hline \pm1 & 0 \\ \pm2 & 0 \\ \pm4 & \pm720 \end{array} . None of these is the square of a positive integer.

    . . Q.E.D.

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  5. #5
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    2 Soroban
    Imho this statement: "Since is an integer, the cubic expression must be an integer" is wrong. For instans, P=11, {n^2} = 9<br />
then
    "cubic expression" must be \frac{{11}}{9}<br />
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  6. #6
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    Quote Originally Posted by Soroban View Post

    Assume P is a square.

    Since P contains the factor n, it must contain the factor n^2.

    This isn't true. For example, consider the perfect square 36. It is divisible by 18, but not by 18^2.

    If n is prime, that is a different matter.
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