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Math Help - Squares of natural numbers in an arithmetic progression

  1. #1
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    Squares of natural numbers in an arithmetic progression

    Hi, I've got a problem solving this task:

    "Prove that if there is a square of a natural number in an arthmetic progression of natural numbers, then there is an infinite number of squares of natural numbers in this progression"

    I assume that the difference of this progression must also be natural (particularly 0) so that all the members are natural only.

    I found out that any square of natural number x can be expressed as a sum of the first x odd numbers, which builds up an arithmetic progression too (starting from 1, with the difference 2):

    x^2 = 1 + 3 + 5 + 7 + ... + (2x-1), or
    \sum_{n=1}^{x} (2x-1)

    But I'm stuck at this point. Any suggestions on how to proceed?
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  2. #2
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    Quote Originally Posted by pinkparrot View Post
    Hi, I've got a problem solving this task:

    "Prove that if there is a square of a natural number in an arthmetic progression of natural numbers, then there is an infinite number of squares of natural numbers in this progression"

    I assume that the difference of this progression must also be natural (particularly 0) so that all the members are natural only.

    I found out that any square of natural number x can be expressed as a sum of the first x odd numbers, which builds up an arithmetic progression too (starting from 1, with the difference 2):

    x^2 = 1 + 3 + 5 + 7 + ... + (2x-1), or
    \sum_{n=1}^{x} (2x-1)

    But I'm stuck at this point. Any suggestions on how to proceed?
    Hint: if a+kd=n^2, then: (n+md)^2=a+(k+m^2d + 2mn)d.
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