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Number Theory: Difference of Perfect Squares

  1. Sep 15, 2015 #1

    S.R

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    1. The problem statement, all variables and given/known data:
    Determine whether there exists an integer x such that x^2 + 10 is a perfect square.

    2. Relevant equations:
    N/A

    3. The attempt at a solution:
    Assume x^2 + 10 = k^2 (a perfect square).

    Solve for x in terms of k:
    x = ±sqrt(k^2 - 10)

    Since k is an integer and k^2 - 10 > 0, k > sqrt(10) > 3.

    From here (and multiple other approaches), I'm not sure how to continue. Any help in proving whether such an integer value of x exists would be appreciated.
     
  2. jcsd
  3. Sep 15, 2015 #2
    Hmmm....I've got something in mind ,
    You have :
    ##k^2=x^2+10##
    And thus ,
    ##k^2-x^2=10##
    ##(k-x)(k+x)=10##
    And since 10=5×2 ,
    Therefore , we can say that
    Either ##k-x=2 ## and ##k+x=5##
    Or ##k+x=2## and ##k-x=5##
    Which will give you fractional values of x and k , which don't satisfy the condition .
    Therefore no integer should satisfy this equation.
     
  4. Sep 15, 2015 #3

    S.R

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    Ah, ok. Thank-you.
     
    Last edited: Sep 15, 2015
  5. Sep 15, 2015 #4

    Mark44

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    No, this doesn't follow. If a * b = 0, then you can definitely say something about a being 0 or b being 0, but if a * b = m, with m ≠ 0, then you effectively don't know anything about a and b.
     
  6. Sep 15, 2015 #5

    Mark44

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    So k2 - x2 = 10, where both k and x are integers.

    Since k is an integer, we can write it as k = x + a, for some integer a. What can you say about ##k^2 - x^2 = (x + a)^2 - x^2##? Can you show that this difference can never be 10?


     
  7. Sep 15, 2015 #6
    Does k have to be an integer or can k be rational? For n>=5 the difference between n squared and n+1 squared is 2n+1>10.
     
  8. Sep 15, 2015 #7

    Mark44

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    I believe that "perfect square" as used in the OP means the square of an integer.
     
  9. Sep 15, 2015 #8

    S.R

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    I attempted this approach, but I guess I'll try again here.

    (x+a)^2 - x^2 = 10
    x^2 + (2a)x + a^2 - x^2 = 10
    (2a)x + a^2 = 10
    a(2x + a) = 10

    From here, I can presumably set a equal to factors of 10, right?
     
  10. Sep 15, 2015 #9

    haruspex

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    I'm puzzled by your response here. The only thing I see wrong with mooncrater's approach is that it overlooked another possible factorisation. Isn't your x+a method effectively the same?
    S.R, there are certain facts about integers that are handy to remember. Perfect squares only take a few values modulo 8. What are they?
     
  11. Sep 15, 2015 #10

    S.R

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    I'm not sure what you mean by "perfect squares only take a few values modulo 8"?
     
  12. Sep 15, 2015 #11

    haruspex

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    For now, just consider modulo 4, since it is adequate for this problem.
    Take an even number, 2n, and square it. What is the result modulo 4? (You understand what is meant by modulo, yes?). Do the same for 2n+1.
     
  13. Sep 15, 2015 #12

    Mark44

    Staff: Mentor

    My main objection is that if a * b = 0, you definitely know something about a or b. OTOH, if a * b = N, there are an infinite number of pairs of numbers that work. In that post we had a * b = 10, from which mooncrater concluded that the two numbers have to be 5 and 2. Of course -5 and -2 work, as do 10 and 1, -10 and -1, 1/2 and 20, and many, many more. If he had included the other pair of positive integers, and mentioned that he was limiting the search to positive integers only, I don't think I would have said anything.
     
  14. Sep 15, 2015 #13

    S.R

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    (2n)^2 = 4n^2
    (4n^2) mod 4 = 0 (assuming n is an integer).

    (2n+1)^2 = 4n^2 + 4n + 1.
    (4n^2 + 4n + 1) mod 4 = 1 (assuming n is an integer).

    Not particularly sure how this fact helps, though?

    EDIT: Oh, if we assume k is a perfect square, then k mod 4 = 0 or k mod 4 = 1 (depending on whether k is even or odd).

    1) Assume x is even
    Let x = 2n, where n is an integer
    x^2 + 10 = 4n^2 + 10

    (4n^2 + 10) mod 4 = 2 ≠ 0 and thus, x^2 + 10 is not a perfect square if x is even.

    2) Assume x is odd
    Let x = 2n+1, where n is an integer
    x^2 + 10 = 4n^2 + 4n + 11

    (4n^2 + 4n + 11) mod 4 = 3 ≠ 1 and thus, x^2 + 10 is not a perfect square if x is odd.

    Since x^2 + 10 is not a perfect square if x is even nor odd, we can conclude that no such integer x exists such that x^2 + 10 is a perfect square.

    Is this correct?
     
    Last edited: Sep 15, 2015
  15. Sep 15, 2015 #14

    haruspex

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    Yes.
     
  16. Sep 15, 2015 #15

    S.R

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    Thank-you for the help!
     
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