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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    28 Sep '11 16:00
    Are there integers for which x^2 + y^2 = 4z + 3?

    If yes, give an example of such a triplet of numbers.
    If no, explain why such a triplet doesn't exist.
  2. 28 Sep '11 16:18
    No, consider the equation mod 4
  3. Standard member TomCr
    woodpusher
    04 Oct '11 21:17
    Originally posted by talzamir
    Are there integers for which x^2 + y^2 = 4z + 3?

    If yes, give an example of such a triplet of numbers.
    If no, explain why such a triplet doesn't exist.
    There are no integers that satisfy.
    Because 1 is the only integer that multiplied by itself yield an odd number, and 4z + 3 must be an odd number, either x or y must be 1.
    So, because x and y are interchangeable, let's assume that x is 1.
    this yields:
    1 + y^2 = 4z + 3
    y^2 = 4z +2
    this yields
    (y^2 -2)/4 = z
    this yields
    z = (1/4)y^2 - 1/2

    Any y integer will yield a remainder of 1/2 , so therefore, z cannot be an integer.
  4. 04 Oct '11 21:38 / 1 edit
    Originally posted by TomCr
    There are no integers that satisfy.
    Because 1 is the only integer that multiplied by itself yield an odd number, and 4z + 3 must be an odd number, either x or y must be 1.
    So, because x and y are interchangeable, let's assume that x is 1.
    this yields:
    1 + y^2 = 4z + 3
    y^2 = 4z +2
    this yields
    (y^2 -2)/4 = z
    this yields
    z = (1/4)y^2 - 1/2

    Any y integer will yield a remainder of 1/2 , so therefore, z cannot be an integer.
    Because 1 is the only integer that multiplied by itself yield an odd number

    3*3=9, 5*5=25 etc etc ad infinitum
  5. 04 Oct '11 21:42
    Originally posted by talzamir
    Are there integers for which x^2 + y^2 = 4z + 3?

    If yes, give an example of such a triplet of numbers.
    If no, explain why such a triplet doesn't exist.
    x=5 y=6 z=13
  6. 04 Oct '11 22:05
    Originally posted by TomCr
    There are no integers that satisfy.
    Because 1 is the only integer that multiplied by itself yield an odd number, and 4z + 3 must be an odd number, either x or y must be 1.
    So, because x and y are interchangeable, let's assume that x is 1.
    this yields:
    1 + y^2 = 4z + 3
    y^2 = 4z +2
    this yields
    (y^2 -2)/4 = z
    this yields
    z = (1/4)y^2 - 1/2

    Any y integer will yield a remainder of 1/2 , so therefore, z cannot be an integer.
    also y^2 = 4z+2 simplifies to (y^2/4) -2 = z
  7. 04 Oct '11 23:27
    Sadly we are not allowed complex numbers:

    x = i
    y = 0
    z = -1
    works
  8. Standard member TomCr
    woodpusher
    05 Oct '11 00:54
    Originally posted by tomtom232
    x=5 y=6 z=13
    5x5 + 6x6 = 4x13 + 3
    25 + 36 = 52 + 3
    61 is not 55.

    5,6,13 does not work.
  9. 05 Oct '11 05:23
    New to the site but wasn't an almost identical question asked earlier? Just consider the equation mod 4.

    Squares of integers are either 0 (if even) or 1 (if odd) mod 4 so the sum of two squares can only be 0, 1 or 2 mod 4 but not 3.