Posers and Puzzles

Posers and Puzzles

  1. Joined
    25 Aug '06
    Moves
    0
    11 Jan '08 14:30
    Prove:

    the only integer solutions of ab+5cd=3, ad=bc are:

    a=1, b=3, c=d=0
    a=-1, b=-3, c=d=0
    a=3, b=1, c=d=0
    a=-3, b=-1, c=d=0
  2. Standard membercelticcountry
    Copyright ©2001-2006
    Eastbourne
    Joined
    20 Sep '04
    Moves
    16434
    15 Jan '08 10:20
    Originally posted by David113
    Prove:

    the only integer solutions of ab+5cd=3, ad=bc are:

    a=1, b=3, c=d=0
    a=-1, b=-3, c=d=0
    a=3, b=1, c=d=0
    a=-3, b=-1, c=d=0
    NO.
  3. SubscriberPonderable
    chemist
    Linkenheim
    Joined
    22 Apr '05
    Moves
    532626
    15 Jan '08 12:39
    If a=0 then b=0 OR c=0 (because of (2))
    If a=0 and b=0 then c=3/5d There are really lots of integers which statisfy this one.
  4. Joined
    07 Sep '05
    Moves
    35068
    15 Jan '08 13:59
    Originally posted by Ponderable
    If a=0 then b=0 OR c=0 (because of (2))
    If a=0 and b=0 then c=3/5d There are really lots of integers which statisfy this one.
    You mean cd = 3/5 - in which case there are no integer solutions.
  5. B is for bye bye
    Joined
    09 Apr '06
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    27526
    15 Jan '08 18:07
    Originally posted by Ponderable
    ...If a=0 and b=0 then c=3/5d There are really lots of integers which statisfy this one.
    List one.
  6. SubscriberPonderable
    chemist
    Linkenheim
    Joined
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    532626
    16 Jan '08 18:26
    I admit to be wrong 🙁
  7. Joined
    24 Apr '05
    Moves
    3061
    17 Jan '08 21:471 edit
    Originally posted by David113
    Prove:

    the only integer solutions of ab+5cd=3, ad=bc are:

    a=1, b=3, c=d=0
    a=-1, b=-3, c=d=0
    a=3, b=1, c=d=0
    a=-3, b=-1, c=d=0
    I guess I'll have a go at this. I'll consider 4 different cases that collectively exhaust possibilities.

    First, consider c=0 and d is nonzero. For this case we have ab=3; ad=0; d nonzero. This doesn't have any solutions because we require a=0 to satisfy the second equation, but that in turn would preclude any integer solutions to the first equation. No integer solutions for this case.

    Second, consider c is nonzero, d=0. For this case we have ab=3; bc=0; c nonzero. Like the first case, this doesn't have solutions because we require b=0 to satisfy the second, and then we don't have any way to satisfy the first. No integer solutions for this case.

    Third, consider c=d=0. Then we have ab=3, which yields the four integer solutions you listed and no others.

    So we are left with considering cases where c and d are both nonzero integers. I will try to show that this, like the first two cases above, has no solutions. We have

    (1) ab=3-5cd and (2) ad=bc; c and d nonzero.

    Substituting (2) into (1) we get the following equations, which both must hold:

    (3) (cb^2)/d = 3-5cd and (4) (da^2)/c = 3-5cd.

    These respectively yield

    (5) b^2 = 3(d/c) - 5d^2 and (6) a^2 = 3(c/d) - 5c^2

    Adding (5) and (6) together, we get

    (7) a^2 + b^2 = 3[(c/d) + (d/c)] - 5(c^2 + d^2), which can be rearranged to

    (8) a^2 + b^2 = [3/(cd) - 5](c^2 + d^2)

    I think it's clear that this has no integer solutions because the LHS would have to be greater than or equal to zero, whereas the RHS can only be strictly negative for c and d nonzero integers.

    So I think the above cases exhaust possibilities, and they yield only the four solutions you listed.
  8. Joined
    25 Aug '06
    Moves
    0
    18 Jan '08 11:49
    Here is my solution.

    ab+5cd=3, ad-bc=0

    (ab+5cd)^2+5(ad-bc)^2=9

    a^2*b^2+10abcd+25c^2*d^2+5a^2*d^2-10abcd+5b^2*c^2=9

    a^2*b^2+5a^2*d^2+5b^2*c^2+25c^2*d^2=9

    (a^2+5c^2)(b^2+5d^2)=9

    At least one of the LHS factors is 1 or 3, but 3 is not of the form m^2+5n^2, so one of the factors is 1, in which case c=0 or d=0, etc.
  9. Joined
    29 Jan '07
    Moves
    3612
    28 Jan '08 20:20
    yes, very good, i can see it now
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