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A problem

  1. Feb 24, 2005 #1
    heres a problem i stumbled:

    find three rational numbers a,b,x such that:

    x^2 + 5 = a^2
    x^2 - 5 = b^2
     
  2. jcsd
  3. Feb 24, 2005 #2

    dextercioby

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    Can u find the integer ones...?

    Daniel.
     
  4. Feb 25, 2005 #3
    no i can't :-\
     
  5. Feb 25, 2005 #4

    dextercioby

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    How about
    [tex] (x+a)(x-a)=-5 [/tex]
    ,which comes from the first.
    [tex] (x+b)(x-b)=+5 [/tex]

    Solve each equation into the integers.

    Daniel.
     
  6. Feb 25, 2005 #5
    how will i do that there's an infinant amount of rational numbers which if ill multiply will give me 5 or -5.
     
  7. Feb 25, 2005 #6

    dextercioby

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    The #of integers is definitely finite.There are only 4 of them,grouped in 2 pairs for equation...

    Daniel.
     
  8. Feb 25, 2005 #7
    how can i solve this, i have two equations
    [tex] (x+a)(x-a)=-5 [/tex]
    [tex] (x+b)(x-b)=+5 [/tex]

    with three variables.
     
  9. Feb 25, 2005 #8

    dextercioby

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    True,but in my prior post i suggested to try to solve this equations by factoring 5 into a product of divisors...Then each paranthesis would get a value...And then each equation would be solved independently,as it would become a sistem 2-2...

    Daniel.
     
  10. Feb 25, 2005 #9

    NateTG

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    Let's assume that [itex]a,b[itex] and [itex]x[/itex] are integers.
    [tex](x+a)(x-a)=5[/tex]
    the only integer factorizations of 5 are
    [tex]-1 \times -5[/tex]
    and
    [tex]1 \times 5[/tex]
    Which imply that
    [tex]|x|=3[/tex]
    and
    [tex]|a|=2[/tex]

    Now, the only integer factorizations of [tex]-5[/tex] are
    [tex]-1 \times 5[/tex]
    and
    [tex]1 \times -5[/tex]
    so
    [tex](x+b)(x-b)=-5[/tex]
    implies that
    [tex]|x|=2[/tex]
    and
    [tex]|b|=3[/tex]
    So we have [tex]|x|=2[/tex] and [tex]|x|=3[/tex] which is contradictory. Therefore there is no solution in the integers.
     
    Last edited: Feb 25, 2005
  11. Feb 25, 2005 #10
    very nice proof, anyone got any ideas on how to find non integer solutions?
     
  12. Feb 25, 2005 #11

    dextercioby

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    I can't think of other method than graphical sollution...There are hyperbolas.So to solve the system,would mean to intersect them...

    Daniel.
     
  13. Feb 25, 2005 #12
    Another way to look at: NatTG So we have and which is contradictory. Therefore there is no solution in the integers,

    is to recognize that (x+1)^2-x^2 = 2x+1>10 for x=6. Thus in the equation a^2=b^2+10 we have to only try values a=0,1,2....5.

    --------------------------------------------------------------------------------
     
    Last edited: Feb 25, 2005
  14. Feb 25, 2005 #13
    i found one solution:

    x^2 + 5 = a^2
    x^2 - 5 = b^2

    x^2 = a^2 - 5
    x^2 = b^2 + 5

    a^2 - 5 = b^2 + 5
    a^2 - b^2 = 10
    (a+b)(a-b)=10

    let: (a-b) = r (rational number)
    then:
    a-b = r
    a+b = 10/r

    b = 10/r - a

    a = r + b
    a = r + 10/r - a
    2a = 10/r + r

    a = r + b

    b = 10/r - a
    b = 10/r - r - b
    2b = 10/r - r

    x^2 - 5 = b^2
    x^2 = b^2 + 5
    4x^2 = 4b^2 + 20

    2b = 10/r - r
    4b^2 = 100/r^2 -20 + r^2

    4x^2 = 100/r^2 -20 + r^2 + 20
    4x^2 = 100/r^2 + r^2

    let r = p/q (p,q both integer)

    4x^2 = (100q^2/p^2) + (p^2/q^2)
    4x^2 = (100q^4 + p^4)/(p^2*q^2)
    x^2 = (100q^4 + p^4)/4(p^2*q^2)

    now we gotta find two integer's p,q such to make (100q^4 + p^4) a square
    i tried a few numbers and found q = 2 p = 3
    so: 100q^4 + p^4 = 1600 + 81 = 1681 = 41^2

    x^2 = 41^2/4(3^2*2^2)
    x = 41/2*3*2
    x = 41/12

    x^2 + 5 = a^2
    1681/144 + 5/1 = a^2
    (1681+720)/144 = a^2
    a^2 = 2401/144
    a = 49/12

    b^2 = a^2 - 10
    b^2 = 2401/144 - 10/1
    b^2 = 2401/144 - 1440/144
    b^2 = 961/144
    b = 31/12
     
  15. Feb 25, 2005 #14

    cronxeh

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    2 equations, 3 unknowns. Infinite number of solutions.

    We can continue this thread till we run outa numbers
     
  16. Feb 26, 2005 #15

    matt grime

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    Yes, but the restriction to rational answers makes it more limited than you think, cronxeh.

    Here, paradoxically may be a way to make it simpler by adding in another variable.

    If there is a rational solution, then we may clear denominators and we're solving

    x^2+5d^2=a^2
    x^2-5d^2=b^2,

    but this simplifies to findinf a b such that b^2+5d^2 and b^2+10d^2 are both perfect squares.

    The thing about differences between squares is that they're sums of consecutive odd integers. I'm not sufficiently interested in the answer to figure out if that leads to a classification of all possible solutions, but it may do.
     
  17. Feb 28, 2005 #16
    We might try and look at this problem from the standpoint of a different factoring. Firstly we use the form advised by matt grime, and look at 10d^2 = a^2-b^2 =(a-b)(a+b). Assume 5 divides (a-b) and 2 divides (a+b), and then since a-b and a+b have the same parity: 10 divides (a-b)=10*u^2 and presume the square 4 divides (a+b) = 4*v^2 This results in
    [tex]a=5u^2+2v^2; b=|v^2-5u^2|. [/tex]

    However, I was unable to find any new results. It is obvious that x^2+5d^2 =a^2 can be multipled by any constant to get a new result, a multiple of 41^2 =5(12)^2 = 49^2. In the case of factors cited above, I have:

    [tex]x^2 + 5(2uv)^2 =(5u^2+2v^2)^2[/tex] arriving at [tex]x^2=25u^4+4v^2[/tex]

    I then get the result: [tex]82^2 = 25*4^4+4*3^4 [/tex], but this can be reduced by division by 4 to:[tex]41^2 =25*4^3+3^4 =100*4^2+3^4=100*2^4+3^4.[/tex] Which is the result and form obtained by Anzas, or [tex]41^2+5*12^2=49^2[/tex].

    Thus I suspect that there is only the one real answer given above, and every other answer is simply a multiple of that one.
     
    Last edited: Feb 28, 2005
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