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B Math olympiad basic number theory problem

  1. Apr 16, 2017 #1
    so this is the question:
    let a and b be real numbers such that 0<a<b. Suppose that a3 = 3a -1 and b3 = 3b -1. Find the value of b2 -a.
    initially my line of thinking was that just solve the equation x3 - 3x +1 = 0
    and take the roots which are more than 0 and then after that i got stuck

    ok that was the question i also have the full worked solution to it but the thing is that i am almost 100% sure i wouldn't think of such a smart solution in the exam so i want some of you all to solve it post your method to see if anyone has different method.
    if you want me to post the solution just ask me to post it below and i would gladly do so. but i am pretty sure many of you should have various other methods to this and no calculator is allowed on this exam and this meant for grade 9 and 10 so please don't use some super complicated methods thanks
     
  2. jcsd
  3. Apr 16, 2017 #2

    PeroK

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    Here's how I did it. There are three roots of the equation, call these ##c, a, b##, where ##c## is the negative root. From the coefficients:

    ##c + a + b = 0## hence ##c = -(a+b)##

    ##ab + ac + bc = -3## hence (substituting ##c## above) ##a^2 + ab + b^2 = 3##

    Now, if we let ##a = b^2 - \alpha## we get:

    ##b(b^3 + (1-2\alpha)b + (3-\alpha)) = 4 - \alpha^2##

    Which has a solution of ##\alpha = 2##.
     
  4. Apr 16, 2017 #3
    but if a = 2 then the equation a3 = 3a -1 is not satisfied
     
  5. Apr 16, 2017 #4

    PeroK

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    ##\alpha = b^2 - a = 2##
     
  6. Apr 16, 2017 #5
    wait your answer is correct this was the method provided
    if t is a solution then t2 -2 is also one
    then hence a , b , c = a2 -2 , b2 -2 , c2 then by matching the solution based on negative or positive then the answer follows
    is this solution super hard to think of and by the way your solution is far more better at least some think i can think of
     
  7. Apr 16, 2017 #6
    i don't understand how you went from step 3 to 4
    please help me thanks
     
  8. Apr 16, 2017 #7

    PeroK

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    Just a bit of algebra.
     
  9. Apr 16, 2017 #8
    ya i tried it but not getting what you get pls show your working thanks because i like your method over the given one
     
  10. Apr 16, 2017 #9

    PeroK

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    If ##a = b^2 - \alpha##, then what is ##a^2##?
     
  11. Apr 16, 2017 #10
    b2 - a =x
    i dont how to type your a so i used x instead

    b2 - x =a
    substituing in
    (b2 - x)2 + (b2 -x)b + b2 = 3
    x2 + b4 - 2b2x +b(b2 -x) + b2 =3
    then where do you go from here
     
  12. Apr 16, 2017 #11

    PeroK

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    The next step is:

    ##b^4 - 2b^2x + b^3-xb + b^2 = 3 - x^2##

    Now use ##b^3 = 3b - 1##
     
  13. Apr 16, 2017 #12
    ooh okay i will give it a try
    ok i got how to get just realized how dumb i was sorry for that
    b(b3 -b(1-2x) +3-x) = 4 - x2
    but how do you solve it?
     
  14. Apr 16, 2017 #13

    PeroK

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    I was trying to get a factor of ##b^3 - 3b + 1## and I saw that ##x=2## does that and solves the equation.

    You have a sign wrong there. It should be:

    ##b(b^3 + b(1-2x) + 3-x) = 4-x^2##
     
  15. Apr 16, 2017 #14
    oh ya about the sign careless
    but back to solving i noted that observation but how could you do it rigorously is there a method if there isn't i am okay with this method because it seems fair enough but if there is a method then please do tell me because the original solution has another method of letting b = 2sinΘ which i would never have thought of
     
  16. Apr 16, 2017 #15

    PeroK

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    The slow way is to solve the quadratic in ##x## and show, by whatever means, that ##x=2## is the only valid solution.
     
  17. Apr 16, 2017 #16
    ooh okay i will give it a try and get back to you thanks but should i use b as the variable or x as the variable
     
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