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A problem that puts me in difficulty

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex] a,b,c,d [/itex] be real number that satisfy the property that [itex] a + b = c + d [/itex] and [itex] a^2 + b^2 = c^2 + d^2 [/itex]. Show that [itex] a^n + b^n = c^n + d^n [/itex] for any [itex] n [/itex], a natural number ( with [itex] n > 0 [/itex] ) .


    2. Relevant equations

    [tex] x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + x^1y^{n-2} + y^{n-1}) [/tex]


    3. The attempt at a solution

    I have two ideas but I couldn't succed to apply them properly:

    1) First, I'm thinking at a mathematical induction, but it seems that in this case the proof via induction isn't so straightforward.

    2) Second, I'm thinking of writing that : [itex] a^n - c^n = d^n - b^n [/itex] and, then then using the formula:

    [tex] x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + x^1y^{n-2} + y^{n-1}) [/tex]

    But this dosen't seem very helpful either. The very first move I thought of making is writing that:

    [itex] a^n - c^n = d^n - b^n \to (a-c)(..... )= (d-b)(......) [/itex] and get rid of the first paranthesis. Unfortunately, I cannot do this beacause I don't know for sure if (a-c) and (d-b) are diffrent from 0.




    Please, can sombeody guide me to the solution ? Are my ideas good ?
     
  2. jcsd
  3. Nov 17, 2013 #2

    ehild

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    The two properties [itex] a + b = c + d [/itex] and [itex] a^2 + b^2 = c^2 + d^2 [/itex] involve some simple relations between a,b,c,d. What are these? (Write out the equations as differences and factorize the second one. )

    ehild
     
  4. Nov 17, 2013 #3
    Well, I'd get that:

    [tex] (a - c)(a+c) = (d-b)(d+b) [/tex]

    Now, if [itex] a - c = 0 [/itex] then [itex] d- b = 0[/itex] , and [itex] a = c [/itex] and [itex] b = d [/itex] . In this case, the problem statement holds. So, let's asume that a -c and d - b are both different from 0 .
    Then, I get that: [itex] a + c = d + b [/itex] .

    Also, by squaring [itex] a + b = c + d [/itex] and using the problem statement I get that [itex] ab = cd [/itex]

    So, I found two new relations:
    [tex] ab = cd [/tex]
    [tex] a + c = d + b [/tex]

    Is this correct ?
     
  5. Nov 17, 2013 #4
    Wait a minute:

    a + c = d + b
    a + b = d + c

    By summing the two relations we have that: 2a + b + c = 2d + b + c , so a = d .

    If a = d then b = c and the statement is proved :) .
     
  6. Nov 17, 2013 #5

    ehild

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    and you also know that a-c=d-b. **

    well done! The other case is discussed in your next post. :smile:

    ehild
     
  7. Nov 17, 2013 #6

    ehild

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    Excellent! You have solved the problem :thumbs:

    ehild
     
  8. Nov 17, 2013 #7
    Thank you very much ! You really helped me. You gave me the right hint at the right time. All the best! Maybe we will see eachother again around here! :) :)
     
  9. Nov 17, 2013 #8

    ehild

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    See you soon:smile:

    ehild
     
  10. Nov 17, 2013 #9

    Ray Vickson

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    Another way: from ##a+b = c+d## and ##a^2 + b^2 = c^2 + d^2## we have
    [tex] (a+b)^2 - a^2 - b^2 = (c+d)^2 - c^2 - d^2 \\
    \text{ or}\\
    2a b = 2 cd \rightarrow ab = cd[/tex]
    Therefore,
    [tex] a^2 + b^2 - 2 ab = c^2 + d^2 - 2cd \\
    \text{ or}\\
    (a-b)^2 = (c-d)^2 \rightarrow a-b = \pm(c-d)[/tex]
    Just consider separately the two cases ##a+b = c+d, a-b = c-d## and ##a+b = c+d, a-b = d-c##.
     
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