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Homework Help: Find number n,m

  1. Jul 17, 2016 #1
    1. The problem statement, all variables and given/known data

    (22016+ 5)m + 22015 = 2n + 1


    find every n and m pairs
    as they are positive integers


    3. The attempt at a solution
    (22016+ 5)0 + 22015 = 22015 + 1

    so one pair is m= 0 , n = 2015
    if m =1 the equation is meaningless
    if m> 1 so there are really amount of powers that cant be handled by n as 1 in the end of the equation is needed to cancel out 5^m so 5^m -1 cant be equal to 2^n

    so the equation has only one answer
    am i right ?
     
  2. jcsd
  3. Jul 17, 2016 #2

    haruspex

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    False, e.g. m=1, n=2.

    Also, m=0, n=2015 is not a solution since it says positive integers, so 0 is not allowed.
     
  4. Jul 17, 2016 #3
    sorry non negative integers so 0 is allowed

    and about m =1 n = 2 that is the only one that is not true for that equation and i considered m = 1 specifically so m>1 is only valid
     
  5. Jul 17, 2016 #4

    haruspex

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    Ok, but I don't see in your post a proof that there is no solution for m>1. Can you explain in more detail?
     
  6. Jul 17, 2016 #5
    (22016+ 5)m + 22015 = 2n + 1
    22016 i'll call this "a"
    if m = 2 a2 + 10a + 24 + a/2 = 2^n
    24 is not 2^n n is natural number there is no n
    m = 3 a3 + 124 + 15a2+ 75 a + a /2 = 2^n
    124 is not 2^n n is natural number there is no n
    and as the m is increasing and all the power is increasing n won't be found

     
  7. Jul 17, 2016 #6

    haruspex

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    Right, but how do you know that a2+ 10a + 24 + a/2 is not of the form 2^n?
     
  8. Jul 17, 2016 #7
    a(a + 10 + 1/2 ) + 24
    24 has only 2^3 times 3
    so 22106 cant be equalized by 2^3
     
  9. Jul 17, 2016 #8

    haruspex

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    Ok, that deals with m=2, but I still don't see that it works for m in general. You would need to show that 5m-1 cannot have a factor that is some large power of 2.
     
  10. Jul 17, 2016 #9
    can it be done by induction?
     
  11. Jul 17, 2016 #10

    haruspex

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    Maybe, but two usual approaches to this sort of problem are (what I call) modulo arithmetic; and magnitude analysis.
    E.g., for large m, what roughly must n equal? Must it be a bit more or a bit less than that? Since it must be an integer, what are the consequences of its being a whole 1 more (or 1 less)?
     
  12. Jul 17, 2016 #11
    as 5 goes to some power 2 mast go power with much more than that maybe for 5 to 20 2 to 50 there are no possible for then to equal
     
  13. Jul 17, 2016 #12
    and also if i pretend graphic of those functions they cant be equal
     
  14. Jul 17, 2016 #13

    haruspex

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    You will need a far more rigorous argument than that!
    Try pulling the 22016 term outside the parentheses, leaving (1+5/22016) inside. What can you then say about the approximate relationship between n and m?
     
  15. Jul 21, 2016 #14

    Hey I was thinking a lot so I found this out look 2^2016 + 5 decided by five in remind is 1 so the last number is or 1 or 6. And remind stays the same now 2^2015-1 divide by 5 and remind is 2 so on the right side where we have these numbers. Divide whole equation by 5 gives a us remind 3 now on the left side where we have 2^n this number has period 4. 2^1 / 5 remind is 2 2^2/5 remind 4. 2^3. Remind 3. 2^4 /5. Remind is 1. So if we want reminds to be equal each other we have to choose only. 2^3+4k. Where k is a natural number. This is the first thing.
     
  16. Jul 21, 2016 #15
    The second thing is this look if we devide 2^2016+5 by 3 remind is 0 and 2^2015 - 1 by 3 remind is 1. So 2^n devided by 3 must remind 1 . Period of 2^n is 4 so if the power of 2 is even remind after division by 3 is 1 and if power of 2 is odd after division remind will be 2
    So the equation to be equal power n must be even but 2^3+4k is odd so odd can't be equal even so there can't be found n and m .

    Are my conclusion right ??
     
  17. Jul 21, 2016 #16
    Speaking about ends

    22016 ends with 6
    22015 ends with 8
    Thus
    (22016 + 5)m + 22015 ends with 9
    n = 4k + 3
     
  18. Jul 21, 2016 #17

    haruspex

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    You can certainly conclude many facts about what n must be modulo various bases, but I don't see any of these lines leading to m=0 being the only solution. (2^3+4k is even.)

    The only way I can see to obtain the desired answer is the magnitudes method I mentioned in post #10 and laid out the first steps for in post #13.
     
  19. Jul 22, 2016 #18
    i really don't understand what can be done that way
     
  20. Jul 22, 2016 #19

    haruspex

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    (22016+ 5)m=22016m(1+5.2-2016)m
    Now apply the binomial theorem to expand the part in parentheses. Try keeping the first three terms, and we'll see how it goes. If you know the 'big O' notation you can use that for the third and subsequent terms.
    Based on this, what can you say immediately about n as a function of m, asymptotically?
     
  21. Jul 22, 2016 #20
    whats that?
     
  22. Jul 22, 2016 #21

    haruspex

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  23. Jul 22, 2016 #22

    haruspex

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    Here's an illustration of the method I'm suggesting, but applied to a trivial problem: 2m+1=2n.
    The answer is obvious, and you could prove it using divisibility arguments, but here's how it would go with magnitude arguments...
    2m<2n
    m<n
    m+1 <= n
    2m+1<=2n=2m+1
    2m<=1
    m<=0
    See if you can do something similar with the problem in this thread.
     
  24. Jul 24, 2016 #23

    haruspex

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    A couple of things I should make clear:
    - I do not know whether there are any other solutions (do you know whether there are supposed to be?)
    - Even if there are not, I do not know whether my suggested magnitudes argument works here. I just know that it can be useful.
    That said, unless this is at a fairly advanced level, I suspect that there are no other solutions.

    Edit: I think I can show that if m>0 then m>22000, or something like that.
     
    Last edited: Jul 24, 2016
  25. Aug 2, 2016 #24
    i tried your way but i couldn't do that kind of things
    another way of solution are reminds that i proposed i don't have another
     
  26. Aug 2, 2016 #25

    haruspex

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    Here's where I got to:
    ##2^{2016m}(1+5.2^{-2016})^m+2^{2015}=2^n+1##
    Clearly n > 2016m, so n>= 2016m+1
    ##2^{2016m}(1+5.2^{-2016})^m+2^{2015}>=2^{2016m+1}##
    Now, for x, r > 0, ## (1+x)^r<e^{rx}##
    ##2^{2016m}e^{5m2^{-2016}}+2^{2015}>2^{2016m+1}##
    ##e^{5m2^{-2016}}+2^{2015-2016m}>2##
    If m>0 then the second term less than 1/2:
    ##e^{5m2^{-2016}}>3/2##
    ##m>\frac 15 2^{2016}\ln(\frac 32)##
    It looks like it should possible to cycle that back into the first equation above to show that n>=2016m+2. Might be a proof of no solutions along those lines.
     
    Last edited: Aug 4, 2016
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