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Math Help - unusual integer equation

  1. #1
    Senior Member abhishekkgp's Avatar
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    unusual integer equation

    Find all the positive integers x, \, y such that x^y=(y+1)^4.

    I could show that no solution is there if 5 \leq y+1<x.
    But the remaining cases still bug me.
    any help.
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    Re: unusual integer equation

    Seems like there is quite a collection for y = 0.
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    Re: unusual integer equation

    Quote Originally Posted by TKHunny View Post
    Seems like there is quite a collection for y = 0.
    there is. but we are concerned with only positive integers.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: unusual integer equation

    Quote Originally Posted by abhishekkgp View Post
    Find all the positive integers x, \, y such that x^y=(y+1)^4.

    I could show that no solution is there if 5 \leq y+1<x.
    But the remaining cases still bug me.
    any help.

    y=4
    x=y+1

    (5,4) is a solution. I think it is the only one.(for x,y>0)
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    Re: unusual integer equation

    Quote Originally Posted by abhishekkgp View Post
    there is. but we are concerned with only positive integers.
    Always with the missing ONE word!
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    Senior Member abhishekkgp's Avatar
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    Re: unusual integer equation

    Quote Originally Posted by Also sprach Zarathustra View Post
    y=4
    x=y+1

    (5,4) is a solution. I think it is the only one.(for x,y>0)
    just found another one (x,y)=(3,8)
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    Re: unusual integer equation

    Quote Originally Posted by abhishekkgp View Post
    Find all the positive integers x, \, y such that x^y=(y+1)^4.

    I could show that no solution is there if 5 \leq y+1<x.
    But the remaining cases still bug me.
    any help.
    Write (a) x^{y/4}=y+1. Clearly, x\geq2 and so y+1=x^{y/4}\geq2^{y/4}. Note that since x^y is a fourth power, then y/4 must be an integer.*

    It can be shown# that 2^n>4n+1 for any integer n\geq5, so 4n+1\leq2^n implies n<5. Now, taking n=y/4 we have y/4<5.

    The only possible value for y are 4, 8, 12, and ,16. The corresponding equations for (a) are x=5, x^2=9, x^3=13, and x^4=17. Only the first two are solvable. Therefore, the only solutions to the originial equation are those that Also sprach Zarathustra and abhishekkgp gave.

    *Not necessary for the solution, only to reduce the number cases.
    #If n=5, then 2^5>4\cdot5+1. Adding 4 gives 4+2^n<2^n+2^n=2^{n+1} and 4n+1+4=4(n+1)+1. Using 2^n>4n+1 we have 2^{n+1}>4(n+1)+1.
    Last edited by melese; July 7th 2011 at 08:11 AM. Reason: repeated words
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    Senior Member abhishekkgp's Avatar
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    Re: unusual integer equation

    Quote Originally Posted by melese View Post
    Write (a) x^{y/4}=y+1. Clearly, x\geq2 and so y+1=x^{y/4}\geq2^{y/4}. Note that since x^y is a fourth power, then y/4 must be an integer.*
    i don't think this is correct for if we take x=4, y=2 then x^y=2^4 is still a fourth power but y/4 is not an integer.
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    Re: unusual integer equation

    Quote Originally Posted by abhishekkgp View Post
    i don't think this is correct for if we take x=4, y=2 then x^y=2^4 is still a fourth power but y/4 is not an integer.
    I cannot be sure I'm not deceived again.

    Since y>0, x\geq2. So, 2^y\leq x^y and by the Binomial Theorem, (1+y)^4<1+y^4. Therefore, 2^y\leq y^4.

    But 2^y>y^4 for y>16 (by induction or otherwise). It follows that y\leq16.

    The possible equations are:
    x^1=(1+1)^4; x^2=(2+1)^4; x^3=(3+1)^4; x^4=(4+1)^4;...; x^{15}=(15+1)^4; x^{16}=(y+1)^4. Then the only solutions I've found are (16,1), (9,2), (5,4), (3,8).
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    Member Goku's Avatar
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    Re: unusual integer equation

    could you try y=4z and x=(4z+1)^1/z, i don,t think that they are any more +ve integer solutions, I wrote a program and as y goes to infinity x approaches 1, any ideas..
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    Senior Member abhishekkgp's Avatar
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    Re: unusual integer equation

    Quote Originally Posted by melese View Post
    I cannot be sure I'm not deceived again.

    Since y>0, x\geq2. So, 2^y\leq x^y and by the Binomial Theorem, (1+y)^4<1+y^4. Therefore, 2^y\leq y^4.

    But 2^y>y^4 for y>16 (by induction or otherwise). It follows that y\leq16.

    The possible equations are:
    x^1=(1+1)^4; x^2=(2+1)^4; x^3=(3+1)^4; x^4=(4+1)^4;...; x^{15}=(15+1)^4; x^{16}=(y+1)^4. Then the only solutions I've found are (16,1), (9,2), (5,4), (3,8).
    i had known that there are only for solutions to that equations.. i have not yet reviewed your solution but m sure u r right this time!
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    Senior Member abhishekkgp's Avatar
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    Re: unusual integer equation

    Quote Originally Posted by melese View Post
    (1+y)^4<1+y^4. .
    but its the other way around. Even if this is a typo i am not getting how you got y<16..
    Last edited by abhishekkgp; July 7th 2011 at 07:23 PM.
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    Re: unusual integer equation

    Quote Originally Posted by abhishekkgp View Post
    but its the other way around. Even if this is a typo i am not getting how you got y<16..
    That was not a typo but a serious mistake. We have 2^y\leq x^y=(1+y)^4.

    For y>16, the inequality 2^y>(1+y)^4 can be shown by induction on y. We can check that 2^{17}>(1+17)^4. Assuming that 2^y>(1+y)^4 multiply both sides by 2. Then 2^{y+1}>(1+y)^4\cdot2. Now it's left to show that (1+y)^4\cdot2>(2+y)^4. Note that \displaystyle\frac{(2+y)^4}{(1+y)^4}=(\frac{2+y}{1  +y})^4=(1+\frac{1}{1+y})^4.
    Since (1+\frac{1}{1+y})^4 gets smaller as y>16 gets larger it follows that (1+\frac{1}{1+y})^4<(1+\frac{1}{1+16})^4<2 and therefore \frac{(2+y)^4}{(1+y)^4}<2 - this completes the inductive part.

    Since we want 2^y\leq(1+y)^4, we must take y\leq16. Otherwise y>16 and the inequality 2^y>(1+y)^4 holds.
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  14. #14
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    Re: unusual integer equation

    Quote Originally Posted by melese View Post
    That was not a typo but a serious mistake. We have 2^y\leq x^y=(1+y)^4.

    For y>16, the inequality 2^y>(1+y)^4 can be shown by induction on y. We can check that 2^{17}>(1+17)^4. Assuming that 2^y>(1+y)^4 multiply both sides by 2. Then 2^{y+1}>(1+y)^4\cdot2. Now it's left to show that (1+y)^4\cdot2>(2+y)^4. Note that \displaystyle\frac{(2+y)^4}{(1+y)^4}=(\frac{2+y}{1  +y})^4=(1+\frac{1}{1+y})^4.
    Since (1+\frac{1}{1+y})^4 gets smaller as y>16 gets larger it follows that (1+\frac{1}{1+y})^4<(1+\frac{1}{1+16})^4<2 and therefore \frac{(2+y)^4}{(1+y)^4}<2 - this completes the inductive part.

    Since we want 2^y\leq(1+y)^4, we must take y\leq16. Otherwise y>16 and the inequality 2^y>(1+y)^4 holds.
    that's great!!
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