5
$\begingroup$

First I tried taking logarithm on both sides, but that just made it messy.

Next I tried putting $x=y=z=t=k(say)$. That made the problem :

$k^k + k^k + k^k = k^{2005}$

$=>3k^k=k^{2005}$

$=>3 = k^{2005-k}$

How can I proceed?

Even by taking every variable equal, I am not able to solve the problem.

How to solve the problem without making unnecessay assumptions ?

  • 0
    Can you provide some more background to the question? What math contest was it from? Aimed at what grades/level? – 2017-01-15
  • 1
    Are negatives and zero allowed? – 2017-01-15
  • 4
    I deduced there are an infinite amount of solutions: simply set $y = 2005$, then follows that $y^z + z^t = 2005^z + z^t = 0$. Now choose $z = -2005$, then $t = -2005$ and we have no restrictions on x then. Thus every ordered 4-tuple $(x,2005,-2005,-2005)$ with $x$ an integer is a solution. Probably there are other solutions, but at least you know there are an infinite amount now. – 2017-01-15
  • 0
    @YiyuanLee The question states "Find all $integers$" – 2017-01-15
  • 0
    @Nirbhay Since when did negative integers and $0$ get kicked out from the _integers_? – 2017-01-15
  • 2
    @YiyuanLee that was exactly what I was trying to tell you...0 and negatives are allowed as they are too integers. – 2017-01-15
  • 0
    I see. In that case you can just use the solution provided by Math_QED. I was wondering that perhaps only positives should be allowed because the question would'nt have been that hard otherwise. My bad. – 2017-01-15
  • 0
    @IanMiller The question is from a journal/magazine ... link http://imomath.com/pcpdf/f1/f40.pdf ... Q."341.1" – 2017-01-15
  • 0
    For others, it is question 341.1 in the linked article. @Nirbhay Do you know what the translation of "Trần Tuyết Thanh" is? Google didn't help me. – 2017-01-15
  • 0
    @IanMiller: Probably the name of the author. – 2017-01-15
  • 0
    Ah. Thanks, that makes sense. – 2017-01-15
  • 0
    @IanMiller Please answer the question or atleast post a hint... – 2017-01-15
  • 0
    @Nirbhay I'm sorry, I don't know how to prove it. I highly suspect there are no answers other than the trivial one mentioned above but I can't prove that assertion. I've exhaustively searched $x,y,z,t\in[1,100]$ and found nothing. – 2017-01-15
  • 1
    Here's another family of solutions: $(x,y,z,t)=(0,a^k,-a,-ka)$ with $a\in 2\mathbb N-1$ and $k\in2\mathbb Z-1$. – 2017-01-15
  • 0
    @IanMiller In $x, y, z, t \in [1,100]$ there is the family $(1,y,-y,-y)$ which satisfies the equation trivially as well $1^y + y^{-y} -y^{-y} = 1^{2005}$. If you include $0$ in that range then a second set $(0,y,-y,-y) y \neq 0$ can be found as well. – 2017-01-24

1 Answers 1

2

The comments have illustrated a range of possible solutions, without anyone really getting close to demonstrating that we have covered all possible solutions.

For a solution in positive integers, we can set $y=t=1$ reducing the equality to:

$x^1+1^z+z^1 = x^{2005} \implies x+1+z = x^{2005} $

and we have the solutions $(x,y,z,t) = (x, 1, x^{2005}-x-1, 1)$ which will be all positive for $x>1$.

Capturing other families of solutions suggested in comments:

@Math_QED : $(x,y,z,t)=(x,2005,-2005,-2005)$

@DejanGovc : $(x,y,z,t)=(0,a^k,-a,-ka)$ ($a,k>0$ and odd)
@DejanGovc + : $(x,y,z,t)=(\pm1,a^k,-a,-ka)$ with $a,k>0$ and odd

@Dylan : $(x,y,z,t)=(\{0,1\},y,-y,-y)$

  • 0
    Regarding the family of solutions I suggested. I just noticed that we should assume that $k$ is also positive, since otherwise $a^k$ will not be an integer (unless $a=1$). Also, there seem to be two further families of solutions: $(x,y,z,t)=(\pm1,a^k,-a,-ka)$ with $a,k\in\mathbb N$ odd. – 2017-01-24