Equation

Honore · **Joined:** Wed, 13 Apr 2005 18:14:33 UTC **Posts:** 93

$x,y \in{Z^{+}}$ and $\displaystyle{x^{x^{x^x}}-(19-y^x)y^{x^y}+74=0}$

My solution approach:

When it is organized as $\displaystyle{x^{x^{x^x}}+74=(19-y^x)y^{x^y}}$

I can say that 1<x<3

otherwise the left side will be astronomically high and another restriction seems to be that $1{\leq}y{\leq}4$ so that the right side of the equation can remain positive.

Therefore, the possible

values can be found for x=2

having the left side $65610=6561{\cdot}10$ and in order to simplify by ten y=3

looks as a good attempt so $6561{\cdot}10=(19-3^2){\cdot}3^{2^3} \Rightarrow 6561=3^8$

Hopefully there is a much better solution. Any comments? Thanks and regards.

outermeasure · **Posted:** Thu, 14 Apr 2011 13:17:05 UTC

Honore wrote:

$x,y \in{Z^{+}}$ and $\displaystyle{x^{x^{x^x}}-(19-y^x)y^{x^y}+74=0}$

My solution approach:

When it is organized as $\displaystyle{x^{x^{x^x}}+74=(19-y^x)y^{x^y}}$

I can say that 1<x<3

otherwise the left side will be astronomically high and another restriction seems to be that $1{\leq}y{\leq}4$ so that the right side of the equation can remain positive.

Therefore, the possible

values can be found for x=2

having the left side $65610=6561{\cdot}10$ and in order to simplify by ten y=3

looks as a good attempt so $6561{\cdot}10=(19-3^2){\cdot}3^{2^3} \Rightarrow 6561=3^8$

Hopefully there is a much better solution. Any comments? Thanks and regards.

Huh? So what if the LHS is very large? And what's wrong with x=1?

Also, are you using $\mathbb{Z}^+$ to mean the set of nonnegative integers or the set of positive integers?

Honore · **Joined:** Wed, 13 Apr 2005 18:14:33 UTC **Posts:** 93

Thank you very much for your comments and questions. Yes,

and

are positive integers. For x=1

no solution seems possible from the set of positive integers for

because I get y^2-19y+75=0

.
I have thought over your first question and all I can do is to get help from computer as follows:

Shadow · **Posted:** Thu, 14 Apr 2011 17:13:37 UTC

Honore wrote:

Thank you very much for your comments and questions. Yes,

and

are positive integers. For x=1

no solution seems possible from the set of positive integers for

because I get y^2-19y+75=0

.
I have thought over your first question and all I can do is to get help from computer as follows:

What? The discriminant of that quadratic is positive, so there are solutions.

Honore · **Joined:** Wed, 13 Apr 2005 18:14:33 UTC **Posts:** 93

Yes it is but not integer.
http://www.wolframalpha.com/input/?i=y^2-19y%2B75%3D0

P.S. I also forced the integer limits in the above Fortran program starting from 32767 to 1 for both variables and it took 4 minutes to give the same result.

Shadow · **Posted:** Thu, 14 Apr 2011 17:18:00 UTC

Honore wrote:

Yes it is but not integer.
http://www.wolframalpha.com/input/?i=y^2-19y%2B75%3D0

P.S. I also forced the integer limits starting from 32767 to 1 for both variables and it took 4 minutes to give the same result.

Oh I see, well then yes, that's easy from the fact that the discriminant is not an integer.

outermeasure · **Posted:** Thu, 14 Apr 2011 17:44:59 UTC

Honore wrote:

Yes it is but not integer.
http://www.wolframalpha.com/input/?i=y^2-19y%2B75%3D0

P.S. I also forced the integer limits in the above Fortran program starting from 32767 to 1 for both variables and it took 4 minutes to give the same result.

That does not prove anything. You have only showed (of course, assuming there isn't any bugs, hardware or software) that there are no solutions other than x=2,y=3, in the set $x,y\in\{1,2,\dots,32767\}$ , not $x,y\in\mathbb{Z}^+$ .

Shadow · **Posted:** Thu, 14 Apr 2011 17:53:04 UTC

outermeasure wrote:

Honore wrote:

Yes it is but not integer.
http://www.wolframalpha.com/input/?i=y^2-19y%2B75%3D0

P.S. I also forced the integer limits in the above Fortran program starting from 32767 to 1 for both variables and it took 4 minutes to give the same result.

That does not prove anything. You have only showed (of course, assuming there isn't any bugs, hardware or software) that there are no solutions other than x=2,y=3, in the set $x,y\in\{1,2,\dots,32767\}$ , not $x,y\in\mathbb{Z}^+$ .

As a caveat to what outermeasure has said, I would note that just because you have large data points doesn't mean a thing. In computations on elliptic curves, even the simplest integral problems can have enormous smallest, non-trivial values, so you cannot assume anything based on such tiny amounts of data as you have there.

Honore · **Joined:** Wed, 13 Apr 2005 18:14:33 UTC **Posts:** 93

Thanks for all comments and critiques. The reason why I have had to bring up the problem here is that I have been unable to find any better solution though I believe there must be at least one and hopefully someone could show how.

outermeasure · **Posted:** Fri, 15 Apr 2011 10:53:24 UTC

Honore wrote:

Thanks for all comments and critiques. The reason why I have had to bring up the problem here is that I have been unable to find any better solution though I believe there must be at least one and hopefully someone could show how.

The equation $x^{x^{x^x}}+74=(19-y^x)y^{x^y}$ tells you y^x<19

. There are only 22 pairs (x,y) of positive integers with y>1 satisfying that. Now check by hand:

y=1, RHS=18 is too small.
$x=1<y\leq 18$ . LHS=75, so y|75 must be one of 1,3,5,15, but none of them are possible because we must also have (19-y)|75.
y=x=2, mod 4 => not a solution.
y=2,x=3, mod 3 => not a solution.
y=2,x=4, again mod 4 => not a solution.
y=3,x=2, $LHS=2^{2^{2^2}}+74=2^{16}+74=65536+74=65610$ , $RHS=(19-3^2)3^{2^3}=10\times 3^8=65610$ is a solution.
y=4,x=2, again mod 4 => not a solution.

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