Solve in natural numbers

**segasai** · January 4th 2010, 06:08 AM

Hi,

I wonder whether somebody can advise me how to solve in natural numbers (or integers) the following equation:
$3^x+4^y=5^z$
I have spent a couple of weeks thinking about it without any considerable progress

. The obvious guess is that the only solution is (x,y,z)=(2,2,2). I've looked at the equation modulo different numbers 3,4,5,9,16,25 ... etc, but so far that wasn't very conclusive in proving that (2,2,2) is the only solution. I've tried a lot of other different things with the equation with the same negative result. Honestly I've run out of ideas of how to solve the equation.

Any help is appreciated.

**Bruno J.** · January 4th 2010, 03:46 PM

In my opinion, if the statement is true (I feel that it probably is), then it must be quite hard to prove.

It would follow directly from Beal's conjecture but unfortunately it is not proven.

**pomp** · January 4th 2010, 09:03 PM

Originally Posted by segasai

Hi,

I wonder whether somebody can advise me how to solve in natural numbers (or integers) the following equation:
$3^x+4^y=5^z$
I have spent a couple of weeks thinking about it without any considerable progress

. The obvious guess is that the only solution is (x,y,z)=(2,2,2). I've looked at the equation modulo different numbers 3,4,5,9,16,25 ... etc, but so far that wasn't very conclusive in proving that (2,2,2) is the only solution. I've tried a lot of other different things with the equation with the same negative result. Honestly I've run out of ideas of how to solve the equation.

Any help is appreciated.

If you're solving over the integers (or your convention is that 0 is a natural number) then there's the additional solution (0,1,1).

I ran a quick check up to z = 1000 and couldn't find any additional solutions. Good luck with the proof!

**Shanks** · January 4th 2010, 09:26 PM

Originally Posted by pomp

If you're solving over the integers (or your convention is that 0 is a natural number) then there's the additional solution (0,1,1).

I ran a quick check up to z = 1000 and couldn't find any additional solutions. Good luck with the proof!

How do you do this?

**pomp** · January 4th 2010, 09:28 PM

Originally Posted by Shanks

How do you do this?

I done it in Haskell: http://en.wikipedia.org/wiki/Haskell...mming_language)

**Sudharaka** · January 5th 2010, 12:34 AM

Dear segasai,

I have attached a solution that I have achieved for this problem. Please go through it and reply me soon about your views regarding the solution. If you have any problem regarding the attachment please don't hesitate to reply me. I value your reply as I enjoyed solving the problem. I have corrected this attachments they are in the order page 2, page 3, page 1, page 4
Hope this helps.

**Shanks** · January 5th 2010, 01:21 AM

(0,0,0) is a solution? clearly faalse!

**Sudharaka** · January 5th 2010, 01:25 AM

Hi everyone,

Sorry if I confused you. (0,0,0) is not a solution. Shame on me. And thank you very much Shanks for telling me that. But could anyone tell me whether my idea is correct or not neglecting this error. I have corrected the attachments.

Thanks again Shanks.

**DavidEriksson** · January 5th 2010, 02:56 AM

Given that $x,y,z \in \mathbf{N}$ we know that $x>0$ .

Then $3^x+4^y \equiv_3 1$ and $5^z \equiv_3 (-1)^z$ . Therefore $z=2a$ and we can write $3^x=5^{2a}-4^y=(5^a+2^y)(5^a-2^y)$ .

We see that both factors can't be a power of 3 since there sum isn't divisible by 3. Therefore we've the following cases:
1) $5^a+2^y=3^x$ and $5^a-2^y=1$
2) $5^a-2^y=3^x$ and $5^a+2^y=1$

For the first case, using mod3 on the equality $5^a-2^y=1$ implies that $y$ must be even and $a$ must be odd. This is true since $5^a-2^y \equiv_3 (-1)^a-(-1)^y\equiv_3 1$ .

For $y>2$ we see that $5^a+2^y \equiv_8 5$ (using that a is odd) but clearly $3^x \equiv_8 1 \mbox{ or } 3$ so there is no solution for $y>2$

We know that $y$ is even so we need to consider $y=0$ and $y=2$ .

For y=0 we obtain $5^a+1=3^x$ which clearly has no solutions (use mod4).

For y=2 we obtain $5^a-2^2=1$ and this implies that $a=1$ which is $z=2$ and a solution is therefore $(2,2,2)$

For the second case, $5^a+2^y=1$ clearly can't be true so there are no new solutions.

Therefore $(2,2,2)$ is the only solution, as expected.

If we add the possibilty for x,y,z to be 0 we've to consider $x=0$
This gives us $1+4^y=5^z$
It's easy to see that $y=z=1$ is a solution.

We have to consider $y>1$
z must be odd since $1+4^y \equiv_3 2$ and $5^z \equiv_3 2^z$

But using mod8 we see that $1+4^y \equiv_8 1$ and $5^z \equiv_8 (-3)^z$ so clearly z must be even.

We've reached a contradiction so there are no more solutions than (2,2,2) and (0,1,1) for $x,y,z \in \{0,1,.. \}$

**Shanks** · January 5th 2010, 03:05 AM

I thought, Sudharaka's solution is correct for the case t'=11.
Sudharaka, could you explain to us why you set t'=11, and can other case be reduced to the case t'=11(here I don't think so).

**segasai** · January 5th 2010, 03:43 AM

Thank you, Sudharaka for the suggested solution.
But as Shanks pointed, the assumption t'=11 is incorect. In fact I checked numerically, it is really incorrect. But I believe that your solution can be fixed ... even if t' is not const.

**Swlabr** · January 5th 2010, 03:44 AM

Originally Posted by DavidEriksson

For $y>2$ we see that $5^a+2^y \equiv_8 5$ (using that a is odd) but clearly $3^x \equiv_8 1 \mbox{ or } 3$ so there is no solution for $y>2$

Why does this hold? I can't see how you get $\equiv_8 5$ . I just get $\equiv_8 5 + 4^{y'}$ where $y' = \frac{y}{2}$ .

Other than this and your omission of the x=0 case, it all looks good!

This means I can stop running my maple program now. Pity. It has been on the go for over 6000 seconds now...

**DavidEriksson** · January 5th 2010, 03:57 AM

Originally Posted by Swlabr

Why does this hold? I can't see how you get $\equiv_8 5$ . I just get $\equiv_8 5 + 4^{y'}$ where $y' = \frac{y}{2}$ .

Other than this and your omission of the x=0 case, it all looks good!

This means I can stop running my maple program now. Pity. It has been on the go for over 5000 seconds now...

Hi!
When using "natural numbers" 0 is usually excluded, but i can add the case to the solution.

We know that a is odd and y is even. For $y>2$ it's easy to see that $8 \mid 2^y$ so $2^y \equiv_8 0$ for $y>2$

Since a is odd, $a=2k+1$ and thus $5^a=5^{2k+1} \equiv_8 5 \cdot 1$

**Defunkt** · January 5th 2010, 04:40 AM

It seems that Sudharaka's solution is incorrect. He claims that the equation $9x_1 + 16y_1=25z_1$ has an integer solution, and while that claim may be correct, we have no idea that this solutions will be of the form $(x_1,y_1,z_1) = (3^{x_0}, 4^{y_0}, 5^{z_0})$ with $x_0,y_0,z_0$ integers...

**Bingk** · January 5th 2010, 05:50 AM

Hi ... I've been disturbed by the proof by Sudharaka, because it somehow excludes the solution (0,1,1).

So, I've been looking into it, and I spotted what I think is a problem.

$x'$ , $y'$ , and $z'$ are integers, so when you let $x' = x_0 + 2$ , $y' = y_0 + 2$ and $z' = z_0 + 2$ , then $x_0$ , $y_0$ , and $z_0$ can be negative (but not less than -2).

This means that $x_1$ , $y_1$ , and $z_1$ are not necessarily integers, they may be fractions ... but what you have done is limited them to integer solutions ... I guess this means that your contradiction does not mean there are no other solutions to the given problem, but that $4^{x_0}$ can't be an integer ....

Defunkt, I think you're thinking about it the wrong way ... indeed, what he has shown is that the solutions for $9x_1 + 16y_1=25z_1$ cannot be of that form, hence no other solutions exist

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