# A Non-negative Power Puzzle

1. Nov 10, 2006

### K Sengupta

Determine all possible non-negative integer solutions to the equation:

3^x + 5^y = 7^z + 1

[ Comment:

* I have only been able to derive (x,y,z) = (0,0,0); (1,1,1) as two of the solutions to the given problem.

* It can be shown that each of x, y and z must be odd ( whenever x,y and z are positive) and x=4p+1, y=4q+1 for non-negative integers p and q.

* I have not been able to derive any other solution or able to conclusively prove the non-existance of any other solutions apart from the ones mentioned above.

Proposed Methodology

Considering x=0, we obtain; 5^y = 7^z; implying y=z=0.
Similarly, considering y=0 and x=0 in turn, we respectively obtain, (x,z)=0 and (y,z) =(0,0)
Hence, if at least one of x, y and z is 0; then, it follows that :
(x,y,z) = (0,0,0)

For z=1, we obtain; 3^x+ 5^y = 8; which is feasible only if :
(x,y,z) = (1,1,1)

Now, reducing both sides of the given equation to Mod 3; we obtain:
2 ^y = 2 ( Mod3 ).; so that y must be odd.
so, 3^x + 5 = 7^z + 1 ( Mod 8).
Since, 3^x = 1, 3(mod 8) and 7^z = 1,7 (mod 8); it follows that both x and z must be odd.
Now, 3^x + 5^y = 7^z + 1 gives:
3^x + 5 = 7^y +1 (Mod 10).
Since, both x and z are odd; this is possible if only if x = 1+4p and z = 1+4q for non-negative integers p and q.

*** I am unable to proceed further.
** Consequently, I have not been able to derive solutions other than
(0,0,0); (1,1,1) or to prove that no solutions other than these two satisfies the provisions of the problem ].

Last edited: Nov 10, 2006
2. Nov 10, 2006

### AKG

(0,0,0) isn't a solution.

3. Nov 10, 2006

### Office_Shredder

Staff Emeritus
1+1=1+1 seems reasonable enough to me AKG.

Whenever I see something like this, if I know what the answer is, and get completely stuck, it's a good idea to just plug the generic answer in and see how you could have gotten there in the first place.

For your conclusion that y was odd, you were using the assumption that z=1 in the first place, so you don't have a general result for y

4. Nov 11, 2006

### K Sengupta

On Today 06:00 AM; Office_Shredder wrote:

On Yesterday 02:01 PM ; K Sengupta wrote:

I would like to clarify that the term "reduction" in terms of the second paragraph inclusive of the above quote corresponds to reduction to Mod 3 of the original equation, that is, 3^x + 5^y = 7^z + 1 ( and not the equation 3^x+ 5^y = 8).

Consquently, we would indeed obtain 2^y= 2 (Mod 3); implying inter-alia that y must be odd.

5. Nov 11, 2006

### AKG

Oh sorry, I totally misread the question.

6. Nov 18, 2006

### Playdo

Last edited: Nov 18, 2006
7. Nov 19, 2006

### Playdo

Ok, I slept on this I think I have a solution. First write the equation like this.

(1) $$3^x+5^y-7^z-1=0$$

You already know that x=y=z = 0 or 1 are solutions. now to prove there are no more equal positive integer solutions conciser the difference equation

(2) $$\Delta(x+1:x)=(3^{x+1}-3^x)+(5^{x+1}-5^x)-(6^{x+1}-6^x) - 1$$

simplifying gives

(3) $$\Delta(x+1:x)=2*3^x+4*5^x-5*6^x - 1$$

Plug in x = 0 to verify the change between (x,y,z) = (0,0,0) and (x,y,z = (1,1,1)

(4) $$\Delta(1:0)=2+4-5 - 1$$

So equation one is left unchanged by that move. However suppose we move from x=y=z =1 to x=y=z=2 we get

(5) $$\Delta(2:1)=6+20-30 - 1=5$$

therefore x=y=z=2 is not a solution and in fact the discrete analog of the derivative of (1) is increasing. So show that from here on out

(6) For x=y=z>2 $$\Delta(x+2:x+1)>\Delta(x+1:x)$$

Once that is done it implies that (1) is monotone increasing. so lets do it.

(7) $$2*3^{x+1}+4*5^{x+1}-5*6^{x+1}>2*3^x+4*5^x-5*6^x$$

(8) $$4*3^x+16*5^x>25*6^x$$

(9) $$\frac{4*3^x+16*5^x}{25*6^x}>1$$

(10) $$\frac{4*3^x}{25*6^x}+\frac{16*5^x}{25*6^x}>1$$

I am probably missing some details here because I am trying to figure in Tex, but I think you get the idea. But that is how you can analyze it. In fact you can use the difference equation to see when the derivative goes negative and then add up that chunk to see if it hits szero at the next crossing. Plot equation (1) as a function where x=y=z and you will see there are only probably two places where it crosses the origin. That should give you a clue. Then start with z and consider the curves x=y<z.
Then consider the curves x<y=z and x<y<z. For each of these you should be able to use graphical methods and difference equations to solve the problem. it doe seem pretty involved though, and probably there is a trick to it if it is a homework problem.

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