# x^n+y^n=z^(n+1)

• Jul 17th 2010, 01:03 PM
Also sprach Zarathustra
x^n+y^n=z^(n+1)
How many solution in integers have to the following equation:
( $n\in \mathbb{N}$)

$x^n+y^n=z^{n+1}$
• Jul 17th 2010, 01:44 PM
Opalg
Quote:

Originally Posted by Also sprach Zarathustra
How many solution in integers have to the following equation:
( $n\in \mathbb{N}$)

$x^n+y^n=z^{n+1}$

Obviously loads of solutions if n=1. Also if n=2, for example $11^2+2^2 = 125 = 5^3$. In fact, let z be any number with the property that each of its prime factors of the form 4k+3 occurs to an even power. Then z^3 is the sum of two squares.

Another solution, valid for all n, is that $2^n+2^n=2^{n+1}$.
• Jul 17th 2010, 01:48 PM
Also sprach Zarathustra
• Jul 17th 2010, 01:59 PM
Opalg
Quote:

Originally Posted by Also sprach Zarathustra

Apart from that solution x = y = z = 2, I have no idea.
• Jul 17th 2010, 03:36 PM
undefined
For n=3, we have some solutions x=y=2^1, 2^5, 2^9, 2^13, ...

For n=4, we have some solutions x=y=2^1, 2^6, 2^11, 2^16, ...

etc.

Edit: More solutions for n=3 given by (x,y)

9,18
28,84
65,260
70,105
126,630
144,288
162,162
266,665
273,364
456,760
469,603
756,945
793,854

n = 4

17,34
82,246
194,291
486,486

PARI/GP

Code:

f(x,y,z)=for(i=1,x,for(j=i,y,a=i^z+j^z;b=round(a^(1/(z+1)));if(b^(z+1)==a,print(i," ",j))))
Then for example

Code:

f(1000,1000,3) f(1000,1000,4)
• Jul 18th 2010, 10:00 PM
simplependulum
Quote:

Originally Posted by Opalg
Obviously loads of solutions if n=1. Also if n=2, for example $11^2+2^2 = 125 = 5^3$. In fact, let z be any number with the property that each of its prime factors of the form 4k+3 occurs to an even power. Then z^3 is the sum of two squares.

Another solution, valid for all n, is that $2^n+2^n=2^{n+1}$.

Yes, then the general solution for $n = 2$ is given by :

$(x,y,z) = ( p^3 - 3pq^2 , 3p^2 q - q^3 , p^2 + q^2 )$
• Jul 19th 2010, 01:41 AM
Also sprach Zarathustra
Quote:

Originally Posted by simplependulum
Yes, then the general solution for $n = 2$ is given by :

$(x,y,z) = ( p^3 - 3pq^2 , 3p^2 q - q^3 , p^2 + q^2 )$

You are very close to the solution...