How many solution in integers have to the following equation:

($\displaystyle n\in \mathbb{N}$)

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

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- Jul 17th 2010, 12:03 PMAlso sprach Zarathustrax^n+y^n=z^(n+1)
How many solution in integers have to the following equation:

($\displaystyle n\in \mathbb{N}$)

$\displaystyle x^n+y^n=z^{n+1}$ - Jul 17th 2010, 12:44 PMOpalg
Obviously loads of solutions if n=1. Also if n=2, for example $\displaystyle 11^2+2^2 = 125 = 5^3$. 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 $\displaystyle 2^n+2^n=2^{n+1}$. - Jul 17th 2010, 12:48 PMAlso sprach Zarathustra
And what about n>3?

- Jul 17th 2010, 12:59 PMOpalg
- Jul 17th 2010, 02:36 PMundefined
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))))`

Code:`f(1000,1000,3)`

f(1000,1000,4)

- Jul 18th 2010, 09:00 PMsimplependulum
- Jul 19th 2010, 12:41 AMAlso sprach Zarathustra