How many solution in integers have to the following equation:

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Results 1 to 7 of 7

- Jul 17th 2010, 01:03 PM #1

- Jul 17th 2010, 01:44 PM #2
Obviously loads of solutions if n=1. Also if n=2, for example . 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 .

- Jul 17th 2010, 01:48 PM #3

- Jul 17th 2010, 01:59 PM #4

- Jul 17th 2010, 03:36 PM #5
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, 10:00 PM #6

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- Jul 19th 2010, 01:41 AM #7