$2^n=7x^2+y^2$ solutions My problem is related to the equation from above. It actually is a very particular one. I noticed that for every positive integer $n$ there's ONE SINGLE solution $(x_1,y_1)$ so that $x_1$ and $y_1$ are ODD positive integers ( I didn't prove it, I tested it with a program through roughly 30 tests).
If that's true , then how can I prove it? I tried to make a proof by absurd reduction ( assuming the contrary of sentence , that is , there could be at least one more solution for some values of n but not necessarly any n and run into a contradiction) which didn't work, and personally I can't "feel" why it can't be more than one solution. I hope you could help me to prove it. Thanks in advance!
P.s.: $n>=3$ . 

2^n    ≡ y^2 mod 7  ⇒ y^2 < 7
It can be seen that y^2 can only have values 1, 2 and 4. Only 1 and 4 are acceptable if y is integer, that is y= (+ or -) 1 or y =(+ or -) 2. With these values of y some powers of n give integers for x, for example y=2, x =6 for n=8 or x=y=2 for n=5. 
This is the reason for existing only one single x and y for every n.
