Can 15x² + y² = 2^2000 have solutions with both x and y odd?

[Anzas]

are there two odd numbers $$x,y$$
that are the solutions of the equation $$15x^2+y^2=2^{2000}$$

 

[Muzza]

For various reasons, the diophantine equation 15a + b = 2^2000 has the solutions

a = 2^2000 + 2^2000 * t,
b = -14*2^2000 - 15*2^2000 * t,

where t is some integer. Now, if x^2 = a and y^2 = b (with x, y natural), then it's necessary (but obviously not sufficient) that both a and b be positive. This places some severe restrictions on t, in fact, if you try to solve the system

a >= 0
b >= 0

you'll find that t = -1 is the only possibility, but then a = 0, which isn't odd.

 

[Hurkyl]

For various reasons, the diophantine equation 15a + b = 2^2000 has the solutions

a = 2^2000 + 2^2000 * t,
b = -14*2^2000 - 15*2^2000 * t,

I'd be curious to know those reasons...

 

[Muzza]

Damn it, it should be a = 2^2000 + t, b = -14*2^2000 - 15t... I messed up trying to distribute a multiplication over a parenthesis. This breaks the "solution". :(

 

[Anzas]

so i suppose the question remains open?

 

[Zurtex]

Are you sure it has any solutions?

I've messed about with this and found if there is a solution then y must be one of the 4 forms:

$$y = 1 + 30c \quad \text{or} \quad y = 11 + 30c \quad \text{or} \quad y = 19 + 30c \quad \text{or} \quad y = 29 + 30c \quad c \in \mathbb{Z}$$