Are There Odd Solutions for x and y in the Equation 15x² + y² = 2²⁰⁰⁰?

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The discussion centers on the diophantine equation 15x² + y² = 2²⁰⁰⁰, specifically investigating whether there are odd integer solutions for x and y. The derived solutions for a and b are a = 2²⁰⁰⁰ + 2²⁰⁰⁰ * t and b = -14*2²⁰⁰ - 15*2²⁰⁰ * t, with t being an integer. Analysis reveals that the only feasible value for t, which maintains positivity in both a and b, is t = -1, leading to a = 0, thus invalidating the odd solution requirement. The discussion concludes with the identification of potential forms for y, specifically y = 1 + 30c, y = 11 + 30c, y = 19 + 30c, or y = 29 + 30c, where c is an integer.

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are there two odd numbers x,y
that are the solutions of the equation 15x^2+y^2=2^{2000}
 
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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.
 
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... :confused:
 
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". :(
 
so i suppose the question remains open? :confused:
 
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}
 
Last edited:

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