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, exploring whether there are odd integer solutions for x and y. It is established that the derived equations for a and b impose strict conditions on the integer t, leading to the conclusion that t = -1 is the only viable option, resulting in a = 0, which cannot yield odd solutions. The conversation also highlights that if a solution exists, y must conform to specific forms related to modular arithmetic. Ultimately, the question of whether any odd solutions exist remains unresolved. The complexity of the problem suggests further exploration is necessary.
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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}
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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