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I want to find the possible solution of the following problem:

$T_k= 21600$ seconds

$1281.49 < T < 21600 $(T in seconds)

$x$ and $y$ don't have unit

$x$ and $y$ have to be whole number

$x > 0$ and $y > 0$

$$0.5(2x+1)*T-0.5(2y+1)*T_k=0$$

I don't know which technique use.

Someone can guide me?

Thanks

1 Answers1

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This is equivalent to $$T(2x+1)=21600(2y+1) \\\frac{2x+1}{2y+1}=\frac{21600}{T}$$ Now, $1281.49\lt T\lt 21600 \implies 1\lt \frac{21600}{T} \lt 16.86$ i.e. $$1\lt \frac{2x+1}{2y+1} \lt 16.86 $$ We can choose infinitely many $x,y$ such that this inequality is true. As an example, consider $x=n, y=n-1, T=21600\left(\frac{2n-1}{2n+1}\right)$ for $n\in\mathbb N$ and $n\ge 2$.

Vishu
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