Integer solutions to a equation

Question

Let and be positive integers bigger than n. If:

Prove that:

The first 3 pairs (k,c) and their ratio:

(3,2) - 1.5

(17,12) - 1.41666...

(99,70) - 1.414285714...

Answer 1

You have $$2+ \frac{1}{c^2}= \Big(\frac{k}{c}\Big)^2$$ Now notice that $n$ goes to $\infty$ implies $c$ goes to $\infty$ as $c>n$. thus we if we take $n\to \infty$ in the above equation LHS goes to 2, i.e. $$\lim_{n\to \infty} \frac{k}{c} = \sqrt{2}$$

Answer 2

We get $c<k$ at the first sight.

And from the equation we get

$2c^2+1=k^2, k>c>n>0$

Or $(k-√2c)(k+√2c)=1$.

Now $n \to \infty \Rightarrow k+√2c \to \infty $ as $k>c>n$. This means $k-√2c \to 0$ to result to $1$ on multiplication with $k+√2c$.

Means, $\lim \limits_{n \to \infty} \frac{k}{√2c}=1 \Rightarrow \lim \limits_{n \to \infty} \frac{k}{c}=√2$.

Answer 3

Since

$k, c > n, \tag 1$

we have

$c > 0; \tag 2$

thus

$k^2 = 1 + 2c^2 \tag 3$

implies

$\dfrac{k^2}{c^2} = \dfrac{1}{c^2} + 2;\tag 4$

now

$c \to \infty \; \text{as} \; n \to \infty, \tag 5$

whence

$\displaystyle \lim_{n \to \infty} \dfrac{k^2}{n^2} = 2; \tag 6$

therefore, since

$\dfrac{k}{c} > 0, \tag 7$

$\displaystyle \lim_{n \to \infty} \dfrac{k}{c} = \sqrt 2. \tag 8$