Intereting Posts

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fundamental group of the complement of a circle
The elliptic integral $\frac{K'}{K}=\sqrt{2}-1$ is known in closed form?
Show that $y_1$ and $y_2$ are not Linearly Independent
How do we prove that $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$?
iid variables, do they need to have the same mean and variance?
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number of derangements
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Fourier Series.
Calculating the divisors of the coordinate functions on an elliptic curve
Show that in a discrete metric space, every subset is both open and closed.
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Modular arithmetic with polynomial

Let $k$ be a “representable” positive integer, in the sense

that $k=|x^2-2y^2|$ for some integers $x,y$. Does it necessarily

follow that $k$ can also be represented with small parameters,

i.e. $k=|u^2-2v^2|$ with $|u|\leq\sqrt{k},|v|\leq\sqrt{k}$ ?

My (mostly useless) thoughts : if $N$ is very big and $a_N,b_N$

are the unique integers such that $a_N+b_N\sqrt{2}=(\sqrt{2}-1)^N(x+y\sqrt{2})$,

then $|a_N+b_N\sqrt{2}|$ can be arbitrarily small, but unfortunately

that doesn’t make the individual $a_N$ and $b_N$ small.

Also, we know that $k$ is representable iff all the prime divisors of

$k$ are such that $2$ is a quadratic residue to them, but that doesn’t seem

to help either.

- Algebraic integers of a cubic extension
- Conditions on integral cubic polynomial with cyclic group and prime coefficients
- The units of $\mathbb Z$
- Pre-requisites needed for algebraic number theory
- Is there an efficient algorithm to compute a minimal polynomial for the root of a polynomial with algebraic coefficients?
- Proving whether ideals are prime in $\mathbb{Z}$

- Units of p-adic integers
- Poles of a sum of functions
- How to determine the density of the set of completely splitting primes for a finite extension?
- Finding four numbers
- Ramification index in number fields extension
- Are there any good algebraic geometry books to recommend?
- Consecutive sets of consecutive numbers which add to the same total
- units of a ring of integers
- Order of operations (BODMAS)
- General misconception about $\sqrt x$

Appendix 1: all pairs $(u,v)$ in the tree depicted satisfy $u \geq 2v.$ As a result,

$$ k = u^2 – 2 v^2 \geq 4 v^2 – 2v^2 = 2 v^2, $$

so $$2 v^2 \leq k$$

and $$ \color{blue}{ v \leq \sqrt {\frac{k}{2}}}. $$

Appendix 2:

we may demand

$$ v \leq \frac{u}{2}. $$

Therefore

$$ 2 v^2 \leq \frac{u^2}{2}, $$

$$ -2 v^2 \geq – \frac{u^2}{2}, $$

$$ k = u^2 -2 v^2 \geq u^2 – \frac{u^2}{2} = \frac{u^2}{2}, $$

$$ u^2 \leq 2 k, $$

$$ \color{blue}{ u \leq \sqrt {2k}}. $$

preliminary: I already think you are roughly correct. The Conway topograph method deals most directly with $u+v$ when both are positive. The largest variables come from

$$ u = 2n + 1, \; \; v = n, \; \; u^2 – 2 v^2 = 2 n^2 + 4 n + 1 $$

Note that this “branch” of the tree illustrates both inequalities well, $ u \leq \sqrt {2k} $ and $ v \leq \sqrt {\frac{k}{2}}. $

I have answered several questions with these diagrams, also the book describing the method is at CONWAY. The point is that any (positive) number represented occurs in the first tree on the positive side of the river:

Just noticed the absolute values in the original question. If you are willing to represent $-k$ instead of $k,$ you get the bounds you wanted. This happens in an upside-down tree below the river where we have $u^2 – 2 v^2 = -k$ for positive $k,$ and with $u,v > 0$ and $v \geq u.$ Similar arguments to the above give your desired bounds,

$$ u \leq \sqrt k, \; \; \; v \leq \sqrt k. $$

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