# Proving that sequence elements satisfy inequality involving $\mod{1}$

I'm trying to prove that

$n(3-\sqrt{8}) \; (\!\!\! \mod{1}) < m(3-\sqrt{8}) \; (\!\!\! \mod{1})\;\; \forall \; m < n \;|\;n,m \in \mathbb{N}$ $\leftrightarrow n \in \{a_k:(a_k=6a_{k-1}-a_{k-2} \;| \;a_1=1 \wedge a_2=6)\}=\{1,6,35,204,1189,...\}$.

This shows that the sequence $a_k$ (OEIS page), gives the $x$'s such that $y=x(1+\sqrt{8})$ is closer to being an integer than when any lower integer $x$ is used.

$a_k$ has a couple of interesting properties that seem to tie it to the problem stated above:

• $a_k$ gives the values of $x$ in solutions to the Diophantine equation $y^2=1+8x^2$.
• $a_k=\lceil a_{k-1}(3+\sqrt{8}) \rceil$

Unfortunately, I think I'm in way over my head (I hardly even know modular arithmetic), but I just can't let this one go (I really should, exams are coming up!), so any insights into the connection with the items on the list, possible proof strategies, mod. arithmetic tricks that might be useful, etc., would be greatly appreciated!

I posted another question concerning the same basic problem, but the character of the problem/formulation has changed quite a bit, so I deemed it acceptable to post this one. If you disagree, please do let me know.

You had an if and only if in your question. I did one direction. Not sure about the other one yet.

LATER: Got it. Stuck on the end. If and only if is true.

ORIGINAL: It's true. Let your $n$ solve $w^2 - 8 n^2 = 1$ with $w > 0.$ Then $w > n \sqrt 8,$ then $w + n \sqrt 8 > 2 n \sqrt 8.$ Since $$(w + n \sqrt 8)(w - n \sqrt 8) = 1,$$ we get $$0 < w - n \sqrt 8 = \frac{1}{w + n \sqrt 8} < \frac{1}{2 n \sqrt 8}.$$ Now $$3n - n \sqrt 8 \equiv w - n \sqrt 8 \pmod 1.$$ With common notation $$\{ t \} = t - \lfloor t \rfloor,$$ we have $$\{ n (3 - \sqrt 8) \} < \frac{1}{2 n \sqrt 8},$$ then $$n \{ n (3 - \sqrt 8) \} < \frac{1}{2 \sqrt 8} \approx 0.177.$$

Suppose $m < n$ and $r^2 - 8 m^2 = A$ gives the smallest $A > 0$ possible. So $A \geq 1.$ $$0 < r - m \sqrt 8 = \frac{A}{r + m \sqrt 8} .$$ $$r + m \sqrt 8 > 2 m \sqrt 8,$$ so $$0 < r - m \sqrt 8 < \frac{A}{ 2 m \sqrt 8} ,$$ $$r + m \sqrt 8 < 2 m \sqrt 8 + \frac{A}{ 2 m \sqrt 8} = \frac{A + 32 m^2}{2 m \sqrt 8},$$ $$r - m \sqrt 8 = \frac{A}{r + m \sqrt 8} > \frac{2 A m \sqrt 8}{A + 32 m^2}.$$ We have again $$3m - m \sqrt 8 \equiv r - m \sqrt 8 \pmod 1.$$ $$\{ m (3 - \sqrt 8) \} > \frac{2 A m \sqrt 8}{A + 32 m^2}.$$ $$m \{ m (3 - \sqrt 8) \} > \frac{2 A m^2 \sqrt 8}{A + 32 m^2} = \frac{4 A m^2 \sqrt 2}{A + 32 m^2} = \frac{ A \sqrt 2}{8 + \frac{A}{4 m^2} }.$$

If $A > 1$ then $A \geq 4.$ Then $$m \{ m (3 - \sqrt 8) \} > \frac{ 4 \sqrt 2}{8 + \frac{4}{ m^2} } \geq \frac{ 4 \sqrt 2}{9 } \approx 0.628.$$ Since this $m < n$ we are finished with this case.

I'm not completely sure how well $A = 1$ works, let us see. Note that $A = 1$ means that this $m$ is an earlier one in the sequence $1,6,35..$ $$m \{ m (3 - \sqrt 8) \} > \frac{ A \sqrt 2}{8 + \frac{A}{4 m^2} }.$$ With $A=1$ we have $$m \{ m (3 - \sqrt 8) \} > \frac{ 1 \sqrt 2}{8 + \frac{1}{4 m^2} } \geq \frac{ 1 \sqrt 2}{8.25 } \approx 0.17142.$$ This is on the wrong side of $0.177.$ However, since $m$ is earlier in the main sequence, we know $5 m < n$ and $(1/m) > (5/n).$ We have the pair $$\{ m (3 - \sqrt 8) \} \geq \frac{ 1 \sqrt 2}{8.25 m} > \frac{ 5 \sqrt 2}{8.25 n} \approx \frac{0.857099}{n}$$ $$\{ n (3 - \sqrt 8) \} < \frac{1}{2 n \sqrt 8} \approx \frac{0.17677}{n}.$$ So this case works as well.

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Did some experiments, the if and only if works. The $n$ do best because we have $w^2 - 8 n^2 = 1.$

The big thing is that the ratio of consecutive $n$ values is below $6.$ So, if we take some other $m$ between two of these, we can demand $$m < 6 n$$ comparing with the previous $n.$ If, in addition, $$r^2 - 8 m^2 = A \geq 8,$$ we find $$\{ m (3 - \sqrt 8) \} > \frac{2 A m \sqrt 8}{A + 32 m^2} > \frac{12 A n \sqrt 8}{A + 1152 n^2}.$$ In turn, with $A \geq 8,$ $$\{ m (3 - \sqrt 8) \} > \frac{96 n \sqrt 8}{8 + 1152 n^2} > \frac{1}{2 n \sqrt 8},$$ meaning that $m$ does not do as well as the $n$ value below it.

Finally, if $A = 4,$ we are saying $r^2 - 8 m^2 = 4.$ It happens, number theory, that this can happen only when $m = 2n.$ That is, $w^2 - 8 n^2 = 1,$ followed by $(2w)^2 - 8 \cdot (2n)^2 = 4.$ A full proof of this is a long slog in quadratic forms; in case you are not familiar, we can find all solutions to $u^2 - 2 v^2 = 1,$ and all the $v$ must be even (check mod 8), they are exactly double your $n$ values. The relief, though, is that then $$\{ m (3 - \sqrt 8) \} = \{2 n (3 - \sqrt 8)\} = 2 \{ n (3 - \sqrt 8)\}$$ because $\{ n (3 - \sqrt 8)\} < 1/2$

• I discovered that the consequence of this being that the sequence gives the nearest-integer solutions actually only follows if $n(3-\sqrt{8}) \; (\!\!\! \mod{1}) < 1 - [m(3-\sqrt{8}) \; (\!\!\! \mod{1})]$ is also true. Is there a way to see that this follows trivially from the already proven? Commented Jan 10, 2016 at 10:28