# Is my approach correct to this equation?

The problem is the following:

Does $$a \in \mathbb{R}$$ exist such that $$[a + \sqrt{2n + 1}] = [a + \sqrt{2n + 2}]$$ for all $$n \in \mathbb{N}$$? ($$[x]$$ denotes the whole part of $$x$$).

Note: I will also use $$\{x\}$$ to denote the fractional part of $$x$$. Also that $$[x] + \{x\} = x$$.

My approach is the following:

There is no such that $$a$$.

Proof: For the sake of simplicity, let's assume $$a = k + \alpha$$, $$k \in \mathbb{Z}$$ and $$0 \le \alpha \lt 1$$. Because of $$[x + k] = [x] + k$$ for any $$k \in \mathbb{Z}$$, and substituting $$a = k + \alpha$$, our equation $$[k + \alpha + \sqrt{2n + 1}] = [k + \alpha + \sqrt{2n + 2}]$$ becomes $$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}]$$. Now our task is to find $$\alpha \in [0, 1)$$ that satisfies the equation for any $$n \in \mathbb{N}$$.

First, let's find a formula for $$\alpha$$ for given $$n$$. There are two cases:

1. $$[\sqrt{2n + 1}] = [\sqrt{2n + 2}] = m$$, $$m \in \mathbb{N}$$. This happens when $$2n + 2$$ isn't a perfect square $$\iff$$ $$2n + 2 \ne p^2 \iff n \ne \frac {p^2} {2} - 1$$, $$p \in \mathbb{N}$$.

$$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}] \iff [\alpha + [\sqrt{2n + 1}] + \{\sqrt{2n + 1}\}] = [\alpha + [\sqrt{2n + 2}] + \{\sqrt{2n + 2}\}] \iff [\alpha + \{\sqrt{2n + 1}\} + m] = [\alpha + \{\sqrt{2n + 2}\} + m] \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha + \{\sqrt{2n + 2}\}]$$.
From this we get that $$\alpha \in [0, 1 - \{\sqrt{2n + 2}\}) \tag {1'}$$
Edit #1 (no longer actual beacuse of Edit #2): took out the maximum function from $$(1')$$ for obvious reason: the fractional part function is monotonically growing on its period, so if $$[x] = [y]$$, then $$\{x\} \lt \{y\}$$ only if $$x \lt y$$.
Edit #2: $$(1')$$ is only a partial solution for the 1. case, it is indeed a solution if both sides of the latest equation is equal to $$0$$. But they can both equal to $$1$$, so we have to consider that $$\alpha \in [1 - \{\sqrt{2n + 1}\}, 1)$$. 1. case overall:
$$\alpha \in [0, 1 - \{\sqrt{2n + 2}\}) \cup [1 - \{\sqrt{2n + 1}\}, 1) \tag 1$$

2. $$[\sqrt{2n + 1}] \ne [\sqrt{2n + 2}] \implies [\sqrt{2n + 1}] = m, [\sqrt{2n + 2}] = m + 1$$, $$m \in \mathbb{N}$$. This happens when $$2n + 2$$ is a perfect square $$\iff$$ $$2n + 2 = p^2 \iff n = \frac {p^2} {2} - 1$$, $$p \in \mathbb{N}$$. Now $$[\sqrt{2n + 2}] = \sqrt{2n + 2} \implies \{\sqrt{2n + 2}\} = 0$$.

$$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}] \iff [\alpha + [\sqrt{2n + 1}] + \{\sqrt{2n + 1}\}] = [\alpha + [\sqrt{2n + 2}] + \{\sqrt{2n + 2}\}] \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha + \{\sqrt{2n + 2}\}] + 1 \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha] + 1 \iff [\alpha + \{\sqrt{2n + 1}\}] = 1$$
From this we get that $$\alpha = 1 - \{\sqrt{2n + 1}\} \tag 2$$

Now we prove the following: (this is where it gets really ambiguous)
There does exists $$n_1, n_2 \in \mathbb{N}$$, $$n_1 \ne \frac {{p_1}^2} {2} - 1$$, $$n_2 = \frac {{p_2}^2} {2} - 1$$, $$p_1$$, $$p_2 \in \mathbb{N} \iff [\sqrt{2n_1 + 1}] = [\sqrt{2n_1 + 2}]$$, $$[\sqrt{2n_2 + 1}] \ne [\sqrt{2n_2 + 2}]$$ such that $$1 - \{\sqrt{2n_1 + 2}\} \le 1 - \{\sqrt{2n_2 + 1}\} \lt 1 - \{\sqrt{2n_1 + 1}\} \tag 3 \iff$$ there is no $$\alpha$$ satisfying both $$(1)$$ and $$(2)$$.
$$\implies \{\sqrt{2n_1 + 1}\} \lt \{\sqrt{2n_2 + 1}\} \le \{\sqrt{2n_1 + 2}\} \tag 4$$
(this is where it gets really-really-really ambiguous)

If we substitute $$n_1 = 3$$ and $$n_2 = 1$$ in $$(4)$$ we get the following:
$$\{\sqrt{7}\} \lt \{\sqrt{3}\} \le \{\sqrt{8}\} \implies \approx 0.64 \lt 0.73 \le 0.82$$ wich is indeed true. So we found two exceptions for $$n$$ and we should be done with the proof.

This part is strongly wrong! I made a mistake. $$(4)$$ is wrong. In fact, I can't find $$n_1$$ and $$n_2$$ satisfying $$(3)$$. Now I believe that I am wrong and there does exist a desired $$a$$.

Note: I made a lot of edits. My proof changed since I posted the question. Please take a look at the edits.

• @TheSimpliFire appears the quoted part in yellow should read "every" $n$ rather than "any" $n,$ as here the "any" is ambiguous. Now that I think of it, "all" is pretty good, then it reads "for all" – Will Jagy Jan 4 at 20:17
• @TheSimpliFire I am afraid that one of us didn't understand something properly. – Krisztián Kiss Jan 4 at 20:18
• @KrisztiánKiss I recommend you change the quoted phrase from "for any" to "for all" – Will Jagy Jan 4 at 20:21
• @WillJagy thank you. – Krisztián Kiss Jan 4 at 20:23

Your idea of looking at $$\alpha=\{a\}$$ is a good one.
If $$\alpha \lt 2-\sqrt 3\approx 0.268$$ we choose $$n=1$$ and note that $$\sqrt {2n+1}+\alpha=\sqrt 3+\alpha$$ has floor $$1$$ while $$\sqrt {2n+2}=\sqrt 4=2$$
If $$\alpha \ge 2-\sqrt 3$$ we want to choose $$n$$ so that $$\lfloor\alpha+\sqrt {2n+2}\rfloor = k \gt \lfloor \alpha +\sqrt {2n+1} \rfloor\\ \lfloor \alpha^2+2\alpha\sqrt{2n+2}+2n+2 \rfloor=k^2\gt \lfloor \alpha^2+2\alpha\sqrt{2n+1}+2n+1\rfloor \\ \lfloor \alpha^2+2\alpha\sqrt{2n+2}+1 \rfloor=k^2\gt \lfloor \alpha^2+2\alpha\sqrt{2n+1}\rfloor$$ We note that when $$2n$$ is a square $$\sqrt {2n+1}\approx \sqrt{2n}(1+\frac 1{2n})$$ (rounding up the Taylor series) so if we choose $$\sqrt{2n} \gt \frac 1\alpha$$ and $$2n$$ a square it will fail.
You can choose $$a=\frac{1}{2}$$. Then the equality holds for all $$n \in \mathbb{N}$$. We can prove it by contradiction. Assume there exists $$n$$ such that we don’t have equality. Then we can find $$q \in \mathbb{N}$$ such that $$\displaystyle \frac{1}{2} + \sqrt{2n+1} < q \leq \frac{1}{2} + \sqrt{2n+2}$$ This is equivalent to $$\displaystyle 8n+4 < (2q-1)^2 \leq 8n+8$$ In particular this implies that one of the four numbers $$8n+5, 8n+6, 8n+7, 8n+8$$ must be equal to $$(2q-1)^2$$. What is the remainder when we divide $$(2q-1)^2$$ by $$8$$? Well, we have $$\displaystyle (2q-1)^2 = 4q^2-4q+1 = 4q(q-1)+1.$$ Since one of the numbers $$q-1$$ and $$q$$ is even, the first summand is divisible by $$8$$ and hence $$(2q-1)^2\mod 8 =1$$. But the other four numbers obviously leave remainders $$5,6,7,$$ and $$8,$$ respectively. So there we have our contradiction, meaning that no such $$q$$ can exist.