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PEN Math problems from the collection "Problems in Elementary Number Theory"

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Peter

#1 h May 25, 2007, 12:25 AM • 1 Y

Y by Adventure10

Find all real numbers $\alpha$ for which the equality $\lfloor \sqrt{n}+\sqrt{n+\alpha}\rfloor =\lfloor \sqrt{4n+1}\rfloor$ holds for all positive integers $n$ .

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mszew

#2 h Oct 29, 2007, 6:31 PM • 2 Y

Y by Adventure10, Mango247

Notice that if $\alpha <0$ then it is not true for $n=1$
$[\sqrt {1} + \sqrt {1 + \alpha}] < 2=[\sqrt {5}]$
Same for $3 \le \alpha$
$[\sqrt {1} + \sqrt {1 + \alpha}]=[1 + \sqrt {1 + \alpha}] >2=[\sqrt {5}]$

$1 \le \alpha$
$1/2 \le \alpha /2$
$n^2 < n^2 + n \alpha$
$n < \sqrt {n^2+n \alpha}$
$n+1/2 \le \alpha /2+ \sqrt {n^2+n \alpha}$
$2n+1 \le \alpha +2 \sqrt {n^2+n \alpha}$
$4n+1 \le 2n+\alpha+ 2 \sqrt {n^2+n \alpha}$
$\sqrt {4n + 1} \le \sqrt {n} + \sqrt {n + \alpha}$

$1\le \alpha < 3$
$\alpha /2< 3/2$
$n^2+n \alpha < n^2 +3 n < n^2$
$\sqrt {n^2+n \alpha} < n$
$\alpha /2+\sqrt {n^2+n \alpha}< 3/2+n$
$\alpha /2+\sqrt {n^2+n \alpha}< n+1+\sqrt {4n + 1}$
$\alpha +2 \sqrt {n^2+n \alpha}< 2n+2+2 \sqrt {4n + 1}$
$2n+\alpha+ 2 \sqrt {n^2+n \alpha}< 4n+1+2 \sqrt {4n + 1}+1$
$(\sqrt {n} + \sqrt {n + \alpha})^2 < (\sqrt {4n + 1}+ 1)^2$
$\sqrt {n} + \sqrt {n + \alpha}- \sqrt {4n + 1} < 1$
then $0 \le \sqrt {n} + \sqrt {n + \alpha}- \sqrt {4n + 1} < 1$ for $1\le \alpha <3$

Idem for $0\le \alpha \le 1$ but $0 \le \sqrt {4n + 1}-\sqrt {n} + \sqrt {n + \alpha}< 1$

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Rust

#3 h Oct 30, 2007, 11:45 AM • 2 Y

Y by Adventure10, Mango247

Peter wrote:

Find all real numbers $\alpha$ for which the equality
$\lfloor \sqrt {n} + \sqrt {n + \alpha}\rfloor = \lfloor \sqrt {4n + 1}\rfloor$
holds for all positive integers $n$ .

Because $(\sqrt x )'' < 0$ we get $\sqrt x + \sqrt {x + \alpha } < 2\sqrt {x + \alpha/2} = \sqrt {4x + 2\alpha}$ .
If $\alpha>2$ for $n=k^2-1$ we get
$\sqrt {k^2 - 1} + \sqrt {k^2 - 1 + \alpha } < 2k$ .
It is equavalent to $(\alpha -2)k^2 < \alpha ^2$ and false for big $k$ .
Therefore solution is $9 - 6\sqrt 2 \le \alpha \le 2$ .
If it must be true for $n = 0$ solution is $1\le \alpha \le 2.$

This post has been edited 1 time. Last edited by Rust, Oct 30, 2007, 1:21 PM

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mszew

#4 h Oct 30, 2007, 12:30 PM • 1 Y

Y by Adventure10

Rust: Do you have a counterexample for $\alpha >2$ ?
Could you please find the mistake in my previous message?

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Rust

#5 h Oct 30, 2007, 1:23 PM • 2 Y

Y by Adventure10, Mango247

If $\alpha > 2$ , then for $n = k^2 - 1, \ k > \frac {\alpha }{\sqrt {\alpha -2}}$ give contradition.

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Woett

#6 h Mar 14, 2021, 2:36 PM

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I agree with Rust that the equality holds for all $n$ if, and only if, $9 - 6\sqrt{2} \le \alpha \le 2$ . And note that $9 - 6\sqrt{2} \approx 0.51$ .

Proof. For $n = 1$ we can readily check that the equality holds for all $\alpha$ with $0 \le \alpha < 3$ , so in particular for $\alpha \in [9 - 6\sqrt{2}, 2]$ . We may therefore assume $n \ge 2$ . To prove our claim, we must show both the necessity and the sufficiency of both the lower and upper bound.
To prove the necessity of the lower bound, we look at $n=2$ . It follows that $\alpha$ must be such that $\sqrt{2+\alpha} \ge 3 - \sqrt{2}$ . Squaring both sides and subtracting $2$ gives $\alpha \ge 9 - 6\sqrt{2}$ .
To prove the necessity of the upper bound, assume by contradiction $\alpha = 2 + \epsilon$ with $\epsilon > 0$ and choose $n > \frac{1}{\epsilon}$ such that $n+1$ is a square, say $n+1 = k^2$ . Then we have the following inequalities:

$\begin{align*} 4n(n + 2 + \epsilon) &> 4n^2 + 8n + 4 \\ 2\sqrt{n}\sqrt{n+2+\epsilon} &> 2n + 2 > 2n + 2 - \epsilon \\ 2n+2+\epsilon + 2\sqrt{n}\sqrt{n+2+\epsilon} &> 4n + 4 \\ (\sqrt{n} + \sqrt{n+2+\epsilon})^2 &> 4n + 4 \\ \sqrt{n} + \sqrt{n+2+\epsilon} &> 2\sqrt{n+1} = 2k \end{align*}$
On the other hand, $\sqrt{4n+1} < \sqrt{4n+4} = 2k$ , contradicting the equality.

To prove that these bounds are also sufficient, first we claim that $\sqrt{n} + \sqrt{n+\alpha} \ge \sqrt{4n+1}$ holds for all $n \ge 2$ , whenever $\alpha \ge 9 - 6\sqrt{2}$ . It suffices to prove this for $\alpha = 9 - 6\sqrt{2}$ , so let us assume that for the moment. By the quadratic formula we see $8(2\alpha-1) - (\alpha-1)^2 = -\alpha^2 + 18\alpha - 9 = 0$ . For $n \ge 2$ we therefore have the following inequalities:

$\begin{align*} 4n(2\alpha-1) - (\alpha-1)^2 &\ge 0 \\ 8\alpha n &\ge 4n + 1 - 2\alpha + \alpha^2 \\ 4n^2 + 4\alpha n &\ge 4n^2 + 4n + 1 - 4\alpha n - 2\alpha + \alpha^2 \\ 4n(n + \alpha) &\ge (2n+1)^2 - 2(2n+1)\alpha + \alpha^2 = ((2n+1) - \alpha)^2 \\ 2\sqrt{n}\sqrt{n+\alpha} &\ge 2n + 1 - \alpha \\ 2n + \alpha + 2\sqrt{n}\sqrt{n+\alpha} &\ge 4n + 1 \\ (\sqrt{n} + \sqrt{n+\alpha})^2 &\ge 4n + 1 \\ \sqrt{n} + \sqrt{n+\alpha} &\ge \sqrt{4n+1} \end{align*}$
With the above inequality in mind, we can finish the proof by showing that $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le \lfloor \sqrt{4n+1} \rfloor$ holds for all $n$ . Assume $k^2 \le n < (k+1)^2$ . We then have to distinguish between $n < k^2 + k$ or $n \ge k^2+k$ .
If $k^2 \le n \le k^2 + k - 1$ , then $\lfloor \sqrt{4n+1} \rfloor = 2k$ and $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le \lfloor 2\sqrt{n+1} \rfloor = 2k$ , since $\sqrt{n+1} < k + \frac{1}{2}$ .
On the other hand, if $k^2 + k \le n \le k^2 + 2k$ , then $\lfloor \sqrt{4n+1} \rfloor = 2k+1$ . And $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le 2k+1$ , since $\sqrt{n} + \sqrt{n + 2} < 2\sqrt{n+1} \le 2k+2$ .

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