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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
ISI UGB 2025 P2
SomeonecoolLovesMaths   4
N a few seconds ago by MathsSolver007
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
4 replies
SomeonecoolLovesMaths
Today at 11:16 AM
MathsSolver007
a few seconds ago
IMO ShortList 1998, combinatorics theory problem 5
orl   47
N 4 minutes ago by mathwiz_1207
Source: IMO ShortList 1998, combinatorics theory problem 5
In a contest, there are $m$ candidates and $n$ judges, where $n\geq 3$ is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most $k$ candidates. Prove that \[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]
47 replies
orl
Oct 22, 2004
mathwiz_1207
4 minutes ago
Cyclic equality implies equal sum of squares
blackbluecar   34
N 4 minutes ago by Markas
Source: 2021 Iberoamerican Mathematical Olympiad, P4
Let $a,b,c,x,y,z$ be real numbers such that

\[ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 \]
Show that $a^2+b^2+c^2=x^2+y^2+z^2$.
34 replies
blackbluecar
Oct 21, 2021
Markas
4 minutes ago
Common tangent to diameter circles
Stuttgarden   5
N 6 minutes ago by zuat.e
Source: Spain MO 2025 P2
The cyclic quadrilateral $ABCD$, inscribed in the circle $\Gamma$, satisfies $AB=BC$ and $CD=DA$, and $E$ is the intersection point of the diagonals $AC$ and $BD$. The circle with center $A$ and radius $AE$ intersects $\Gamma$ in two points $F$ and $G$. Prove that the line $FG$ is tangent to the circles with diameters $BE$ and $DE$.
5 replies
Stuttgarden
Mar 31, 2025
zuat.e
6 minutes ago
2020 EGMO P2: Sum inequality with permutations
alifenix-   29
N 7 minutes ago by Markas
Source: 2020 EGMO P2
Find all lists $(x_1, x_2, \ldots, x_{2020})$ of non-negative real numbers such that the following three conditions are all satisfied:

[list]
[*] $x_1 \le x_2 \le \ldots \le x_{2020}$;
[*] $x_{2020} \le x_1  + 1$;
[*] there is a permutation $(y_1, y_2, \ldots, y_{2020})$ of $(x_1, x_2, \ldots, x_{2020})$ such that $$\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3.$$[/list]

A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2, 1, 2)$ is a permutation of $(1, 2, 2)$, and they are both permutations of $(2, 2, 1)$. Note that any list is a permutation of itself.
29 replies
alifenix-
Apr 18, 2020
Markas
7 minutes ago
IMO 2018 Problem 2
juckter   97
N 10 minutes ago by Markas
Find all integers $n \geq 3$ for which there exist real numbers $a_1, a_2, \dots a_{n + 2}$ satisfying $a_{n + 1} = a_1$, $a_{n + 2} = a_2$ and
$$a_ia_{i + 1} + 1 = a_{i + 2},$$for $i = 1, 2, \dots, n$.

Proposed by Patrik Bak, Slovakia
97 replies
juckter
Jul 9, 2018
Markas
10 minutes ago
prove that a_50 + b_50 > 20
kamatadu   8
N 11 minutes ago by Markas
Source: Canada Training Camp
The sequences $a_n$ and $b_n$ are such that, for every positive integer $n$,
\[ a_n > 0,\qquad\ b_n>0,\qquad\ a_{n+1}=a_n+\dfrac{1}{b_n},\qquad\ b_{n+1} = b_n+\dfrac{1}{a_n}. \]Prove that $a_{50} + b_{50} > 20$.
8 replies
kamatadu
Dec 30, 2023
Markas
11 minutes ago
EGMO P4 infinite sequence
aditya21   29
N 12 minutes ago by Markas
Source: EGMO 2015, Problem 4
Determine whether there exists an infinite sequence $a_1, a_2, a_3, \dots$ of positive integers
which satisfies the equality \[a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}} \] for every positive integer $n$.
29 replies
aditya21
Apr 17, 2015
Markas
12 minutes ago
IMO 2014 Problem 1
Amir Hossein   133
N 13 minutes ago by Markas
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
133 replies
Amir Hossein
Jul 8, 2014
Markas
13 minutes ago
Sequences and limit
lehungvietbao   16
N 14 minutes ago by Markas
Source: Vietnam Mathematical OLympiad 2014
Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and
\[ \begin{cases}  {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\   x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all $n=1,2,3,\ldots$.
Prove that they are converges and find their limits.
16 replies
lehungvietbao
Jan 3, 2014
Markas
14 minutes ago
Real triples
juckter   67
N 14 minutes ago by Markas
Source: EGMO 2019 Problem 1
Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and

$$a^2b + c = b^2c + a = c^2a + b.$$
67 replies
juckter
Apr 9, 2019
Markas
14 minutes ago
Social Club with 2k+1 Members
v_Enhance   24
N 26 minutes ago by mathwiz_1207
Source: USA December TST for IMO 2013, Problem 1
A social club has $2k+1$ members, each of whom is fluent in the same $k$ languages. Any pair of members always talk to each other in only one language. Suppose that there were no three members such that they use only one language among them. Let $A$ be the number of three-member subsets such that the three distinct pairs among them use different languages. Find the maximum possible value of $A$.
24 replies
v_Enhance
Jul 30, 2013
mathwiz_1207
26 minutes ago
A strong inequality problem
hn111009   1
N 32 minutes ago by Tung-CHL
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
1 reply
hn111009
Today at 2:02 AM
Tung-CHL
32 minutes ago
Altitude configuration with two touching circles
Tintarn   3
N 32 minutes ago by NumberzAndStuff
Source: Austrian MO 2024, Final Round P2
Let $ABC$ be an acute triangle with $AB>AC$. Let $D,E,F$ denote the feet of its altitudes on $BC,AC$ and $AB$, respectively. Let $S$ denote the intersection of lines $EF$ and $BC$. Prove that the circumcircles $k_1$ and $k_2$ of the two triangles $AEF$ and $DES$ touch in $E$.

(Karl Czakler)
3 replies
Tintarn
Jun 1, 2024
NumberzAndStuff
32 minutes ago
Floor of square root
v_Enhance   43
N Apr 25, 2025 by Ilikeminecraft
Source: APMO 2013, Problem 2
Determine all positive integers $n$ for which $\dfrac{n^2+1}{[\sqrt{n}]^2+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
43 replies
v_Enhance
May 3, 2013
Ilikeminecraft
Apr 25, 2025
Floor of square root
G H J
Source: APMO 2013, Problem 2
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v_Enhance
6877 posts
#1 • 14 Y
Y by abk2015, Davi-8191, integrated_JRC, itslumi, MathLuis, centslordm, guillermo.dinamarca, HWenslawski, GeoKing, Adventure10, Mathefishian, bjump, ItsBesi, and 1 other user
Determine all positive integers $n$ for which $\dfrac{n^2+1}{[\sqrt{n}]^2+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
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v_Enhance
6877 posts
#2 • 26 Y
Y by ssilwa, qua96, mathisfun7, emiliorosado, abk2015, raknum007, mathisreal, Promi, rashah76, mathleticguyyy, Muaaz.SY, itslumi, MathLuis, centslordm, Kimchiks926, guillermo.dinamarca, HWenslawski, Ahmad_Alo, samrocksnature, Halykov06, Lamboreghini, Adventure10, aidan0626, Mango247, Mathefishian, and 1 other user
This was a silly problem; there are no such $n$.

Let $n = m^2 + k$ where $0 \le k \le 2m$. Then we require \[
	\frac{(m^2+k)^2+1}{m^2+2} = m^2 + (2k-2) + \frac{(k-2)^2+1}{m^2+2}
	\] to be an integer. Now $(k-2)^2 + 1 < 4m^2 + 8$, and hence $\tfrac{(k-2)^2+1}{m^2+2}$ must be either $1$, $2$ or $3$. We will examine each case.

If $(k-2)^2+1 = m^2+2$, then $(k-2)^2 = m^2+1$, which is impossible unless $m=0$ and $k-2 = \pm 1$, so no solution in this case.

If $(k-2)^2+1 = 2m^2+4$, then $(k-2)^2 - 2m^2 = 3$. Modulo $3$ we get that $(k-2)^2 + m^2 \equiv 0 \pmod{3}$ which is enough to force $k-2 \equiv m \equiv 0 \pmod{3}$, and yet this implies $9 \mid (k-2)^2 - 2m^2 = 3$, contradiction.

Finally, if $(k-2)^2 + 1 = 3m^2 + 6$, then $(k-2)^2 - 3m^2 = 5$. Again, taking modulo $3$, we get $(k-2)^2 \equiv 2 \pmod{3}$ which is not possible.

So, no solutions for $n$.
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IDMasterz
1412 posts
#3 • 2 Y
Y by centslordm, Adventure10
It was.. pretty silly... though it was just a bounding argument really.
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hatchguy
555 posts
#4 • 5 Y
Y by Joy_Gomuj, centslordm, lrjr24, Adventure10, Mango247
Another way to solve $m^2 + 2 | a^2 +1 $.

Notice that if a prime $p$ divides $a^2+1$ then $p$ is of the form $4k+1$. So all prime divisors of $m^2 +2 $ are of the form $4k + 1$. However, this would imply that $m^2 +2 \equiv 1 \pmod 4 \implies m^2 \equiv 3 \pmod 4$, which can't happen.

EDIT: V_enhance is right, wrong way to finish the problem.
This post has been edited 1 time. Last edited by hatchguy, May 4, 2013, 4:51 AM
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v_Enhance
6877 posts
#5 • 6 Y
Y by centslordm, SSaad, Adventure10, Mango247, Mathefishian, and 1 other user
hatchguy wrote:
Notice that if a prime $p$ divides $a^2+1$ then $p$ is of the form $4k+1$. So all prime divisors of $m^2 +2 $ are of the form $4k + 1$. However, this would imply that $m^2 +2 \equiv 1 \pmod 4 \implies m^2 \equiv 3 \pmod 4$, which can't happen.
If $a$ is even, then it's possible that $m^2 + 2 \equiv 2 \pmod{4} \implies m^2 \equiv 0 \pmod{4}$, which is definitely permissible. In particular, $(m,a) = (12, 27)$ is a solution here.
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Nguyenhuyhoang
207 posts
#6 • 3 Y
Y by centslordm, Adventure10, Mango247
From v_Enhance's solution, we have that $k \leq 2m$, and after some calculations, we have $3k+1 \vdots p^2+2$. Combine these two and use the inequality method, we easily found that there is no solution.
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JSGandora
4216 posts
#7 • 4 Y
Y by emiliorosado, centslordm, Adventure10, Mango247
Let $m^2$ be the greatest square less than or equal to $n$, then $m^2\leq n<(m+1)^2$. Now let $n=m^2+r$ where $0\leq r<2m+1$. So $[\sqrt{n}]^2=m^2$ and thus
\begin{align*}\frac{n^2+1}{[\sqrt{n}]^2+2} &=\frac{(m^2+r)^2+1}{m^2+2}\\
&=\frac{m^4+2m^2r+r^2+1}{m^2+2} \\
&=m^2+\frac{2m^2r+r^2+1-2m^2}{m^2+2} \\
&=m^2+(2r-2)+\frac{r^2+1-4r+4}{m^2+2} \\
&=m^2+2r-2+\frac{r^2-4r+5}{m^2+2}
\end{align*}
We place a bound on the numerator:
\[r^2-4r+5\leq (2m+1)^2-4(2m+1)+5=4m^2-4m+2< 4(m^2+2).\]
Therefore in order for the expression to be an integer, we must have
\begin{align*}r^2-4r+5=(r-2)^2+1&=\{m^2+2, 2m^2+4, 3m^2+6\} \\ 
(r-2)^2&=\{m^2+1, 2m^2+3, 3m^2+5 \}
\end{align*}
The first case $m^2+1$ is impossible since there are no two consecutive numbers that are squares. The second case is impossible by taking modulo $8$ since the quadratic residues modulo $8$ are $0, 1,$ and $4$, then $2m^2+3$ can be congruent to $3, 5, 3$, none of which are quadratic residues. And the last case take modulo $3$ and see that $3m^2+5$ is congruent to $2$ modulo $3$ which is not a quadratic residue. Therefore there are no such integers $n$. $\blacksquare$
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fclvbfm934
759 posts
#8 • 3 Y
Y by centslordm, Adventure10, Mango247
Let $n = m^2 + k$, where $0 \le k \le 2m$. Therefore, $[\sqrt{n}] = m$. We are therefore trying to see when $\frac{m^4 + 2m^k + k^2 + 1}{m^2 + 2}$ is an integer. We rewrite the fraction as $m^2 + 2k-2 + \frac{k^2 - 4k + 5}{m^2 + 2}$. So we want $m^2 + 2 | k^2 - 4k + 5$. But notice that since $k \le 2m$, we have $k^2 - 4k + 5 \le k^2 + 5 \le 4m^2 + 5$. Therefore, if $\frac{k^2 - 4k + 5}{m^2 + 2}$ is an integer, then it is either $1, 2, 3$. We now go through the casework:

Case 1: the integer is 1:
Therefore, $k^2 - 4k + 5 = m^2 + 2$ which leads to $(k-2)^2  = m^2 + 1$. The only squares the differ by $1$ are $0$ and $1$. which would give us $m = 0$, contradiction. There are no solutions for this case.

Case 2: the integer is 2:
We then have $(k-2)^2 = 2m^2 + 3$. Let $a = k-2$ and $b = m$. Then, we have $a^2 - 2b^2 = 3 \Rightarrow a^2 \equiv 2b^2 \pmod{3}$. Since $\left( \frac{2}{3} \right) = -1$, we see that $3|a, b$ but then $9|a^2 - 2b^2 \Rightarrow 9|3$, yet another contradiction. There are no solutions for this case.

Case 3: the integer is 3:
We then have $(k-2)^2 = 3m^2 + 5$ which would give us $(k-2)^2 \equiv 2 \pmod{3}$, yet another contradiction.

Therefore, there are no solutions at all.
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Taussig
23 posts
#9 • 2 Y
Y by centslordm, Adventure10
At v enhance, what was your reasoning behind defining m and k? I don't understand. Please explain.
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leminscate
109 posts
#10 • 3 Y
Y by centslordm, Adventure10, Mango247
Basically, it's so that you don't have the floor function of the square root hanging around, as it is annoying to deal with. So you let $[\sqrt{n}]=m$, i.e. $m^2 \leq n \leq (m+1)^2-1 = m^2+2m$.
We can then let $n=m^2+k$ where $0 \leq k \leq 2m$. The rest of the solution is then standard divisibility ideas and bounding.
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Taussig
23 posts
#11 • 3 Y
Y by centslordm, Adventure10, Mango247
leminscate wrote:
Basically, it's so that you don't have the floor function of the square root hanging around, as it is annoying to deal with. So you let $[\sqrt{n}]=m$, i.e. $m^2 \leq n \leq (m+1)^2-1 = m^2+2m$.
We can then let $n=m^2+k$ where $0 \leq k \leq 2m$. The rest of the solution is then standard divisibility ideas and bounding.

Interesting, thank you for pointing that out. Would the k be just some unknown constant?
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SMOJ
2663 posts
#12 • 3 Y
Y by centslordm, Adventure10, Mango247
yes
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ind
133 posts
#13 • 3 Y
Y by centslordm, Adventure10, Mango247
if we divide by the usual method of long division we get remainder as$p^2-4p+5=0$
where$ p$ is the $r $used by gandora
but $p$ is not an integer.
We can even generalise it...
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bobthesmartypants
4337 posts
#14 • 3 Y
Y by centslordm, Adventure10, Mango247
My solution:

Let $n=x^2+2k$ where $0\le k\le 2x$; then $\lfloor \sqrt{n}\rfloor^2=x^2$ so we want to find when $\dfrac{(x^2+k)^2+1}{x^2+2}$ is integral.

After polynomial division this is equivalent to finding when $\dfrac{(k-2)^2+1}{x^2+2}$ is integral. Since $(k-2)^2+1 \le (2x-2)^2+1 < 4(x^2+2)$, we have three cases:

Case 1: $(k-2)^2+1=x^2+2$. In this case $(k-2)^2=x^2+1$ clear contradiction.

Case 2: $(k-2)^2+1=2(x^2+2)$. In this case $(k-2)^2-3=2x^2$; taking mod 8 gives the LHS equal to 1, 5, or 6 mod 8 and the only possible one is 6 mod 8. But then $2x^2\equiv 6\pmod{8}\implies x^2\equiv 3\pmod{4}$ contradiction.

Case 3: $(k-2)^2+1=3(x^2+2)$. In this case $(k-2)^2-5=3x^2$. Taking mod 9, the LHS can be 4, 5, 8, or 2 mod 9 which are all impossible, contradiction.

Thus there are no solutions.

EDIT: oops v_Enhance pointed out above that taking mod 3 for cases 2 and 3 suffice...
This post has been edited 1 time. Last edited by bobthesmartypants, Mar 8, 2016, 7:16 PM
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songssari
102 posts
#15 • 3 Y
Y by centslordm, Adventure10, Mango247
v_Enhance wrote:
This was a silly problem; there are no such $n$.

Let $n = m^2 + k$ where $0 \le k \le 2m$. Then we require \[
	\frac{(m^2+k)^2+1}{m^2+2} = m^2 + (2k-2) + \frac{(k-2)^2+1}{m^2+2}
	\]to be an integer. Now $(k-2)^2 + 1 < 4m^2 + 8$, and hence $\tfrac{(k-2)^2+1}{m^2+2}$ must be either $1$, $2$ or $3$. We will examine each case.

If $(k-2)^2+1 = m^2+2$, then $(k-2)^2 = m^2+1$, which is impossible unless $m=0$ and $k-2 = \pm 1$, so no solution in this case.

If $(k-2)^2+1 = 2m^2+4$, then $(k-2)^2 - 2m^2 = 3$. Modulo $3$ we get that $(k-2)^2 + m^2 \equiv 0 \pmod{3}$ which is enough to force $k-2 \equiv m \equiv 0 \pmod{3}$, and yet this implies $9 \mid (k-2)^2 - 2m^2 = 3$, contradiction.

Finally, if $(k-2)^2 + 1 = 3m^2 + 6$, then $(k-2)^2 - 3m^2 = 5$. Again, taking modulo $3$, we get $(k-2)^2 \equiv 2 \pmod{3}$ which is not possible.

So, no solutions for $n$.
Or I think you can look at the second case modulo 8...
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Vrangr
1600 posts
#16 • 4 Y
Y by AlastorMoody, centslordm, Adventure10, Mango247
Solution
This post has been edited 7 times. Last edited by Vrangr, May 3, 2020, 9:39 AM
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Mathotsav
1508 posts
#17 • 3 Y
Y by centslordm, Adventure10, Mango247
v_Enhance wrote:
Determine all positive integers $n$ for which $\dfrac{n^2+1}{[\sqrt{n}]^2+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
Solution:
Let $n=k^2+l$ with $l \leq 2k$.
So we have $k^2+2|(k^2+l)^2+1$ or $k^2+2|k^4+2k^2l+l^2+1$. As $k^2+2|k^4-4$ and $k^2+2|2k^2l+4l$ we get that $k^2+2|l^2+4l+5=(l+2)^2+1$. If $k$ is odd then $k^2+2$ is $3$ mod $4$ but since $(l+2)^2+1$ has no divisor of the form $4k+3$ this gives a contradiction. So let $l=2m$. Thus $4m^2+2|(l+2)^2+1$. So $l$ is odd so let $l=2s-1$ for some $n$. So we have $2m^2+1|2s^2+2s+1$. We get that $s \leq n$ or $s \leq 2m$. Thus $2s^2+2s+1 \leq 8m^2+8m+1 \leq 9m^2+1$ for $m \geq 8$. So for $m \geq 8$ if $2s^2+2s+=k(2m^2+1)$ then $k$ is odd, obviously $k \neq 1$ as $s^2+s$ can never be equal to $m^2$ for positive integers $m,s$, and $k \leq 4$. So we have $k=3$. So we get that $3|2s^2+2s+1$ which is a contradiction by mod $3$ check. Thus $m<8$ if $2m^2+1|2s^2+2s+1$. One can easily check that no solutions exist for $m<8$. Hence no solutions for $m$ exist. Thus no solutions for the original problem exist
This post has been edited 1 time. Last edited by Mathotsav, Oct 12, 2019, 6:56 PM
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Lukaluce
268 posts
#18 • 1 Y
Y by centslordm
v_Enhance wrote:
hatchguy wrote:
Notice that if a prime $p$ divides $a^2+1$ then $p$ is of the form $4k+1$. So all prime divisors of $m^2 +2 $ are of the form $4k + 1$. However, this would imply that $m^2 +2 \equiv 1 \pmod 4 \implies m^2 \equiv 3 \pmod 4$, which can't happen.
If $a$ is even, then it's possible that $m^2 + 2 \equiv 2 \pmod{4} \implies m^2 \equiv 0 \pmod{4}$, which is definitely permissible. In particular, $(m,a) = (12, 27)$ is a solution here.

You probably wish to say "If $a$ is odd".
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IAmTheHazard
5001 posts
#19 • 1 Y
Y by centslordm
The answer is that there is no such $n$.
We first let $n=a^2+b$, where $0 \leq b \leq 2a$. Then, we need:
$$\frac{(a^2+b)^2+1}{a^2+2}=a^2+2(b-1)+\frac{(b-2)^2+1}{a^2+2}$$to be an integer. This is equivalent to saying $\tfrac{(b-2)^2+1}{a^2+2}$ is an integer, call this $k$. We now consider the following cases on which integer $k$ is:
Case 1: $k=1$. Then we need $(b-2)^2=a^2+1$. However, the only consecutive perfect squares are $0,1$, which implies $(b-2)^2=1$ and $a=0$. But since $0 \leq b \leq 2a$, we also need $b=0$ which is a contradiction.
Case 2: $k=2$. Then we need $(b-2)^2=2a^2+3$. Now take $\pmod{3}$, which shows that the LHS is either $0,1$ and the RHS is either $0,2$. This implies that $b \equiv 2 \pmod{1}$ and $a \equiv 0 \pmod{3}$, so $a=3m$ and $b=3n+2$ for some positive integers $m,n$. When we substitute, the equation becomes:
$$9n^2=18m^2+3$$wherupon taking $\pmod{9}$ implies that no solutions exist.
Case 3: $k=3$. Then we need $(b-2)^2+1=3a^2+6$. Taking $\pmod{3}$, the LHS is either $1,2$ and the RHS is $0$, so no solutions exist in this case.
Case 4: $k \geq 4$. This implies $(b-2)^2+1=ka^2+2k$. But for $a \geq 2$ we have:
\begin{align*}
(b-2)^2+1&\geq (2a-2)^2+1\\
&\geq (2a)^2+1\\
&\geq 4a^2+1\\
&\geq ka^2+1\\
&> ka^2+2k
\end{align*}so equality never holds (the reason we need $a \geq 2$ is because the first inequality is not true if $a=1$ and $b=0$, for instance). If $a=1$ the LHS is still less than the RHS, since the LHS is at most $2^2+1=5$ and the RHS is at least $12$. Thus there are no solutions in this case either.
Combining these cases finishes. $\blacksquare$
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MathLuis
1525 posts
#20 • 1 Y
Y by centslordm
v_Enhance wrote:
Determine all positive integers $n$ for which $\dfrac{n^2+1}{[\sqrt{n}]^2+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.

Problem not common :P . Assume that exists solutions:
It is known that $f(x)=x^2+c$ is surjective on $x \in \mathbb Z^+$. So we can let $n=a^2+b$ where $0 \le b \le 2a$ and hence $[\sqrt{n}]=a$ now lets use these informations:
$$a^2+2 \mid a^4+2a^2b+b^2+1 \implies a^2+2 \mid (b-2)^2+1$$$$b-2 \le 2(a-1) \implies (b-2)^2+1 \le 4a^2-8a+5<4a^2+8=4(a^2+2)$$That means $k(a^2+2)=(b-2)^2+1$ where $k \in (1,3)$
Case 1.- $k=1$
$$(b-2)^2=a^2+1 \implies (b-1)(b-3)=a^2 \implies b=4 \implies a^2=3 \; \text{contradiction!!}$$Case 2.- $k=2$
$$(b-2)^2 \equiv 2a^2+3 \pmod 3 \implies 9 \mid (b-2)^2 \; \text{and} \; 9 \mid a^2$$$$(b-2)^2-2a^2=3 \implies 3 \equiv 0 \pmod 9 \; \text{contradiction!!}$$Case 3.- $k=3$
$$(k-2)^2+1 \equiv 0 \pmod 3 \; \text{contradiction!!}$$Hence no such positive integers.
Thus we are done :blush:
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jasperE3
11321 posts
#21 • 2 Y
Y by centslordm, Mango247
If $n=m^2+k,k\in[0,2m]\cap\mathbb Z,m\in\mathbb Z$ then we have:
$$\frac{k^2-4k-3}{m^2+2}+m^2+2k-2\in\mathbb Z.$$By the aforementioned bounds, we obtain $\frac{k^2-4k-3}{m^2+2}\in\{1,2,3\}$.

$\textbf{Case 1: }\frac{k^2-4k-3}{m^2+2}=1$
$\Leftrightarrow(k-2)^2-m^2=1$, and it's well-known that the only squares which are one apart are $0$ and $1$, but there are no solutions here after testing $k\in\{1,3\}$, $m=0$.

$\textbf{Case 2: }\frac{k^2-4k-3}{m^2+2}=2$
We have $(k-2)^2=2m^2+3$, now$\pmod3$ gives that $k\equiv2$ and $m\equiv0$. Now $v_3(\text{LHS})\ge2$ while if $m=3n$, $v_3(\text{RHS})=v_3(3(6n^2+1))=1$, so no solutions.

$\textbf{Case 3: }\frac{k^2-4k-3}{m^2+2}=3$
So $(k-2)^2=3m^2+5$, by$\pmod5$ we have $(k-2)^2\equiv3m^2$, which enforces $k\equiv2$ and $m\equiv0$. Then $v_5(\text{LHS})\ge2$ while, if $m=5n$, $v_5(\text{RHS})=v_5(5(15n^2+1))=1$, contradiction.

No $n$ satisfy this condition. $\square$
This post has been edited 1 time. Last edited by jasperE3, May 22, 2021, 10:47 PM
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srisainandan6
2811 posts
#22 • 1 Y
Y by centslordm
Similar solution to everyone else, written in a hurry though.

Let $n=k^2+m$, where $k,m$ are both nonnegative integers and $m \in [0,2k]$.

We have that $\frac{{(k^2+m)}^2+1}{k^2+2} = k^2+2m-2+\frac{m^2-4m+5}{k^2+2}$ needs to be an integer, so $\frac{m^2-4m+5}{k^2+2}$ needs to be an integer. Since $m \in [0,2k]$, we have that $\frac{m^2-4m+5}{k^2+2}$ can only be $1,2,$ or $3$. I claim that it can't be any.

If the quantity is equal to $1$, we have that ${(m-2)}^2 = k^2+1$, and this clearly holds false.

If the quantity equals to $2$, we have that ${(m-2)}^2=2k^2+3$. Observing this in mod $3$, we see that $m \equiv 2 \mod 3$ and $k \equiv 0 \mod 3$. Let $m=3m'+2$ and $k=3k'$. Substituting these values in we get that $3{(m')}^2=6{(k')}^2+1$, which clearly isn't possible again by observing in modulo $3$.

If the quantity is equal to $3$, we have that ${(m-2)}^2=3k^2+5$. The right hand side is equal to $2 \mod 3$, hence this is absurd.

From this we have that $\frac{m^2-4m+5}{k^2+2}$ can never be an integer, so there exist no solutions for $n$. $\blacksquare$
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starchan
1609 posts
#23
Y by
.
There is no such $n$. Let $k = \lfloor \sqrt{n} \rfloor$ and $n = k^2 + \ell$ for some $0 \leqslant \ell \leqslant 2k$. Clearly $k^2 + 2 \mid (\ell-2)^2+1$. Due to clear size reasons the quotient of the last divisibility cannot exceed $3$. When the quotient exceeds $1$ working mod $3$ yields contradiction and when the quotient is $1$ there is only one possibility; which fails as well.
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DottedCaculator
7353 posts
#24
Y by
Solution
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Fakesolver19
106 posts
#26
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Solution
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megarnie
5606 posts
#27
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There are no such $n$.

Let $n=m^2+c$, where $0\le c\le 2m$.


We have $\frac{(m^2+c)^2+1}{m^2+2}$ is an integer.

Note that $m^2+c\equiv c-2\pmod{m^2+2}$.

So $\frac{(c-2)^2+1}{m^2+2}$ is a positive integer.

Now, \[(c-2)^2+1\le (2m-2)^2+1=4m^2-8m+5<4m^2+8=4(m^2+2)\]
Since $(c-2)^2+1$ and $m^2+2$ are positive, we have \[\frac{(c-2)^2+1}{m^2+2}\in \{1,2,3\}\]
Case 1: $(c-2)^2+1=m^2+2$.
Then $m^2+1$ is a perfect square, which is not possible as $m>0$.

Case 2: $(c-2)^2+1=2m^2+4$.
Then $2m^2+3$ is a perfect square.

Note that $m^2\in \{0,1,4\}\pmod 8$, so $2m^2+3\in \{3,5\}\pmod 8$, both are not QR's.

Case 3: $(c-2)^2+1=3m^2+6$.
Then $3m^2+5$ is a perfect square. Clearly not possible as it is $2\pmod 3$.

We have exhausted all cases, so we are done.
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Iora
194 posts
#28 • 3 Y
Y by Mango247, Mango247, Mango247
I claim that there are no such solutions.
In order to remove not so beautifull integer function, let $n=a^2+b$ where $ 0 \le b \le 2a$. Our question turns into
$$ \frac{a^4+b^2+2a^2b+1}{a^2+2}  \in \mathbb{Z}$$Long division yields
$$A= \frac{ b^2-4b+5}{a^2+2} \in \mathbb{Z}$$Since $b \le 2a$, the maximum value our expression can get is
$$ \frac{ 4a^2-8a+5}{a^2+2}= \frac{ 4a^2+8-8a-3}{a^2+2}= 4 - \frac{8a+3}{a^2+2} \in \{1,2,3 \}$$Hence we will divide into cases

\begin{align*}
     & \text{if} \ A=1 :  b^2-4b+5=a^2+2 \Rightarrow (b-2-a)(b-2+a)=1 \Rightarrow \ \text{ no solutions} \\
     & \text{if} \ A=2:  b^2-4b+5=2a^2+4 \Rightarrow  (b^2-2)^2-2a^2=3, \ \text{taking $\mod 8$, no solutions} \\
     & \text{if} \ A=3: b^2-4b+5=3a^2+6 \Rightarrow (b^2-2)^2-3a^2=5 \ \text{ taking $\mod 3$, no solutions}
 \end{align*}Hence we have proven our claim, therefore we are done.
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asdf334
7585 posts
#29
Y by
Let $n=a^2+r$ with $0\le r\le 2a$; note that $a\ge 1$. Clearly we obtain
\[a^2+2\mid (r-2)^2+1\]and by bounding we obtain either
\[(r-2)^2=a^2+1\]\[(r-2)^2=2a^2+3\]\[(r-2)^2=3a^2+5\]and the first equation fails by DOS, the second by noting that we must have $3\mid a$ implying that $\nu_3(2a^2+3)=1$, and the third by simply taking $\pmod 3$. We are done. $\blacksquare$
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cj13609517288
1916 posts
#30
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The answer is that no $n$ works.

If $x=\left\lfloor\sqrt n\right\rfloor$, then $x^2\le n\le x^2+2x$. Let $n=x^2+r$, then
\[\frac{n^2+1}{\left\lfloor\sqrt n\right\rfloor^2+2}=\frac{(x^2+r)^2+1}{x^2+2}\rightarrow x^2+2\mid (r-2)^2+1.\]
But $(r-2)^2+1<4(x^2+2)$, so
\[\frac{(r-2)^2+1}{x^2+2}\in\{1,2,3\}.\]
Case 1. It is $1$. Then $(r-2)^2=x^2+1$, so $(r-2-x)(r-2+x)=1$. So $r-2-x=r-2+x$, so $x=0$, absurd as $n$ is positive.
Case 2. It is $2$. Then $(r-2)^2=2x^2+3$, but taking mod $8$ yields a contradiction.
Case 3. It is $3$. Then $(r-2)^2=3x^2+5$, but taking mod $3$ yields a contradiction.

As we have reached a contradiction in all cases, we are done.
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Pyramix
419 posts
#31
Y by
We show that there are no such $n$.

Let $[\sqrt{n}]=k$. Then, $n=k^2+l$ where $0\leq l\leq2k\Longrightarrow n^2=k^4+2k^2l+l^2$
Then, $(k^2+2)\mid(k^4+2k^2l+l^2+1)\Longrightarrow(k^2+2)\mid(l^2-4l+5)$
$$\Longrightarrow1+(l-2)^2=m(k^2+2)$$Note that $m\geq1$ and $1+(l-2)^2\leq1+(2k-2)^2=m(k^2+2)=4k^2-8k+5<4(k^2+2)$
So, $m\in\{1,2,3\}$.
For $m=1$, we get $1+(l-2)^2=k^2+2$ which yields no solutions.
For $m=2$, we get $1+(l-2)^2=2k^2+4\Longrightarrow(l-2)^2-2k^2=3$. But $2k^2\equiv0,2\pmod{8}\Longrightarrow(l-2)^2\equiv3,5\pmod{8}$, a contradiction as $(l-2)^2\equiv0,1,4\pmod{8}$.
For $m=3$, we get $1+(l-2)^2=3k^2+6\Longrightarrow(l-2)^2\equiv2\pmod{3}$ which is impossible as $(l-2)^2\equiv0,1\pmod{3}$. Therefore, there are no solutions.
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john0512
4187 posts
#32
Y by
We claim that there are no solutions.

Claim 1: The equation $a^2-2b^2=3$ has no integer solutions. $a^2$ is either 0 or 1 mod 3, and $-2b^2\equiv b^2$ is also either 0 or 1 mod 3. For the total to be a multiple of 3, both have to be 0 mod 3, so $a$ and $b$ are both multiplies of 3, so the LHS is a multiple of 9, contradiction.

Claim 2: The equation $a^2-3b^2=5$ has no integer solutions. $a^2$ is either 0, 1, or 4 mod 5, and $-3b^2\equiv 2b^2$ is either 0, 2, or 3 mod 5. The only way for the total to be 0 mod 5 is if both $a$ and $b$ are multiples of 5, so the LHS is a multiple of 25, contradiction.

The equation is clearly not true for $n=1,2,3$. From now on, suppose $n\geq 4$. Suppose that $n=k^2+s$ for a positive integer $k\geq 2$ and nonnegative integer $s$ such that $s\leq 2k$ (essentially make $k$ as large as possible). Then, $$\frac{n^2+1}{\lfloor \sqrt{n}\rfloor ^2+2}=\frac{(k^2+s)^2+1}{k^2+2}=(k^2+2s-2)+\frac{s^2-4s+5}{k^2+2}.$$Thus, $$s^2-4s+5\equiv 0\pmod{k^2+2}$$$$(s-2)^2\equiv -1\pmod{k^2+2}.$$Then, note that $$(s-2)^2\leq (2k-2)^2<4k^2<4k^2+7,$$so we must have one of $$(s-2)^2=k^2+1$$$$(s-2)^2=2k^2+3$$$$(s-2)^2=3k^2+5.$$The first one clearly cannot happen as a positive square plus 1 is never a square, while the other two cannot happen by Claims 1 and 2, so we are done.
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andyxpandy99
365 posts
#33
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There are no solutions. The key idea is to let $n = m^2+k$ where $0 \leq k \leq 2m$. Note that this implies that $\lfloor\sqrt{n}\rfloor = m$. Note that we require $$m^2+2 \mid n^2+1 = (m^2+k)^2+1$$Note that we can rewrite this as $$(m^2+k)^2+1 \equiv 0 \pmod {m^2+2}$$Equivalently, $$(k-2)^2+1 \equiv 0 \pmod {m^2+2}$$so we are left with $$m^2+2 \mid (k-2)^2+1$$
The key observation is that $4(m^2+2) > (2m-2)^2+1 \geq (k-2)^2+1$. This is trivial to prove. It suffices to check three cases.

Case 1: $(k-2)^2+1 = m^2+2$

Difference of squares gives us $(k-2+m)(k-2-m) = 1$ from which we get $k-2-m = 1$ and $k-2+m = 1$. This yields that $m = 0$ so there are no solutions.

Case 2: $(k-2)^2+1 = 2(m^2+2)$.

Rearranging yields $(k-2)^2 = 2m^2+3$. We will prove that $2m^2+3$ cannot be a square. Note that if $3 \nmid m$ then $2m^2+3 \equiv 2 \pmod 3$ which is bad. So let $m = 3c$ such that $2m^2+3 = 18c^2+3 = 3(6c^2+1)$. However, this means that $3 \mid 6c^2+1$ which is clearly false. Therefore, $2m^2+3$ cannot be a square and there are no solutions for this case.

Case 3: $(k-2)^2+1 = 3(m^2+2)$.

Rearranging yields $(k-2)^2 = 3m^2+5$. Note that since $3m^2+5 \equiv 2 \pmod 3$ we know that $3m^2+5$ cannot be a square and there are no solutions in this case.

We've exhausted all cases and shown that there are no solutions.
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huashiliao2020
1292 posts
#34
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Let $n = m^2 + k$ s.t. m is maximal but k stays a positive integer; we want \[\frac{(m^2+k)^2+1}{m^2+2} = m^2 + (2k-2) + \frac{(k-2)^2+1}{m^2+2}\in\mathbb{Z}.\]Since $(k-2)^2 + 1 < 4m^2 + 8$, we only need to compute if the fraction equals 1,2,3.

Case 1. $(k-2)^2+1 = m^2+2\rightarrow(k-2)^2 = m^2+1$, which is impossible unless $m=0\rightarrow n=0$, contradiction.
Case 2. $(k-2)^2+1 = 2m^2+4\rightarrow (k-2)^2 + m^2 \equiv 0 \pmod{3}\implies k-2 \equiv m \equiv 0 \pmod{3}\implies 9 \mid (k-2)^2 - 2m^2 = 3$, contradiction.
Case 3. $(k-2)^2 + 1 = 3m^2 + 6\rightarrow(k-2)^2 \equiv 2 \pmod{3}$, contradiction. We've exhausted all cases, so there is no solution for such n.
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thdnder
198 posts
#35
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This is so boring problem.
Answer: There is no such $n$.

Let $x = \left\lfloor \sqrt n \right\rfloor$, and let $y = n - x^2$. Then the condition becomes $\frac{(x^2 + y)^2 + 1}{x^2 + 2}$ is an integer and $y \le 2x$. Thus $x^2 + 2 \mid (y - 2)^2 + 1$. Since $(y - 2)^2 + 1 < (2x - 2)^2 + 1 < 4x^2 + 8$, so $\frac{(y - 2)^2}{x^2 + 2} \in \{1, 2, 3\}$.

If $3(x^2 + 2) = (y - 2)^2 + 1$, then $3 \mid (y - 2)^2 + 1$, a contradiction.

If $2(x^2 + 2) = (y - 2)^2 + 1$, then $2x^2 + 3 = (y - 2)^2$. So taking mod 3 forces $3 \mid x , (y - 2)$. Then taking mod 9 gives $9 \mid 3$, a contradiction.

If $(x^2 + 2) = (y - 2)^2 + 1$, then $x^2 + 1 = (y - 2)^2$, so $x = 0$ and $y = 3$. This contradicts $y \le 2x$.

This completes proof. $\blacksquare$
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YaoAOPS
1541 posts
#36
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Let $n = k^2 + m$ for $0 \le m \le 2k$. This then becomes \[ \frac{(k^2 + m)^2 + 1}{k^2 + 2} = \frac{k^4 + 2k^2m + m^2 + 1}{k^2 + 2} = k^2 - 2 + 2m + \frac{m^2 - 4m + 5}{k^2 + 2} \]This implies that $t(k^2 + 2) = m^2 - 4m + 5$ for $t \in \{1, 2, 3\}$.
If $k^2 + 2 = (m - 2)^2 + 1$, then $(k, m) = (0, 3)$, which is invalid.
If $2k^2 + 4 = (m - 2)^2 + 1$, taking $\pmod{8}$ gives no solutions. If $3k^2 + 6 = (m - 2)^2 + 1$, taking $\pmod{3}$ gives no solutions.
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HamstPan38825
8863 posts
#37
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No such $n$ exist. Let $\lfloor \sqrt n \rfloor = k$ and set $n = k^2+r$, where $r \leq 2k$. The divisibility condition $$k^2+2 \mid (k^2+r)^2 + 1 \iff k^2+2 \mid r^2-4r+5$$after some reduction. In particular, this implies that $$r^2-4r+5 \in \{k^2+2, 2k^2+4, 3k^2+6\}.$$
If $r^2-4r+5=k^2+2$, then $k^2+1$ must be a square, which is impossible.

If $r^2 -4r + 5 = 2k^2+4$, then $2k^2+3$ must be a square, which is impossible by mod $8$.

If $r^2-4r + 5 = 3k^2+6$, then $3k^2+5$ must be a square, which is impossible by mod $3$.

Thus there are no valid pairs $(r, k)$ and thus no valid $n$.
This post has been edited 1 time. Last edited by HamstPan38825, Dec 16, 2023, 8:05 PM
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kamatadu
480 posts
#38
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This problem is so annoying that you have to do nothing else apart from doing some hell lot of bounding stuff (at least that is what I did in the way I solved it). Please let me know if there are any calculation mistakes since it is my nature to make sillies in all possible ways.

I claim that there does not exist any such $n$.

Let $k^2 \le n < (k+1)^2$. Then using the floor function inequalities, we get,
\[ \dfrac{k^4+1}{k^2 + 2}\le \dfrac{n^2 + 1}{k^2 + 2} \le \dfrac{(k+1)^4 + 1}{k^2 + 2}. \]
Now we have that $\dfrac{k^4+1}{k^2+2}=k^2 - 2 + \dfrac{5}{k^2+2} > k^2 -2$. This gives us that $\dfrac{n^2 + 1}{k^2 + 2} \ge k^2 - 1$. Also, we have that $\dfrac{(k+1)^2 + 1}{k^2 + 2} = k^2 + 4k + 4 - \dfrac{4k+6}{k^2+2} < k^2 + 4k + 4$. This gives us $\dfrac{n^2 + 1}{k^2 + 2} \le k^2 + 4k + 3$. Combining these two bounds, we get,
\[ k^2 - 1 \le \dfrac{n^2 + 1}{k^2 + 2} \le k^2 + 4k + 3. \]
Thus we get $(k^2-1)(k^2+2) - 1 \le n^2 \le (k^2+4k+3)(k^2+2) - 1$, that is $k^4 + k^2 -3 \le n^2 \le k^4 + 4k^3 + 5k^2 + 8k + 5$. Now we first need to find all the set of possible perfect squares in the interval $[k^4 + k^2 -3, k^4 + 4k^3 + 5k^2 + 8k + 5]$. Now we can note that all the perfect squares in this range are $\left\{(k^2+1)^2, (k^2+2)^2, \ldots, (k^2 + 2k + 2)^2\right\}$. This gives us that $n\in \left\{k^2+1,k^2+2,\ldots, k^2+2k+2\right\}$.

Now we denote $n$ as $(k^2 + i)$ where $1 \le i \le 2k+2$. Since $\dfrac{n^2 + 1}{k^2 + 2}$ is an integer, we get that $k^2 + 2\mid n^2 + 1$, that is,
\[ k^2 + 2 \mid (k^2 + i)^2 + 1 \equiv (-2 + i)^2 + 1. \]
Now note that $1 \le (i-2)^2 + 1 \le 4k^2 + 1$ which gives that the only possible values for $(i-2)^2 + 1$ are $\left\{k^2+2,2k^2+4,3k^2+6\right\}$. Now among these possibilities, $(i-2)^2 + 1 = 2k^2 + 4$ fails due to $\mod 8$. Also, $(i-2)^2 + 1 = 3k^2 + 6$ fails due to $\mod 3$.

Thus we must have $(i-2)^2 + 1 = k^2 + 2 \implies (i-2+k)(i-2-k) = 1$. This is only possible if $k = 0$. So now we are working with $0^2 \le n < 1^2$ which forces $n=0$, contradiction as $n$ was a positive integer.
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Inconsistent
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#39
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The answer is there is no such $n$. Suppose otherwise, then $n = k^2 + s$ where $0 \leq s < 2k + 1$ and $k = \lfloor \sqrt{n} \rfloor$.

Now we have $\frac{(k^2+s)^2 + 1}{k^2 + 2} \in \mathbb{Z}$, so by plugging in $k^2 = -2$ (i.e. long division), we have:

$\frac{(s-2)^2 + 1}{k^2 + 2} \in \mathbb{Z}$.

Now notice we must have: $(s-2)^2 = k^2 +1, 2k^2 + 3, 3k^2 + 5$, since the numerator is positive and $4k^2 + 7 > 4k^2-4k + 1 \geq (s-2)^2$. These cases each fail by $k \geq 1$, modulo $8$, modulo $3$ respectively, so there are no solutions.
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shendrew7
796 posts
#40
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Let $n=a^2+b$, where $0 \leq b \leq 2a$. Then $a^2+2$ must divide
\[a^4+2a^2b+b^2+1 \implies 2a^2b-2a^2+b^2+1 \implies b^2-4b+5.\]
Due to our bound on $b$, we require $\frac{b^2-4b+5}{a^2+2} \in \{1,2,3\}$, which gives the solutions
\[b = \sqrt{2a^2+3}+2, \quad \sqrt{3a^2+5}+2.\]
Testing modulo 3, neither solutions are integers, giving us $\boxed{\text{no solutions}}$. $\blacksquare$
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lifeismathematics
1188 posts
#41
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set $\lfloor \sqrt{n} \rfloor :=k$

so we have $k^2 \leqslant n < (k+1)^2$ , so we have $n:=k^2+t$ where $0 \leqslant t<2k$we have:

$\frac{n^2+1}{(\lfloor \sqrt{n} \rfloor)^2+2}=\frac{k^4+t^2+2k^2t+1}{k^2+2} \in \mathbb{N} \implies \frac{t^2+5-4t}{k^2+2} \in \mathbb{N}$

now we have the bound:

$\frac{2}{k^2+2} \leqslant \frac{(t-2)^2+1}{k^2+2} \leqslant \frac{4k^2+5-8k}{k^2+2}<4$

so we just need to check for $\frac{(t-2)^2+1}{k^2+2} \in \{1,2,3\}$

so we just check that $(t-2)^2=k^2+1,2k^2+3,3k^2+5$ has no solutions , and so no such $n$ works. $\square$
This post has been edited 1 time. Last edited by lifeismathematics, Apr 25, 2024, 5:34 PM
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bjump
1027 posts
#42
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solution
This post has been edited 1 time. Last edited by bjump, May 5, 2024, 12:53 PM
Reason: forgot somth
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RedFireTruck
4223 posts
#43
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Let $n=k^2+a$ where $0\le a\le 2k$ and $k>0$. Then we want $$\frac{(k^2+a)^2+1}{k^2+2}=\frac{k^4+2ak^2+a^2+1}{k^2+2}$$to be an integer. Subtracting $k^2+2a-2$ means we want $\frac{a^2-4a+5}{k^2+2}=\frac{(a-2)^2+1}{k^2+2}$ to be an integer. Since $0\le a\le 2k$, we must have that the fraction is $1$, $2$, or $3$.

$(a-2)^2+1=k^2+2$ gives $(a-2-k)(a-2+k)=1$, which is not possible.

$(a-2)^2+1=2k^2+4$ gives $(a-2)^2=2k^2+3$. Since squares are never $2\pmod{3}$, let $k=3j$ and $a-2=3b$. Then, $9b^2=18j^2+3$ so $3b^2=6j^2+1$. This is clearly not possible because $3\not|1$.

$(a-2)^2+1=3k^2+6$ gives $(a-2)^2=3k^2+5$. Since perfect squares are $0,\pm1\pmod{5}$, let $k=5j$ and $a-2=5b$. Then $25b^2=75j^2+5$ so $5b^2=15j^2+1$. This is clearly not possible because $5\not|1$.

Therefore, there are no solutions for $n$.
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pie854
243 posts
#44
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Let $\lfloor \sqrt n \rfloor=t$ and $r$ be such that $0\leq r< 2t+1$ and $n=r+t^2$. Now we have $$t^2+2 \mid (r+t^2)^2+1 \implies t^2+2\mid (r-2)^2+1.$$Note that $4(t^2+1)>(2t)^2+1>(r-2)^2+1$. If $3(t^2+2)=(r-2)^2+1$ or $2(t^2+2)=(r-2)^2+1$ then $\pmod 3$ and $\pmod 8$, respectively, yield contradictions. If $t^2+2=(r-2)^2+1$ then $t=0, r=3$ which is not possible. Thus no solutions exist.
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Ilikeminecraft
627 posts
#45
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Let $n = a^2 + k$ such that $k < 2a + 1$(so that $\lfloor{\sqrt n}\rfloor = a$). We have that:
\begin{align*}
  \frac{(a^2 +k)^2 + 1}{a^2 + 2} & = \frac{(k - 2)^2 + 1}{a^2 + 2}
\end{align*}However, since $k < 2a + 1,$ we have that $(k - 2)^2 + 1 < 4a^2 - 4a + 2 < 4(a^2 + 2).$ Hence, we have that $\frac{(k - 2)^2 + 1}{a^2 + 2} = 3, 2, 1.$

\begin{enumerate}
\item If it is 1, we have that $(k - 2)^2 + 1 = a^2 + 2,$ and by difference of squares, $(k - a - 2)(k + a - 2) = 1.$ Hence, $k= 3, a = 0, k = 1, a = 0.$ However, neither of these satisfy $k < 2a + 1.$
\item If it is $2,$ we have that $(k - 2)^2 + 1 = 2(a^2 + 2).$ Hence, $(k - 2)^2- 2a^2 = 3.$ By taking modulo 8, this is impossible.
\item If it is $3,$ we have that $(k - 2)^2 + 1 = 3(a^2 + 2).$ Hence, $(k - 2)^2 - 3a^2 = 5.$ By taking modulo 3, this is impossible.
\end{enumerate}
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