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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
Quadratic system
juckter   35
N 14 minutes ago by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
14 minutes ago
IMO Shortlist 2012, Geometry 3
lyukhson   75
N an hour ago by numbertheory97
Source: IMO Shortlist 2012, Geometry 3
In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
75 replies
lyukhson
Jul 29, 2013
numbertheory97
an hour ago
Diophantine
TheUltimate123   31
N an hour ago by SomeonecoolLovesMaths
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
31 replies
TheUltimate123
Mar 29, 2023
SomeonecoolLovesMaths
an hour ago
Cyclic ine
m4thbl3nd3r   1
N 2 hours ago by arqady
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
1 reply
m4thbl3nd3r
5 hours ago
arqady
2 hours ago
Non-homogenous Inequality
Adywastaken   7
N 2 hours ago by ehuseyinyigit
Source: NMTC 2024/7
$a, b, c\in \mathbb{R_{+}}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
7 replies
Adywastaken
5 hours ago
ehuseyinyigit
2 hours ago
FE with devisibility
fadhool   2
N 2 hours ago by ATM_
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
2 replies
fadhool
4 hours ago
ATM_
2 hours ago
Japan MO Finals 2023
parkjungmin   2
N 2 hours ago by parkjungmin
It's hard. Help me
2 replies
1 viewing
parkjungmin
Yesterday at 2:35 PM
parkjungmin
2 hours ago
Iranian geometry configuration
Assassino9931   2
N 2 hours ago by Captainscrubz
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
2 replies
Assassino9931
Today at 9:39 AM
Captainscrubz
2 hours ago
f(m + n) >= f(m) + f(f(n)) - 1
orl   30
N 3 hours ago by ezpotd
Source: IMO Shortlist 2007, A2, AIMO 2008, TST 2, P1, Ukrainian TST 2008 Problem 8
Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

Author: Nikolai Nikolov, Bulgaria
30 replies
orl
Jul 13, 2008
ezpotd
3 hours ago
Classic Diophantine
Adywastaken   3
N 3 hours ago by Adywastaken
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
3 replies
Adywastaken
5 hours ago
Adywastaken
3 hours ago
Add d or Divide by a
MarkBcc168   25
N 3 hours ago by Entei
Source: ISL 2022 N3
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$. Let $x_1=1$, and for $k\geq 1$, define
$$x_{k+1} = \begin{cases}
x_k + d &\text{if } a \text{ does not divide } x_k \\
x_k/a & \text{if } a \text{ divides } x_k
\end{cases}$$Find, in terms of $a$ and $d$, the greatest positive integer $n$ for which there exists an index $k$ such that $x_k$ is divisible by $a^n$.
25 replies
MarkBcc168
Jul 9, 2023
Entei
3 hours ago
Alice and Bob play, 8x8 table, white red black, minimum n for victory
parmenides51   14
N 3 hours ago by Ilikeminecraft
Source: JBMO Shortlist 2018 C3
The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses $4$ rows and $4$ columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can win the game no matter how Bob plays.
14 replies
parmenides51
Jul 22, 2019
Ilikeminecraft
3 hours ago
GEOMETRY GEOMETRY GEOMETRY
Kagebaka   71
N 4 hours ago by bin_sherlo
Source: IMO 2021/3
Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.
71 replies
Kagebaka
Jul 20, 2021
bin_sherlo
4 hours ago
Equation of integers
jgnr   3
N 4 hours ago by KTYC
Source: Indonesia Mathematics Olympiad 2005 Day 1 Problem 2
For an arbitrary positive integer $ n$, define $ p(n)$ as the product of the digits of $ n$ (in decimal). Find all positive integers $ n$ such that $ 11p(n)=n^2-2005$.
3 replies
jgnr
Jun 2, 2008
KTYC
4 hours ago
Integral Solutions
Brut3Forc3   29
N Apr 28, 2025 by cursed_tangent1434
Source: 1976 USAMO Problem 3
Determine all integral solutions of \[ a^2+b^2+c^2=a^2b^2.\]
29 replies
Brut3Forc3
Apr 4, 2010
cursed_tangent1434
Apr 28, 2025
Integral Solutions
G H J
G H BBookmark kLocked kLocked NReply
Source: 1976 USAMO Problem 3
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Brut3Forc3
1948 posts
#1 • 4 Y
Y by ahmedosama, Adventure10, Mango247, and 1 other user
Determine all integral solutions of \[ a^2+b^2+c^2=a^2b^2.\]
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basketball9
1012 posts
#2 • 2 Y
Y by Adventure10, Mango247
Click to reveal hidden text
This post has been edited 1 time. Last edited by basketball9, Apr 4, 2010, 2:14 PM
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xpmath
2735 posts
#3 • 2 Y
Y by Adventure10, Mango247
solution
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harekrishna
173 posts
#4 • 2 Y
Y by Adventure10, Mango247
I am not sure about xpmath's solution. Shouldn't $ {a_1} ^2 + {b_1}^2 + { c_1} ^ 2 = 4 (a_1 b_1)^2 $
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Arrange your tan
970 posts
#5 • 4 Y
Y by CircleGeometryGang, sabkx, Adventure10, Mango247
basketball9 wrote:
We divide both sides by $ a^2b^2$....

[1] So we get $ 1/a^2 + 1/b^2 + c^2 = 1$

[2] so $ 1/a^2 + 1/b^2 = 1 - c^2/a^2b^2$

[1] a) The third term should be $\frac{c^2}{a^2b^2}$.

b) While this equation would have been mathematically correct (see [1] a above), it is misleading,
because the direct next step after dividing by $a^2b^2$, but befoe reordering terms, would be
$\frac {1}{b^2} + \frac {1}{a^2} + \frac {c^2}{a^2b^2} = 1$.


[2] If you type it out horizontally, it has to have grouping symbols around the denominator such as

$\frac{1}{a^2} + \frac{1}{b^2} = 1 - c^2/(a^2b^2)$

Otherwise, type out the fractions in a vertical style and avoid the needed use of grouping symbols:

$\frac{1}{a^2} + \frac {1}{b^2} = 1 - \frac{c^2}{a^2b^2}$


(This post is being neutral towards the correctness of your overall method.)

- - -- - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - -
harekrishna wrote:
I don't know about xpmath's solution. Shouldn't ${a_1}^2 + {b_1}^2 + {c_1}^2 = 4(a_1b_1)^2?$ . . . EDITED

Yes, it should be the equivalent to that.
This post has been edited 1 time. Last edited by Arrange your tan, May 18, 2010, 3:28 AM
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andersonw
652 posts
#6 • 2 Y
Y by Adventure10, Mango247
@basketball: I don't see how that shows that c needs to be 0. As long as c<ab, the right hand side is still positive.

xpmath's solution still works, because the right side will always be 0 mod 4, so all of the terms on the left side need to be 0 mod 2 as well (because a perfect square is only 0 or 1 mod 4, of course), and you can still perform the infinite descent. The coefficient of the right hand side does grow larger, but it is still always 0 mod 4.
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varunrocks
1134 posts
#7 • 1 Y
Y by Adventure10
I also do not understand Basketball9's logic.
@harekrishna: xpmath got that 2a_1=a, 2b_1=b, 2c_1=c
so we get:
4(a_1)^2+4(b_1)^2+4(c_1)^2=4(a_1)(b_1)
Dividing through by 4, we get:
(a_1)^2+(b_1)^2+(c_1)^2=(a_1)(b_1)
Which is correcy.
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AIME15
7892 posts
#8 • 2 Y
Y by Adventure10, Mango247
Actually, the equation is

\begin{align*}
a^2+b^2+c^2 & = a^2b^2
\\ 4a_1^2+4b_1^2+4c_1^2 &= 16a_1^2b_1^2
\\ a_1^2+b_1^2+c_1^2&=4a_1^2b_1^2.
\end{align*}
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xpmath
2735 posts
#9 • 2 Y
Y by Adventure10, Mango247
Hmm must've typoed, sorry. The idea is still the same though, like Anderson said.
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varunrocks
1134 posts
#10 • 2 Y
Y by Adventure10, Mango247
Sorry, I also have typoed!
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Zhero
2043 posts
#11 • 7 Y
Y by maXplanK, CircleGeometryGang, Adventure10, Mango247, and 3 other users
Rearrange this as $c^2 + 1^2 = (a^2 - 1)(b^2 - 1)$. By looking at this mod 4, it can easily be seen that $c$ must be even. It is well-known that if $p | c^2 + 1^2$, then $p = 2$, $p \equiv 1 \pmod{4}$, or $p | c, 1$. Since $c^2 + 1^2$ is odd and no prime divides 1, we must have that all prime factors of $c^2 + 1^2$ are congruent to 1 modulo 4. Hence, the product of the prime factors of $a^2 - 1$ must be conrguent to 1 modulo 4. But $a^2 - 1 \equiv 0, 3 \pmod{4}$, so we have a contradiction. Hence, there are no nontrivial solutions.

A rigorized version of xpmath's solution
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SodaKing1
738 posts
#12 • 2 Y
Y by Adventure10, Mango247
Sorry for bringing this back up but could someone explain what infinite descent means? I understand how the first part of the solution but not the descent part. Also if someone could pm me maybe a link to somewhere that explains the concept well.
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NT2048
374 posts
#13 • 2 Y
Y by Adventure10, Mango247
It's not really a fancy concept or anything. We start with this equation:
\[ a^2+b^2+c^2=a^2b^2.\]

We find that $a, b, c$ are even and we let $ a=2a_1, b=2b_1, c=2c_1$. We find $ {a_1}^2+{b_1}^2+{c_1}^2=4{a_1}^2{b_1}^2$.

Using another argument, we can show that $a_1, b_1, c_1$ are even. So let $ a_1=2a_2, b_1=2b_2, c_1=2c_2$.

Then we sub this into the equation and find that $a_2, b_2, c_2$ are even.

And we can repeat this argument an infinite number of times to get that $a_3, a_4 ... a_i ...$ are all even and it would imply that $a, b, c$ have an infinite number of factors of 2. But this is impossible, so solutions can't exist.
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fighter
507 posts
#14 • 2 Y
Y by Adventure10, Mango247
lemma-1: a ,b ,c are even

proof: using mod 4;

now, a = 2*a', b = 2*b', c = 2*c'

then, a'^2 + b'^2 + c'^2 = (a'^2)*(b'^2) which is a recurrence;

so, the solutions are a = b = c = 0
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AllenWang314
661 posts
#15 • 1 Y
Y by Adventure10
Where does this go wrong?
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djmathman
7938 posts
#17 • 1 Y
Y by Adventure10
AllenWang314 wrote:
snippit
This statement is the first place where your solution breaks, and it feels really fishy to me. In particular, I have no idea how non-QR-ness of two integers imply that they both must be even. (Also keep in mind that any two non-QRs mod a prime must multiply to a QR.)
This post has been edited 1 time. Last edited by djmathman, Mar 13, 2018, 9:05 PM
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AllenWang314
661 posts
#18 • 2 Y
Y by Adventure10, Mango247
explanation

Is this correct?
This post has been edited 1 time. Last edited by AllenWang314, Mar 14, 2018, 3:22 AM
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djmathman
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#19 • 1 Y
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Quote:
In particular, one of these is $3$ mod $4$ and must be divisible by a prime $3$ mod $4$.
Not if $a=0$!
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Deligne
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#20
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Assume without loss of generality that $a\geqslant b$. Then $a^2 b^2 =a^2 +b^2 +c^2 \leqslant 2a^2 +c^2$. Therefore $c^2 \geqslant a^2(b^2- 2)$. Since $b^2 -2$ is not a perfect square, we have that $c^2 \neq a^2 (b^2-2)$. Similarly $c^2 \neq a^2 (b^2-1)$. Hence $c^2 \geqslant a^2 b^2$. It follows that $a=b=c=0$.
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huashiliao2020
1292 posts
#21
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xpmath wrote:
solution

Yup, that's what I did, except that a_1^2+b_1^2+c_1^2=4a_1^2b_1^2, which again is 0 mod 4 on RHS, so LHS must be 0 mod 2 for each of a_1, b_1, and c_1 to have their sum of their squares by 0 mod 4. Also, a bit of an explanation. If both a and b are odd, then the RHS is 1 mod 4, and LHS is 1 (a^2) + 1 (b^2) + c^2 mod 4, must be 1 mod 4, absurd, because this implies c^2 must be 3 mod 4. Now if at least of one of a and b is even, then RHS is 0 mod 4, LHS is 0 mod 4 + b^2 + c^2, or b^2+c^2 must be 0 mod 4, absurd (2+2, 1+3 all impossible) unless both b and c are 0 mod 2. Now a and b must both be 0 mod 2, meaning RHS is even and c must also be even for even + even + even = even.
This post has been edited 1 time. Last edited by huashiliao2020, Apr 14, 2023, 4:16 AM
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HamstPan38825
8863 posts
#22
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Write the given equation as $$(a^2-1)(b^2-1) = c^2+1.$$By Fermat Christmas, divisors of the RHS must be $2$ or 1 mod $4$. On the other hand, the factors on the LHS are either multiples of $4$ or $3$ mod $4$, unless $(a, b, c) = (0, 0, 0)$. Indeed this is the sole solution.
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Seungjun_Lee
526 posts
#23
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Since $c \le ab$
Let $c = ab - k$
$a^2 + b^2 + k^2 - 2abk = 0$
Let $(k,a,b)$ is the solution with the minimum $k +a+b$
Let $k \ge a \ge b>0$
Let $f(x) = x^2 - 2abx + a^2 + b^2$
Let two solutions of $f(x) =0$ as $x_1=k$ and $x_2$

$x_2 \in \mathbb{Z}$
$f(a) \ge 0$
Then $3a^2 \ge 2a^2 + b^2 \ge 2a^2b$
Then $b= 1$
$(a-k)^2 + 1 = 0$
Contradiction

Then $b=0$ and $a=c=0$
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cj13609517288
1916 posts
#24
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Take mod $4$. This is either $0+0+0=0$ or $0+0+1=1$. But the second one requires $a$ and $b$ to be both odd, contradiction. The first one falls to infinite descent and has solution $(0,0,0)$.
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p.lazarov06
55 posts
#25
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\begin{align*}
    a^2+b^2+c^2&=a^2b^2\\
    a^2b^2-a^2-b^2+1&=c^2+1\\
    (a^2-1)(b^2-1)&=c^2+1\\
\end{align*}
It's clear that $a$ and $b$ can't be both odd at the same time, because $c^2+1\not\equiv\pmod{4}$. Now if both $a$ and $b$ are at least $2$, so $(a^2-1)(b^2-1)$ will have a prime divisor $p=4k+3$, and by Fermat's Christmas Theorem $p$ will divide both $c$ and $1$ which is impossible. And now by bashing the small values of $a$ and $b$ we get the only solution to be:

\[(a;b;c)=(0;0;0)\]
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surpidism.
10 posts
#26
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Note that perfect squares are $0$ or $1$ $(mod\ 4)$.
$a^2b^2 \equiv 1(mod\ 4)$ is not possible as this lead both $a^2$ and $b^2$ are $1(mod\ 4)$. So, $LHS \not\equiv RHS$ $(mod\ 4)$
Thus, $a^2b^2 \equiv 0 (mod\ 4)$ and $a$, $b$, $c$ are even.
Let $a = 2a_1$, $b = 2b_1$, $c = 2c_1$ and substitute in the original equation.
\begin{align*}
4a_1 + 4b_1 + 4c_1 = 16a_1^2b_1^2\\
 a_1 + b_1 + c_1 = 4a_1^2b_1^2\
\end{align*}Again $a_1$, $b_1$, $c_1$ are even. Let $a_1 = 2a_2$, $b_1 = 2b_2$, $b_1 = 2c_2$ and we find that $a_2$, $b_2$, $c_2$ are also even.
If we repeat this argument infinitely many times, we see that $a$, $b$, $c$ can be divided by $2$ infinitely many times, which is only possible when $a = b = c = 0$.
This post has been edited 1 time. Last edited by surpidism., May 20, 2023, 5:56 AM
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Pyramix
419 posts
#27
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Brut3Forc3 wrote:
Determine all integral solutions of \[ a^2+b^2+c^2=a^2b^2.\]
If $a=0$ or $b=0$ then all three must be 0. If $c=0$ then $(a^2-1)(b^2-1)=1$, which has only solution $a=b=c=0$.
Now, take $a,b,c>0$.

Clearly, $a=b=c$ is not possible unless $a=b=c=0$. Also, $a,b>1$. (i.e., $a^2,b^2\geq4$).

Note that $c^2+1=(a^2-1)(b^2-1)$. So, if an odd prime $p\mid(a^2-1)$, then $p\mid(c^2+1)$. This means $p\equiv1\pmod{4}$. It follows that if $a$ is even, then all divisors of $a^2-1$ are $\equiv1\pmod{4}$. So, $a^2-1\equiv1\pmod{4}$, a contradiction. It follows that $a$ must be odd. Similarly, $b$ must also be odd.
But then $c^2\equiv a^2b^2-a^2-b^2\equiv1-1-1\equiv3\pmod{4}$, a contradiction.

Therefore, the only integer solution to the given equation is $\boxed{(a,b,c)=(0,0,0)}$.
This post has been edited 1 time. Last edited by Pyramix, May 20, 2023, 1:35 PM
Reason: Unnecessary bounds removed.
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shendrew7
796 posts
#28
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Rearranging, we find
\[c^2+1 = (a^2-1)(b^2-1).\]
Fermat's Christmas Theorem tells us all factors of the LHS with magnitude greater than 1 must be 1 or 2 modulo 4, contradiction. Thus $c^2+1 \leq 1 \implies c=0$, giving the only solution $\boxed{(0,0,0)}$. $\blacksquare$
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Mr.Sharkman
500 posts
#29
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Simplifying our expression, we get
$$(a^{2}-1)(b^{2}-1) = c^{2}+1. $$If $c$ is odd, then $$c^{2}+1 \equiv 2 \pmod 4. $$But, if $2|(a^{2}-1)(b^{2}-1), $ then $8|(a^{2}-1)(b^{2}-1),$ so $c$ is even. Now, if $p|c^{2}+1,$ we have that $c^{2} \equiv -1 \pmod p, $ so $$\left(\frac{-1}{p} \right) = 1.$$Hence, $p \equiv 1 \pmod 4.$ Now, $a^{2}-1$ is odd, so $a$ is even. Thus, one of $a-1$ or $a+1$ is $1 \pmod 4,$ and the other is $3 \pmod 4.$ But, these are divisors of $c^{2}+1,$ and all divisors of $c^{2}+1$ are $1 \pmod 4,$ so this is impossible. Thus, there are no solutions, except for when $c=0,$ since then there are no prime factors $p,$ and when $a=b=0.$
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MuradSafarli
109 posts
#30
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(a,b,c)=(0,0,0) clearly works,otherwise
"Let O be odd and E be even. The following combinations (a,b,c) – (O,E,E), (E,O,E), (E,E,O), (O,O,E), (O,E,O), (E,O,O), (O,O,O), (E,E,E) easily yield contradictions modulo 4."
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cursed_tangent1434
625 posts
#31
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We claim that the only triple of solutions is $(a,b,c)=(0,0,0)$ which clearly works. We now show that this is the only solution. First note that if $(a,b,c)$ is a triple of solutions $|a|,|b| \ne 1$ since this reduces the equation to $c^2+1=0$ which has no real solutions. We also observe that if $(a,b,c)$ is a triple of solutions so are all triples of the form $(\pm a,\pm b,\pm c)$. Hence it suffices to solve the equation over positive integers. We rearrange the equation to,
\[c^2+1=(ab)^2-a^2-b^2+1= (a^2-1)(b^2-1)=(a-1)(a+1)(b-1)(b+1)\]If $a$ is odd, one of $a-1$ and $a+1$ is divisible by $4$ and thus, $4\mid c^2+1$ which is impossible for all positive integers $c$. If $a$ is even, one of $a-1$ and $a+1$ is $3 \pmod{4}$. Thus, there exists a prime $p \equiv 3 \pmod{4}$ which divides the right-hand side. But then, $p \mid c^2+1$ which is impossible by Fermat's Christmas theorem. Hence, $a$ cannot take any positive integral value, implying that all solutions are of the claimed forms.
This post has been edited 1 time. Last edited by cursed_tangent1434, Apr 29, 2025, 9:18 AM
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