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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
Easy P4 combi game with nt flavour
Maths_VC   1
N 2 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
2 hours ago
Central sequences
EeEeRUT   14
N 3 hours ago by HamstPan38825
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
14 replies
EeEeRUT
Apr 16, 2025
HamstPan38825
3 hours ago
Elementary Problems Compilation
Saucepan_man02   32
N 4 hours ago by atdaotlohbh
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
32 replies
Saucepan_man02
May 26, 2025
atdaotlohbh
4 hours ago
Random Points = Problem
kingu   5
N 4 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
+1 w
kingu
Apr 27, 2024
happypi31415
4 hours ago
Combo resources
Fly_into_the_sky   1
N 4 hours ago by Fly_into_the_sky
Ok so i never did combinatorics in my life :oops: and i am willing to be able to do P1/P4 combos (or even more)
So yeah how can i start from scratch?
Remark:i don't want compuational combo resources :noo:
1 reply
Fly_into_the_sky
4 hours ago
Fly_into_the_sky
4 hours ago
Very odd geo
Royal_mhyasd   2
N 5 hours ago by Royal_mhyasd
Source: own (i think)
nevermind
2 replies
Royal_mhyasd
Yesterday at 6:10 PM
Royal_mhyasd
5 hours ago
Polynomial Application Sequences and GCDs
pieater314159   46
N 5 hours ago by cursed_tangent1434
Source: ELMO 2019 Problem 1, 2019 ELMO Shortlist N1
Let $P(x)$ be a polynomial with integer coefficients such that $P(0)=1$, and let $c > 1$ be an integer. Define $x_0=0$ and $x_{i+1} = P(x_i)$ for all integers $i \ge 0$. Show that there are infinitely many positive integers $n$ such that $\gcd (x_n, n+c)=1$.

Proposed by Milan Haiman and Carl Schildkraut
46 replies
pieater314159
Jun 19, 2019
cursed_tangent1434
5 hours ago
c^a + a = 2^b
Havu   10
N 5 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
10 replies
Havu
May 10, 2025
Havu
5 hours ago
Own made functional equation
JARP091   0
5 hours ago
Source: Own (Maybe?)
\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]
0 replies
JARP091
5 hours ago
0 replies
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   16
N 6 hours ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
16 replies
OgnjenTesic
May 22, 2025
JARP091
6 hours ago
equal segments on radiuses
danepale   8
N 6 hours ago by zuat.e
Source: Croatia TST 2016
Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.
8 replies
danepale
Apr 25, 2016
zuat.e
6 hours ago
Inequality
SunnyEvan   8
N 6 hours ago by arqady
Let $a$, $b$, $c$ be non-negative real numbers, no two of which are zero. Prove that :
$$ \sum \frac{3ab-2bc+3ca}{3b^2+bc+3c^2} \geq \frac{12}{7}$$
8 replies
SunnyEvan
Apr 1, 2025
arqady
6 hours ago
Inequality conjecture
RainbowNeos   2
N 6 hours ago by RainbowNeos
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
2 replies
RainbowNeos
May 29, 2025
RainbowNeos
6 hours ago
2- player game on a strip of n squares with two game pieces
parmenides51   2
N 6 hours ago by Gggvds1
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 3
Alice and Bob play a game on a strip of $n \ge  3$ squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of $ n = 7$ squares.
IMAGE
The players alternate. In each move, they advance their own game piece by one or two squares in the direction of the opponent’s piece. The piece has to land on an empty square without jumping over the opponent’s piece. Alice makes the first move with her own piece. If a player cannot move, they lose.

For which $n$ can Bob ensure a win no matter how Alice plays?
For which $n$ can Alice ensure a win no matter how Bob plays?

(Karl Czakler)
2 replies
parmenides51
Mar 26, 2024
Gggvds1
6 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
7938 posts
#19 • 1 Y
Y by Adventure10
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
4 posts
#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
8871 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
Y by
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
1926 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
56 posts
#25
Y by
\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
Y by
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
799 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
501 posts
#29
Y by
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
112 posts
#30
Y by
(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
654 posts
#31
Y by
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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