Real triples

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322 posts

#1 h Apr 9, 2019, 11:11 AM • 13 Y

Y by Bst, itslumi, centslordm, chessgocube, Lilathebee, HWenslawski, megarnie, rwll, Adventure10, Mango247, buddyram, ItsBesi, Mathlover_1

Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and

$a^2b + c = b^2c + a = c^2a + b.$

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juckter

322 posts

juckter

#2 h Apr 9, 2019, 11:22 AM • 7 Y

Y by chessgocube, Lilathebee, HWenslawski, panche, rwll, Adventure10, Mango247

First assume that none of the numbers is equal to zero. Then considering the first equation we have

$a(ab - 1) = c(b^2 - 1)$
By the condition we have $ab - 1 = -c(a + b)$ so dividing by $c$ we get

$a(a + b) = 1 - b^2 \implies a^2 + ab = 1 - b^2$
Now add the other two similar relations to find $2(a^2 + b^2 + c^2) = 3 - ab - bc - ca = 2$ , so $a^2 + b^2 + c^2 = 1$ . Then $(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$ implying $a = b = c = \frac{1}{\sqrt{3}}$ , which works.

Now wlog assume $a = 0$ , then we get $c = b^2c = b$ , so in particular $b = c$ and $ab + bc + ca = b^2 = 1$ , giving $b = c = 1$ or $b = c = -1$ , and both work.

In summary the solutions are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right), (0, 1, 1)$ and $(0, -1, -1)$ , along with cyclic permutations.

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Valikk202

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Valikk202

#3 h Apr 9, 2019, 11:41 AM • 5 Y

Y by Kunihiko_Chikaya, chessgocube, Lilathebee, rwll, Adventure10

$\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ also works

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juckter

322 posts

juckter

#4 h Apr 9, 2019, 11:45 AM • 4 Y

Y by chessgocube, Lilathebee, rwll, Adventure10

Right, good catch

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ubermensch

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ubermensch

#5 h Apr 9, 2019, 11:53 AM • 5 Y

Y by chessgocube, Lilathebee, rwll, Adventure10, alexanderhamilton124

Hmm I got a similar solution, but after getting $a^2 + b^2 +c^2=1$ , I used the fact that
$a^2 + b^2 =1-ab => 1-c^2=1-ab=> ab=c^2$ . Similarly, $bc=a^2$ and $ac=b^2$ .
Let $c=max(a,b,c)$ . If $c>0$ , then $a,b>0$ trivially, and if $c<0$ , then $a,b<0$ also trivially. Hence if $c$ is the maximum, we can say that $c \geq a$ and $c \geq b$ , hence $c \geq ab$ if $c>0$ and vice versa if $c<0$ . Hence $a=b=c$ .
Checking $c>0$ and $c<0$ we get 2 solutions: $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ and $(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}})$ .
Putting in $c=0$ gives us trivially 2 solutions(and permutations of them): $(1,1,0)$ and $(-1,-1,0)$ .
Hence solutions are : $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ , $(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}})$ , $(1,1,0)$ , $(-1,-1,0)$

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v_Enhance

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v_Enhance

#6 h Apr 9, 2019, 1:01 PM • 10 Y

Y by SHREYAS333, v4913, snakeaid, chessgocube, Lilathebee, IAmTheHazard, samrocksnature, Dhruv777, Adventure10, Mango247

Answer: $(\pm 1/\sqrt3, \pm 1/\sqrt3, \pm1/\sqrt3)$ where the signs correspond, and $(\pm1, \pm1, 0)$ and permutations where the signs correspond. These work and we prove that is all.

We begin by eliminating the condition via homogenization: the first equality now reads $\begin{align*} a^2b+c\left[ ab+bc+ca \right] &= b^2c+a\left[ ab+bc+ca \right] \\ \iff c^2(b+a) &= c(b^2+a^2) \\ \iff c&=0 \text{ or } a^2+b^2 = c(a+b). \end{align*}$ Cyclic variations hold. So we have two cases.

If any of the variables is zero, say $a = 0$ , then the other two are nonzero. So from $b^2 = bc$ we get $b = c$ giving $(\pm1, \pm1, 0)$ .
Now assume all three variables are nonzero, so $a^2+b^2 = c(a+b)$ . If we sum cyclically we get $2(a^2+b^2+c^2) = 2(ab+bc+ca) \iff (a-b)^2 + (b-c)^2 + (c-a)^2 = 0$ which forces $a=b=c$ and gives the last solution.

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totode

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totode

#7 h Apr 9, 2019, 1:24 PM • 5 Y

Y by chessgocube, Lilathebee, rwll, Dhruv777, Adventure10

Let $k=a^2b+c=b^2c+a=c^2a+b$ , we have:
$2k=b^2c+a+c^2a+b=b^2c+a+c(1-bc-ab)+b=bc(b-c)+a+b+c-abc$ Let $k'=k-(a+b+c)+abc$ , from the last equality (and the cyclic ones)
$k'=bc(b-c)=ca(c-a)=ab(a-b)$ If $(a,b,c)$ is a solution, then $(-a,-b,-c)$ is a solution too. So we can suppose at most one of $a,b,c$ is negative. If $a,b,c>0$ , then $k'>0$ implies $b>c>a>b$ , absurd! and $k'<0$ implies $b<c<a<b$ , absurd! Suppose $a<0$ and $b,c>0$ , then $k'>0$ implies $c<a<b$ and $b>c$ , absurd! And finally $k'<0$ implies $c>a>b$ and $c<b$ , absurd! We deduce that $k'=0$ , which gives $a=b=c$ or $a=0$ and $b=c$ (and its permutations). Eventually we get:
$(a,b,c)=(1,1,0),(1,0,1),(0,1,1),(-1,-1,0),(-1,0,-1),(0,-1,-1),(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}),(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$

This post has been edited 1 time. Last edited by totode, Apr 9, 2019, 4:50 PM
Reason: typo

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Abbas11235

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Abbas11235

#10 h Apr 9, 2019, 3:12 PM • 6 Y

Y by chessgocube, Lilathebee, rwll, ismayilzadei1387, Adventure10, Mango247

If one of them is zero it is easy to get solutions $(0,1,1)$ and $(0,-1,-1)$ and permutations.
If two of them are equal,it is easy to get third one is also equal.So we get two solutions from here. $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ and $(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}})$ .
So assume that none of them is zero and they are different.
If we divides both side of $a^2b+c=b^2c+a$ by $b$ , we will get $a^2+\frac{c}{b}=bc+\frac{a}{b}$ . And similarly,
$a^2+\frac{c}{b}=bc+\frac{a}{b}$ $b^2+\frac{a}{c}=ac+\frac{b}{c}$ $c^2+\frac{b}{a}=ab+\frac{c}{a}$ (...1)
If we subtract these equations two by two .
We will get:
$(a-b)(\frac{1}{c}-(a+b+c))=\frac{c-a}{b}$ $(b-c)(\frac{1}{a}-(a+b+c))=\frac{a-b}{c}$ $(c-a)(\frac{1}{b}-(a+b+c))=\frac{b-c}{a}$ (...2)
Since $a,b,c$ are different,so if we multiple these equations and using the condition $ab+bc+ca=1$ we will get:
$(bc-a^2)(ca-b^2)(ab-c^2)=1$ From the (...1) equations
$\frac{b-c}{a}\frac{a-b}{c}\frac{c-a}{b}=1$ .(...3)
So, $(\frac{a}{b}-1)(\frac{b}{c}-1)(\frac{c}{a}-1)=1$ .
It means that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+1$ .
If we sum the (...1) equations $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+a^2+b^2+c^2$
So, $a^2+b^2+c^2=2$ and $ab+bc+ca=1$
So, $(a+b+c)^2=4$ .
From (...2) equations and (...3) it is easy to get there is no solution in this case.

This post has been edited 4 times. Last edited by Abbas11235, Apr 10, 2019, 2:59 PM

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Steve12345

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Steve12345

#11 h Apr 9, 2019, 3:31 PM • 4 Y

Y by chessgocube, Lilathebee, rwll, Adventure10

redacted.

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jeff10

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jeff10

#12 h Apr 9, 2019, 3:40 PM • 4 Y

Y by chessgocube, Lilathebee, rwll, Adventure10

Solution (slightly longer)

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hwl0304

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hwl0304

#13 h Apr 9, 2019, 7:04 PM • 5 Y

Y by chessgocube, Lilathebee, rwll, Adventure10, ohiorizzler1434

First, we can resolve the case where one of $a,b,c$ is zero; the solution is $(0,\pm 1,\pm 1)$ and permutations.

Next, we can show that if two of them are equal and all of them are nonzero, then all three of them are equal, giving $(\pm \frac{1}{\sqrt3},\pm \frac{1}{\sqrt3},\pm \frac{1}{\sqrt3})$ .

Now, assume $a,b,c$ are nonzero and distinct. Take $a^2b+c=b^2c+a$ and rewrite as $c(1-b^2)=a(1-ab)$ . This can be rewritten as $c(1-b^2)=a(bc+ca)$ , or $1=b^2+ab+a^2$ . Thus, $a^3-b^3=a-b$ , and doing the same thing for the other equalities we get $a^3-a=b^3-b=c^3-c$ . Thus, as $a,b,c$ are distinct, they are the roots of some polynomial $x^3-x+k$ for real $k$ . By vietas, $ab+bc+ac=-1$ , contradiction.

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RnstTrjyn

47 posts

RnstTrjyn

#14 h Apr 10, 2019, 3:53 AM • 4 Y

Y by chessgocube, Lilathebee, Adventure10, Mango247

Fast solution.
$ab=1-bc-ac$
$a^2b+c=b^2c+a \implies a(1-bc-ac)=b^2c+a \implies a^2+ab+b^2=1$ or $c=0$
We can resolve the case when one of $a,b,c$ is zero. The solutions is(+-1;+-1;0).
Now assume that a,b,c are nonezero.
The same way we can prove that $b^2+bc+c^2=1$ and $a^2+ac+c^2=1$ .
From sum of these equations and using that $ab+bc+ac=1$ follows that $a^2+b^2+c^2=1$ .
$a^2=bc$
$b^2=ac$
$c^2=ab$
It follows that $a=b=c$

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trumpeter

3332 posts

trumpeter

#15 h Apr 11, 2019, 1:33 AM • 3 Y

Y by chessgocube, Lilathebee, Adventure10

The answers are $(0,1,1)$ , $(0,-1,-1)$ , $(1,0,1)$ , $(-1,0,-1)$ , $(1,1,0)$ , $(-1,-1,0)$ , $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ , $(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$ . These can be confirmed as follows:

First observe that $(a,b,c)$ works if and only if $(-a,-b,-c)$ works.
Next observe that $(a,b,c)$ works if and only if $(c,a,b)$ and $(b,c,a)$ work.
It suffices to check $(0,1,1)$ and $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ . For the former, we easily have $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b=1.$ For the latter, we have $ab+bc+ca=3a^2=1$ and each of the $a^2b+c$ quantities equal to $a^3+a$ .

Now, we show that these are the only solutions.

First assume $abc=0$ . Since cyclic shifts preserve solutions, WLOG let $a=0$ . Then $bc=1$ and $a^2b+c=c^2a+b$ so $b=c$ . Thus we get the possible solutions $(0,1,1)$ and $(0,-1,-1)$ . Cyclic shifts account for the other four solutions where a variable is $0$ .

Now assume $abc\neq0$ and $a,b,c$ not distinct. Once again by cyclic shifts, WLOG let $b=c$ . Then $a^2b+c=c^2a+b$ so $a^2b=ab^2$ and thus $a=b$ . Then $3a^2=1$ so we get the possible solutions $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ and $(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$ .

Finally assume $abc\neq0$ and $a,b,c$ distinct. Observe that $(a^2b+c)+(a^2c+b)=a(ab+ac)+b+c=a(1-bc)+b+c=a+b+c-abc$ and similarly with the other variables so $a^2c+b=b^2a+c=c^2b+a$ is also true. Then $(a^2b+c)-(a^2c+b)=(b^2c+a)-(b^2a+c)$ so $(2-a^2-b^2)c=(1-ab)(a+b)=c(a+b)^2.$ Since $c\neq0$ , we can divide out to get $2-a^2-b^2=(a+b)^2$ , equivalently $a^2+ab+b^2=1$ . Similarly, we deduce $a^2+ac+c^2=1$ . Then $a^2+ab+b^2=a^2+ac+c^2$ so $(b-c)(a+b+c)=ab-ac+b^2-c^2=0$ and thus $a+b+c=0$ . Then $a,b,c$ are distinct roots of a polynomial $P(x)=x^3+x+k$ for some constant $k$ by Vieta. But $P$ is increasing over $\mathbb{R}$ and thus cannot have multiple distinct roots, contradiction.

Thus the only possible solutions are those in the solution set given above, so the answer is correct.

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#20 h Apr 12, 2019, 1:40 PM • 3 Y

Y by chessgocube, Lilathebee, Adventure10

juckter wrote:

Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and

$a^2b + c = b^2c + a = c^2a + b.$

Just elliminate one variable among $a, b, c$ .

$ab+bc+ca=1 \Longleftrightarrow (a+b)c=1-ab$ . Since $a,\ b$ are real numbers, $a+b\neq 0$ . (Because, consider the graphs the hyperbola $ab=1$ and the line $a+b=0$ on the $a-b$ plane, these graphs no intersection, which was in my mind.) Thus, plugg $c=\frac{1-ab}{a+b}\ \cdots [1]$ into the given second condition, we have :

$a^2b+c=b^2c+a\Longleftrightarrow a(ab-1)=(b^2-1)\cdot \frac{1-ab}{a+b}\Longleftrightarrow (ab-1)(a^2+ab+b^2-1)=0\cdots\ [2]$ .

On the other hand ,

$b^2c+a=c^2a+b\Longleftrightarrow b^2\cdot \frac{1-ab}{a+b}+a=\left(\frac{1-ab}{a+b}\right)^2\cdot a+b$

$\Longleftrightarrow (ab-1)\{b(a^2+ab+b^2)-a\}=(a-b)(a+b)^2\cdots [3]$ .

We are to find the triples $(a,\ b,\ c)$ of real numbers satisfying $[2]$ and $[3]$ simultaneously.

Now, from $[2]$ , we get $ab=1$ or $a^2+ab+b^2=1$ .

Case 1 : $ab=1$

From $[3]$ , we have $(a-b)(a+b)^2=0$ . Since $a+b\neq 0$ , We have $a-b=0$ , yielding $(a,\ b,\ c)=(\pm 1,\ \pm1,\ 0)$ .

Case 2 : $a^2+ab+b^2=1$

From $[3]$ , we have $(a-b)(a^2+3ab+b^2-1)=0$ .

If $a=b$ , then

$a^2+a\cdot a+a^2=1\Longleftrightarrow a^2=\frac 13$ and $[1]$ give $(a,\ b,\ c)=\left(\pm \frac{1}{\sqrt{3}},\ \pm \frac{1}{\sqrt{3}},\ \pm \frac{1}{\sqrt{3}}\right)$ .

If $a^2+3ab+b^2=1$ , then

Solving the system equations of $a^2+3ab+b^2=1$ and $a^2+ab+b^2=1$ and $[1]$ gives $(a,\ b,\ c)=(0,\ \pm 1,\ \pm 1),\ (\pm 1,\ 0,\ \pm 1).$

Therefore, desired triples $(a,\ b,\ c)$ are $(a,\ b,\ c)=(0,\ \pm 1,\ \pm 1),\ (\pm 1,\ 0,\ \pm 1),\ (\pm 1,\ \pm1,\ 0)\ ,\ \left(\pm \frac{1}{\sqrt{3}},\ \pm \frac{1}{\sqrt{3}},\ \pm \frac{1}{\sqrt{3}}\right)$ .

Double sign correspondence.

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DievilOnlyM

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DievilOnlyM

#21 h Apr 12, 2019, 2:00 PM • 4 Y

Y by chessgocube, Lilathebee, Adventure10, Mango247

juckter wrote:

Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and

$a^2b + c = b^2c + a = c^2a + b.$

Another solution

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ihatemath123

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ihatemath123

#67 h Sep 9, 2023, 7:47 PM • 1 Y

Y by centslordm

The only solutions are
$(a,b, c) = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right), \left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right), (0, 1, 1), (0,-1,-1)$ and permutations. If any term is zero, then obviously the only possible solutions are the ones listed above, so we assume all terms are nonzero.

Multiplying the first equation by $a$ , $b$ and $c$ respectively, to get $c = abc + c^2a + c^2b$ and the other two versions of that. Substituting this into each equation and subtracting $abc$ from all three sides, we have
$a^2b + c^2a + c^2b = b^2c + a^2b + a^2c = c^2a + b^2c + b^2a.$ Considering the first two terms, we have
$c a + c b = b^2 + a^2.$ (assuming all the terms are nonzero because we divided by $c$ ). Adding together the two other versions of this, we get
$(a-b)^2 + (b-c)^2 + (c-a)^2 = 0,$ hence $a = b = c$ .

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shendrew7

792 posts

shendrew7

#68 h Oct 8, 2023, 2:34 AM

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We divide the question into 2 cases:

$\textcolor{blue}{\textbf{Case 1:}}$ One of the variables is 0.

Suppose $c=0$ . Then we have $a = b, \quad a^2 b = a \implies a(ab-1) = a(a^2 - 1) = 0,$
giving us $a = 0, \pm 1$ . If $a = 0$ , our first condition is violated, so $a = \pm 1$ gives $\boxed{\text{the permutations of } (1, 1, 0) \text{ and } (-1, -1, 0)}$ .

$\textcolor{blue}{\textbf{Case 2:}}$ None of the variables are 0.

We begin by just using the first two expressions:
$\begin{align*} &\phantom{\implies .} a^2b + c = b^2c = a \\ &\implies a(ab-1) = c(b^2-1) \\ &\implies a(-bc-ca) = c(b^2-ab-bc-ca) \\ &\implies c(a^2 + b^2 - bc - ca) \end{align*}$
Because $c \neq 0$ , we have $a^2 + b^2 - bc - ca = 0$ , and cyclically summing this for the other 2 equations gives $2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = (a-b)^2 + (b-c)^2 + (c-a)^2 = 0,$ implying $a = b = c$ by Trivial Inequality. From the first condition, we get our final two solutions of $\boxed{\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right), \left(-\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}\right)}$ .

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joshualiu315

2513 posts

joshualiu315

#69 h Oct 16, 2023, 1:33 AM

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The answer is $\left (\pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3} \right)$ where signs correspond, and $(\pm 1, \pm 1, 0)$ and permutations where the signs correspond.

We homogenize the first equation as follows:

$\begin{align*} a^2b+c(ab+bc+ca)&=b^2c+a(ab+bc+ca) \\ \implies c^2(b+a) &= c(b^2+a^2) \\ \implies c&=0 \text{ or } a^2+b^2=c(a+b) \end{align*}$

If any variable equals $0$ , say $c$ , we see that $a=b$ from the given equality. Then, we obtain $( \pm 1, \pm 1, 0)$
Otherwise, all of them are non-zero, so we can add up the cyclic variations of the equation we obtained to get

$2(a^2+b^2+c^2) = 2(ab+bc+ca) \implies (a-b)^2 + (b-c)^2 + (c-a)^2 = 0$
meaning $a=b=c$ , which is where our other set of solutions comes from.

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kamatadu

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kamatadu

#70 h Dec 30, 2023, 2:32 PM

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The solutions are $(a,b,c)=(\pm 1,\pm 1,0),,\left(\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}}\right)$ and their permutations.

Note that $a+b\neq 0$ because otherwise $ab +bc+ca=1 \implies ab = 1 \implies -b^2 = 1$ which is impossible.

Now if $ab=1$ , then $c(b+a) = 0 \implies c=0$ . Putting this into the condition, we get that $(a,b,c)=(\pm 1,\pm 1,0)$ which is one of our solutions.

Now if $b=1$ , then $a+c+ac=1$ and $a^2+c=c+a$ which gives the solutions $(a,b,c)=(0,1,1)$ and $(1,1,0)$ .

Now if $b=-1$ , then $-a-c+ac=1$ and $-a^2 + c = c+a$ which gives the solutions $(a,b,c)=(0,-1,-1)$ and $(-1,-1,0)$ .

Now we assume that $a$ , $b$ , $c \neq \pm 1$ and $ab$ , $bc$ , $ca \neq 1$ .

Then we have that $ab+bc+ca=1$ from which we get $c = \dfrac{1-ab}{a+b}$ . We also have $a^b+c = b^2c+a$ from which $c = \dfrac{a(ab-1)}{b^2-1}$ . Setting these two equal, we get that $\dfrac{1-ab}{a+b}=\dfrac{a(ab-1)}{b^2-1}\implies 1-ab = a^2 + b^2\implies c = \dfrac{a^2+b^2}{a+b}\implies c(a+b) = a^2 + b^2$ . Now summing up all the cyclic iterations of this identity, then we get that $a^2+b^2+c^2 = ab+bc+ca = 1$ . Thus $\displaystyle\sum (a-b)^2 = 0\implies a=b=c$ . Putting this in $a^2+b^2+c^2 = 1$ , we get that $a=b=c=\pm \dfrac{1}{\sqrt{3}}$ which gives us our final solution. :yoda:

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eibc

597 posts

eibc

#71 h Dec 31, 2023, 1:20 AM

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I claim that the answers are $(a, b, c) = (\pm \sqrt{3}/3, \pm \sqrt{3}/3, \pm \sqrt{3}/3)$ , and $(\pm1, \pm1, 0)$ and permutations.

To show these are the only answers, homogenize the original equation, so that it becomes
$a^2b + abc + bc^2 + c^2a = b^2c + a^2b + abc + ca^2 = c^2a + ab^2 + b^2c + abc.$ Looking at the first equality, we see that we must have $bc^2 + c^2a = b^2c + ca^2 \implies c(a^2 + b^2 - ab - ac) = 0$ . If $c = 0$ , then the original equations become $ab = 1$ and $a^2b = a = b$ , from which we can easily extract $(a, b, c) = (\pm1, \pm1, 0)$ . For the other equalities, we see that if any of the variables equal $0$ , we get some permutation of $(\pm1, \pm1, 0)$ . Otherwise, we have $a^2 + b^2 = ab + ac$ and permutations; adding these up, we have
$2a^2 + 2b^2 + 2c^2 = 2ab + 2ac + 2bc \implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \implies a = b = c,$ which gives $(a, b, c) = (\pm \sqrt{3}/3, \pm \sqrt{3}/3, \pm \sqrt{3}/3).$

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dolphinday

1318 posts

dolphinday

#72 h Jan 21, 2024, 6:16 PM

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Homogenize the equations so it becomes
$\newline$
$\begin{align*} a^2b + ac^2 + bc^2 + abc \\ = b^2c + + a^2b + a^2c + abc \\ = c^2a + ab^2 + b^2c + abc \end{align*}$
The first two equations yields
$\newline$
$ac^2 + bc^2 = b^2c + a^2c \implies c^2(a + b) = c(a^2 + b^2)$ If $c = 0$ then $a = b = \pm 1$ (and cyclic permutations).
If not, we divide by $c$ and sum to get
$\newline$
$\sum_{\text{cyc}} (a - b)^2 = 0 \implies a = b = c = \pm \frac{\sqrt{3}}{3}$ $\newline$

So then our answers are all permutations of $(\pm 1, \pm 1, 0)$ and $(\pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3})$ .

This post has been edited 1 time. Last edited by dolphinday, Jan 21, 2024, 6:18 PM

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EpicBird08

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EpicBird08

#73 h Apr 27, 2024, 1:55 AM

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The only solutions are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right), \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ , $(1,1,0), (-1, -1, 0)$ , and permutations, which clearly work.

Now we prove that these are the only solutions. Note that no two of $a,b,c$ can equal $0$ because this would violate the given condition. Now suppose that one of them was equal to $0,$ say WLOG $a.$ Then the given conditions become $bc = 1$ and $b^2 c = b = c.$ These imply that $b = c = \pm 1,$ giving us the solutions $(0, 1, 1)$ and $(0, -1, -1).$ By making $b$ or $c$ equal to $0,$ permutations of these follow.

We now assume that all of $a,b,c$ are nonzero. We can homogenize the given condition using $ab + bc + ca = 1$ to get
$a^2 b + abc + c^2 a + c^2 b = b^2 c + abc + a^2 b + a^2 c = c^2 a + abc + b^2 c + b^2 a.$ We will look at the conditions for each pair of these expressions to be equal. Let's just look at the conditions needed for the first two expressions to be equal. Then
$a^2 b + abc + c^2 a + c^2 b = b^2 c + abc + a^2 b + a^2 c.$ Subtracting $a^2 b + abc$ from both sides gives
$c^2 a + c^2 b = a^2 c + b^2 c.$ Dividing both sides by $c$ (since $c \ne 0$ ), we get
$a^2 + b^2 = ca + cb.$ By looking at the other pairs of expressions, we get
$b^2 + c^2 = ba + ca$ and
$c^2 + a^2 = ba + bc.$ Adding all of these equations, we get
$2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca,$ which rearranges into
$(a-b)^2 + (b-c)^2 + (c-a)^2.$ This implies that $a = b = c.$ Finally, plugging into the first condition gives $3a^2 = 1 \rightarrow a = \pm \frac{1}{\sqrt{3}}.$ These give us the solutions $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ and $\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right).$

We have shown that these are the only solutions, so we are done.

This post has been edited 1 time. Last edited by EpicBird08, Apr 27, 2024, 1:55 AM

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de-Kirschbaum

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de-Kirschbaum

#74 h Aug 18, 2024, 11:43 PM

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Consider the following transformation

$a^2b+abc+bc^2+c^2a=b^2c+a^2b+a^2c=c^2a+b^2a+b^2c+abc$ $a^2b+bc^2+c^2a=b^2c+a^2b+a^2c=c^2a+b^2a+b^2c$ Now suppose that $a,b,c \neq 0$ . Then we have
$\begin{align*} bc+ca&=b^2+a^2 \\ ab+ac&=b^2+c^2 \\ ab+bc&=a^2+c^2 \end{align*}$ This gives $a^2=bc, b^2=ac,c^2=ab$ , which means $b^3=c^3=a^3 \implies a=b=c$ . Thus $3a^2=1 \implies a=\pm \frac{1}{\sqrt{3}}$ . So $(a,b,c)=(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$ are solutions.

Note that at most one of $a,b,c$ can be $0$ . WLOG let $a=0$ , then $bc^2=b^2c \implies b=c$ , so $b^2=1 \implies b= \pm 1$ . Thus all permutations of $(1,1,0), (-1,-1,0)$ are also solutions.

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L13832

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L13832

#75 h Aug 22, 2024, 11:57 AM

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juckter wrote:

Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and $a^2b + c = b^2c + a = c^2a + b.$

We have
$a^2b + ac^2 + bc^2 + abc = b^2c + a^2b + a^2c + abc = c^2a + ab^2 + b^2c + abc$ because $ab+bc+ca=1$ .
Case I: $abc\neq 0$
$\begin{align*} a^2b+c=b^2c+a \implies a(ab-1)=c(b^2-1)\implies a(-bc-ca)=c(b^2-1)\implies a^2+b^2+ab=1 \end{align*}$ Similarly we get $b^2+c^2+bc=1$ and $c^2+a^2+ac=1$ . (Motivation: To use C-S)
On adding all the equations we get $2(a^2+b^2+c^2)+ab+bc+ca=3$ , so $a^2+b^2+c^2=1$ .
By C-S the equality-case is achieved at $a=b=c$ , so we get $\boxed{a=b=c=\pm \frac{1}{\sqrt{3}}}$
Case II: $abc=0$
WLOG assume, $a=0$ , then
$\begin{align*} ab+bc+ca=1&\implies bc=1\\ b^2c+a=a^2b+c&\implies b=c \end{align*}$
Hence, the other solutions are $\boxed{(a,b,c)=(0,1,1),(0,-1,-1) \;\text{and it's permutation.}}$

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pie854

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pie854

#76 h Sep 1, 2024, 7:07 PM

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Let us suppose that $a,b,c$ are non zero. In that case, $0=b^2c+a-a^2b-c=a\left (1-ab \right)+c\left (b^2-1 \right)=a\left (bc+ca\right )+c\left (b^2-1\right )=c\left (ab+a^2+b^2-1\right )\implies a^2+b^2+ab=1.$ And similarly, $b^2+c^2+bc=1$ and $c^2+a^2+ca=1$ . Adding all of them we get $a^2+b^2+c^2=1$ . Thus $\frac 12 \left((a-b)^2+(b-c)^2+(c-a)^2\right)=a^2+b^2+c^2-ab-bc-ca=0\implies a=b=c.$ And, $3a^2=1\Rightarrow a=b=c=\pm \frac{1}{\sqrt 3}$ .

Now, if any one of them is $0$ , then we get the other two are both equal to $1$ or to $-1$ .

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VideoCake

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VideoCake

#77 h Jan 22, 2025, 1:17 PM

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Answer. The solution are the triples $(0, 1, 1), (0, -1, -1), (\frac{\sqrt3}{3}, \frac{\sqrt3}{3}, \frac{\sqrt3}{3}), (-\frac{\sqrt3}{3}, -\frac{\sqrt3}{3}, -\frac{\sqrt3}{3})$ and their permutations.
Solution. Note that if one of the numbers is zero, say $a = 0$ , then $a^2b + c = c^2a + b \implies b = c$ and $ab + bc + ca = 1 \implies b = c = \pm1$ . For $(0, 1, 1)$ and $(0, -1, -1)$ we indeed satisfy both equations, as one can check. This means we can now assume that $a, b, c \neq 0$
Note that if two of the numbers are equal, say $x := a = b$ , then $x^3 + c = x^2c + x = xc^2 + x \implies x^2c = xc^2$ and as $x, c \neq 0$ we have $x = c$ . This leads to $a = b = c$ . As the second equation is always going to be true with equal $a, b, c$ , the solution triples are in the form of $(k, k, k)$ with $3k^2 = 1 \Longleftrightarrow k = \pm \frac{\sqrt3}{3}$ . This means that we can now assume that $a, b, c$ are pairwise distinct.

Onto the general case. We know that $ab + bc + ca = 1$ , so
$a^2b + c(ab + bc + ca) = b^2c + a(ab + bc + ca)$ which is equivalent to $bc^2 + c^2a = b^2 + a^2$ for $a, b, c \neq 0$ . Without loss of generality, we can assume that at least one number of the three is positive and that $a < b < c$ . Suppose that $b > 0$ . Then, $a(b + c) = b^2 + c^2 > b(b + c)$ . However, $b + c$ is positive, so $a > b$ , contradiction. Suppose that $b < 0$ . By assumption, we get $a < 0$ . Using similar arguments we can show that $c(a + b) = a^2 + b^2 > 0$ . However, $a + b$ is negative, so $c$ must be negative, contradiction!

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sansgankrsngupta

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sansgankrsngupta

#78 h Feb 18, 2025, 3:58 PM

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SLIGHTLY faster

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Reason: -

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Marcus_Zhang

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Marcus_Zhang

#79 h Mar 1, 2025, 4:07 AM

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Homogenizing ftw.

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eg4334

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eg4334

#80 h Mar 8, 2025, 11:34 PM • 2 Y

Y by blueprimes, Marcus_Zhang

Happy international womens day everyone! Solved with lpieleanu, Arf, blueprimes, Sigma_Pie, resources

we examine the first equation and rewrite into $a^2b + c(ab+bc+ac) = b^2c + a(ab+bc+ac)$ . Things cancel and we are left with either $a=c$ or these pesky zero cases. For simplicity assume all are nonzero. Then all are equal so we have the pairs $\pm (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} )$ . If say $a=0$ then its not hard to find that $b=1, c=1$ or $b=-1, c=-1$ . Then we also have those cyclic permutations.

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cursed_tangent1434

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cursed_tangent1434

#81 h Mar 24, 2025, 4:29 PM

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We claim that the set of all triples are $\pm \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ and $(\pm 1, \pm 1 , 0)$ where the signs correspond and cyclic permutations. It is straightforward to show that these solutions indeed work. We now show that the are the only ones. The idea is to homogenize the given expressions.

We first deal with the edge cases. If $c=0$ note $ab=1$ and $a^2b=a=b$ . Thus, $a,b \ne 0$ so we have $a^2=1$ this implies $a = b = \pm 1$ as desired. Thus, in what follows assume that $a,b,c \ne 0$ . Now,
$a^2b+(ab+bc+ca)c = a^2b+c = b^2c+a = b^2c +(ab+bc+ca)a$ Thus,
$\begin{align*} a^2b+(ab+bc+ca)c & = b^2c +(ab+bc+ca)a \\ bc^2 + ac^2 &= b^2c+a^2c \\ c(a+b) &= a^2+b^2 \end{align*}$ The smart way to finish now is to sum cyclically to victory. However, the following more straightforward finish is also possible. Note,
$c= \frac{a^2+b^2}{a+b}$ or $a+b=0$ . We first exclude the case $a+b=0$ (and similarly the other pairwise sums). Note that if $a+b=0$ then, $ab=1$ . Thus,
$a+c=a^2b+c=b^2c+a$ which implies $b^2=1$ so $b= \pm 1$ and $a=\pm 1$ where the signs alternate. But this implies $ab=-1 \ne 1$ which is a clear violation.
Now, we substitue the above expression for $c$ into the original condition,
$1=ab+bc+ca=ab+\frac{a^2b+b^3}{a+b}+\frac{b^3+ab^2}{a+b}$ which rearranges to,
$(a+b)(a^2+b^2+ab-1)=0$ Since we have excluded the possibility of $a+b=0$ this implies $a^2+b^2+ab-1=0$ . A similar argument yields that $b^2+c^2 + bc -1 =0$ and $c^2+a^2 +ca -1 =0$ . Now, obtaining the pairwise products of these relations we have,
$(a-b)(a+b+c)=a^2-b^2+c(a-b)=0$ Now, if $a+b+c=0$ we note,
$ab-c^2=ab+c(a+b)=1$ so $c^2a+a=a^2b$ . However we also have that $c^2a+b=a^2b+c$ so obtaining the difference of these two expressions we have $a-b=-c$ so
$-b=a+c=b$ which implies that $b=0$ which we have already eliminated. Thus, $a+b+c \ne 0$ and we must have $a=b$ . Similarly we conclude that $b=c$ and $c=a$ , so $a=b=c$ . Plugging this into our original condition we have,
$3a^2=1$ which implies that all solutions are indeed of the claim forms.

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