Solving equations

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tushar3262

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tushar3262

#1 h Nov 24, 2017, 3:15 PM • 2 Y

Y by Adventure10, Mango247

Solve the following equations in real numbers:
$x^3+y^3=1$ $x^4+y^4=1$

This post has been edited 1 time. Last edited by tushar3262, Nov 24, 2017, 3:26 PM

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notharsh

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notharsh

#2 h Nov 24, 2017, 3:25 PM • 2 Y

Y by Adventure10, Mango247

Do you mean $x^4+y^4=1$ ?

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rchokler

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rchokler

#3 h Nov 24, 2017, 3:39 PM • 3 Y

Y by tushar3262, Adventure10, Mango247

For positive integers $m,n$ , the system $x^m+y^m=1$ and $x^n+y^n=1$ , we have $(1,0)$ and $(0,1)$ in all cases, as well as $(-1,0)$ and $(0,-1)$ in all cases where $m$ and $n$ are both even.

This easy to show.

On the other hand, it will be a more challenging problem if we were asked to solve the system above in $\mathbb{C}$ . It is possible to do so in radicals, so I will post the result later today.

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math-Passion

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math-Passion

#4 h Nov 24, 2017, 3:48 PM • 2 Y

Y by Adventure10, Mango247

tushar3262 wrote:

Solve the following equations in real numbers:
$x^3+y^3=1$ $x^4+y^4=1$

Solution

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kvedula2004

989 posts

kvedula2004

#5 h Nov 24, 2017, 3:52 PM • 2 Y

Y by Adventure10, Mango247

@above
What about $0.36+0.64=1$ ? Your solution also only holds in integers, not reals.

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MathLegend27

549 posts

MathLegend27

#6 h Nov 24, 2017, 3:54 PM • 2 Y

Y by Adventure10, Mango247

Oh ok, I see, they have to solve both equations

This post has been edited 1 time. Last edited by MathLegend27, Nov 24, 2017, 4:05 PM

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math-Passion

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math-Passion

#7 h Nov 24, 2017, 4:02 PM • 1 Y

Y by Adventure10

It must work for both equations.

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math-Passion

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math-Passion

#8 h Nov 24, 2017, 4:03 PM • 1 Y

Y by Adventure10

kvedula2004 wrote:

@above
What about $0.36+0.64=1$ ? Your solution also only holds in integers, not reals.

They dont work both inetegrs.

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kvedula2004

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kvedula2004

#9 h Nov 24, 2017, 4:08 PM • 2 Y

Y by Adventure10, Mango247

math-Passion wrote:

kvedula2004 wrote:

@above
What about $0.36+0.64=1$ ? Your solution also only holds in integers, not reals.

They dont work both inetegrs.

How do you know there isnt some exotic triple like $(x,y)=(1+2\sqrt[5]{3},2^{\sqrt{3-e}})$ (yes is doesnt work but my statement still holds)

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math-Passion

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math-Passion

#10 h Nov 24, 2017, 4:14 PM • 2 Y

Y by Adventure10, Mango247

kvedula2004 wrote:

math-Passion wrote:

kvedula2004 wrote:

@above
What about $0.36+0.64=1$ ? Your solution also only holds in integers, not reals.

They dont work both inetegrs.

How do you know there isnt some exotic triple like $(x,y)=(1+2\sqrt[5]{3},2^{\sqrt{3-e}})$ (yes is doesnt work but my statement still holds)

Yes there may be one but as you said, how do we find such a pair that will satisfy both equations.

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math-Passion

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math-Passion

#11 h Nov 24, 2017, 4:15 PM • 2 Y

Y by Adventure10, Mango247

tushar3262 wrote:

Solve the following equations in real numbers:
$x^3+y^3=1$ $x^4+y^4=1$

As I said before one of the terms must be $0$ as any other terms larger than $0$ will not work.

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Math1331Math

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Math1331Math

#12 h Nov 24, 2017, 4:27 PM • 2 Y

Y by Adventure10, Mango247

I'll solve if both $x$ and $y$ are nonnegative other cases are similar

Let $x=\sqrt{\cos{a}}, y=\sqrt{\sin{a}}$

We want to examine the function

$(\sin{a})^{\frac{3}{2}}+(\cos{a})^{\frac{3}{2}}$

$f'=\frac{3}{2}(\sin{a}\cos{a})^{\frac{1}{2}}(\cos^{\frac{1}{2}}{a}-\sin^{\frac{1}{2}}{a})$

Now it is clear to see by derivative test our local and absolute min occur at $(1,0),(0,1)$

This post has been edited 2 times. Last edited by Math1331Math, Nov 24, 2017, 4:28 PM
Reason: .

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Iyerie

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Iyerie

#13 h Nov 24, 2017, 4:30 PM • 1 Y

Y by Adventure10

Partial Solution

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Math1331Math

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Math1331Math

#14 h Nov 24, 2017, 4:48 PM • 1 Y

Y by Adventure10

Iyerie wrote:

Partial Solution

any proof?

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Nikpour

1748 posts

Nikpour

#16 h Nov 24, 2017, 6:57 PM • 1 Y

Y by Adventure10

${{x}^{3}}+{{y}^{3}}=1\Rightarrow {{({{x}^{3}}+{{y}^{3}})}^{4}}=1\Rightarrow {{x}^{12}}+{{y}^{12}}+4{{x}^{3}}{{y}^{3}}+6{{x}^{6}}{{y}^{6}}=1$ ${{x}^{4}}+{{y}^{4}}=1\Rightarrow {{({{x}^{4}}+{{y}^{4}})}^{3}}=1\Rightarrow {{x}^{12}}+{{y}^{12}}+3{{x}^{4}}{{y}^{4}}=1$ $\Rightarrow 6{{x}^{6}}{{y}^{6}}+4{{x}^{3}}{{y}^{3}}-3{{x}^{4}}{{y}^{4}}=0\Rightarrow {{x}^{3}}{{y}^{3}}(6{{x}^{3}}{{y}^{3}}-3xy+4)=0$ $xy=p\Rightarrow p=0\vee 6{{p}^{3}}-3p+4=0$

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rchokler

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rchokler

#17 h Nov 24, 2017, 11:59 PM • 7 Y

Y by tushar3262, T_B, synergy, KKMathMan, Adventure10, Mango247, Sedro

If instead you were to solve in $\mathbb{C}$ , you would get two triple answers at $(1,0)$ and $(0,1)$ and six simple answers $(A_\pm,A_\mp)$ , $(B_\pm,C_\pm)$ , and $(C_\pm,B_\pm)$ where:

$\begin{align*} A_\pm&=-\frac{1}{2}+\frac{1}{2}\sqrt[3]{1+\sqrt{2}}-\frac{1}{2}\sqrt[3]{-1+\sqrt{2}}\pm\frac{i}{2}\sqrt{1+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}}\\ B_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\ &\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\ C_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\ &\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\ \end{align*}$

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Math1331Math

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Math1331Math

#18 h Nov 25, 2017, 12:11 AM • 2 Y

Y by Adventure10, Mango247

lol u do this every time its impressive u can solve this stuff by hand

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BBBBMMMMLLLLE

349 posts

BBBBMMMMLLLLE

#19 h Nov 25, 2017, 1:32 AM • 2 Y

Y by Adventure10, Mango247

rchokler wrote:

If instead you were to solve in $\mathbb{C}$ , you would get two triple answers at $(1,0)$ and $(0,1)$ and six simple answers $(A_\pm,A_\mp)$ , $(B_\pm,C_\pm)$ , and $(C_\pm,B_\pm)$ where:

$\begin{align*} A_\pm&=-\frac{1}{2}+\frac{1}{2}\sqrt[3]{1+\sqrt{2}}-\frac{1}{2}\sqrt[3]{-1+\sqrt{2}}\pm\frac{i}{2}\sqrt{1+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}}\\ B_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\ &\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\ C_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\ &\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\ \end{align*}$

How on earth did you get these?

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integrated_JRC

3465 posts

integrated_JRC

#20 h Nov 25, 2017, 4:48 AM • 2 Y

Y by Adventure10, Mango247

As the first equation has odd exponent (3) and the second one has even exponent (4), both $(x,y)$ can't be negative.

Now, if $x$ and $y$ both are positive reals then $x^m+y^m=x^n+y^n$ yields that $m=n$ , which is not possible here.

Therefore, one of $x$ and $y$ must be $0$ . If one of them is $0$ , then the other one can be $\pm 1$ . But $(-1)^3+0^3=(-1)$ . So, none of $x$ or $y$ is $(-1)$ .

So, only possible solutions are $(x,y)=(0,1)$ & $(1,0)$ .

This post has been edited 2 times. Last edited by integrated_JRC, Dec 2, 2017, 5:27 AM

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rchokler

2975 posts

rchokler

#21 h Nov 25, 2017, 5:15 AM • 2 Y

Y by Adventure10, Mango247

BBBBMMMMLLLLE wrote:

How on earth did you get these?

This is a very long calculation.

First you must derive the equation for $x$ and pull as many linear factors as possible by use of RRT:

$x^3+y^3=1\implies y^{12}=(1-x^3)^4$ and $x^4+y^4=1\implies y^{12}=(1-x^4)^3$ . This means you have:

$(1-x^3)^4-(1-x^4)^3=0\implies 2x^{12}-4x^9-3x^8+6x^6+3x^4-4x^3=0\implies x^3(x-1)^3(2x^6+6x^5+12x^4+16x^3+15x^2+9x+4)=0$
Now we see the multiplicities of the trivial roots. We are left to solve the sextic equation to find the difficult roots.

Basically, we can show that the sextic doesn't factor over $\mathbb{Q}$ , so if solvable by radicals, there is an field extension of $\mathbb{Q}$ over which it either factors into three quadratics or into two cubics.

In this case you will find yourself in luck trying the three quadratics case. Here is one of the quadratic factors:

$2x^2+\left[2-2\sqrt[3]{1+\sqrt{2}}+2\sqrt[3]{-1+\sqrt{2}}\right]x+\sqrt[3]{8+6\sqrt{2}}-\sqrt[3]{-8+6\sqrt{2}}=0$

There are two others as well, and you can solve each with the quadratic formula. Note that the factorization is over the field $\mathbb{Q}\left[\sqrt[3]{1+\sqrt{2}},i\sqrt{3}\right]$ .

However, for two of the factors, applying the quadratic formula gives expressions with complex numbers inside a square root. You must use the formula $\sqrt{z}=\sqrt{\frac{|z|+\Re(z)}{2}}+i\text{ sgn}(\Im(z))\sqrt{\frac{|z|-\Re(z)}{2}}$ to take care of this.

This post has been edited 2 times. Last edited by rchokler, Nov 25, 2017, 5:40 AM

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synergy

425 posts

synergy

#22 h Nov 25, 2017, 2:48 PM • 2 Y

Y by Adventure10, Mango247

rchokler wrote:

If instead you were to solve in $\mathbb{C}$ , you would get two triple answers at $(1,0)$ and $(0,1)$ and six simple answers $(A_\pm,A_\mp)$ , $(B_\pm,C_\pm)$ , and $(C_\pm,B_\pm)$ where:

$\begin{align*} A_\pm&=-\frac{1}{2}+\frac{1}{2}\sqrt[3]{1+\sqrt{2}}-\frac{1}{2}\sqrt[3]{-1+\sqrt{2}}\pm\frac{i}{2}\sqrt{1+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}}\\ B_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\ &\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\ C_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\ &\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\ \end{align*}$

*claps*

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Archeon

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Archeon

#23 h Nov 25, 2017, 4:00 PM • 2 Y

Y by Adventure10, Mango247

Am I oversimplifying?

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Nikpour

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Nikpour

#24 h Nov 25, 2017, 9:08 PM • 2 Y

Y by Adventure10, Mango247

$\begin{array}{l} 6{p^3} - 3p + 4 = 0\\ p = u + v \Rightarrow 6{(u + v)^3} - 3(u + v) + 4 = 0 \Rightarrow 6({u^3} + {v^3}) + (u + v)(18uv - 3) + 4 = 0\\ 18uv - 3 = 0 \Rightarrow v = \frac{1}{{6u}} \Rightarrow 6({u^3} + \frac{1}{{216{u^3}}}) + 4 = 0 \Rightarrow u = \cdots \Rightarrow v = \cdots \Rightarrow p = \cdots \Rightarrow y = \frac{ \cdots }{x}\\ {x^3} + {y^3} = 1 \Rightarrow {x^3} + \frac{{{ \cdots ^3}}}{{{x^3}}} = 1 \Rightarrow x = \cdots \Rightarrow y = \cdots \end{array}$

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math-Passion

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math-Passion

#25 h Nov 25, 2017, 9:10 PM • 1 Y

Y by Adventure10

Archeon wrote:

Am I oversimplifying?

You are wrong, we can also have $(-1,0),(0,-1)$

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T_B

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T_B

#26 h Nov 25, 2017, 9:13 PM • 1 Y

Y by Adventure10

math-Passion wrote:

Archeon wrote:

Am I oversimplifying?

You are wrong, we can also have $(-1,0),(0,-1)$

No you are wrong.. $(-1)^3 + 0^3 \neq 1$

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math-Passion

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math-Passion

#27 h Dec 2, 2017, 1:01 AM • 2 Y

Y by Adventure10, Mango247

T_B wrote:

math-Passion wrote:

Archeon wrote:

Am I oversimplifying?

You are wrong, we can also have $(-1,0),(0,-1)$

No you are wrong.. $(-1)^3 + 0^3 \neq 1$

Sorry, i see my mistake, I looked at the second equation only.

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SS4

901 posts

SS4

#28 h Dec 2, 2017, 1:34 AM • 2 Y

Y by Adventure10, Mango247

If you substitute and simplify you get a 12th degree polynomial... don't think the solutions are too simple.

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integrated_JRC

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integrated_JRC

#29 h Dec 2, 2017, 5:27 AM • 2 Y

Y by Adventure10, Mango247

jrc1729 wrote:

As the first equation has odd exponent (3) and the second one has even exponent (4), both $(x,y)$ can't be negative.

Now, if $x$ and $y$ both are positive reals then $x^m+y^m=x^n+y^n$ yields that $m=n$ , which is not possible here.

Therefore, one of $x$ and $y$ must be $0$ . If one of them is $0$ , then the other one can be $\pm 1$ . But $(-1)^3+0^3=(-1)$ . So, none of $x$ or $y$ is $(-1)$ .

So, only possible solutions are $(x,y)=(0,1)$ & $(1,0)$ .

Is my solution correct ? Please let me know.

This post has been edited 1 time. Last edited by integrated_JRC, Dec 2, 2017, 5:27 AM

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rchokler

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rchokler

#30 h Mar 8, 2018, 5:02 AM • 2 Y

Y by Adventure10, Mango247

I would like to add one note about the radical expressions. There is actually a very easy way to find them if you use Vieta's formulas. Basically, yo can let $a=x+y$ and $b=xy$ and turn this problem into a system in $a,b$ .

$x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3ab$ and $x^4+y^4=(x^2+y^2)^2-2x^2y^2=((x+y)^2-2xy)-2(xy)^2=(a^2-2b)^2-2b^2=a^4-4a^2b+2b^2$ .

This gives the system:
$a^3-3ab=1$
$a^4-4a^2b+2b^2=1$

This new system can be solved in $\mathbb{C}$ routinely as follows:

The first equation becomes $b=\frac{a^3-1}{3a}$ . Substitution gives:
$a^4-4a^2b+2b^2=1\implies a^4-\frac{4a(a^3-1)}{3}+\frac{2(a^3-1)^2}{9a^2}=1\implies a^6-8a^3+9a^2-2=0\implies(a-1)^3(a^3+3a^2+6a+2)=0$
Now it is easy to solve for all pairs $(a,b)$ (Cardano for the cubic factor). Then for each of these, solving for $(x,y)$ by radicals is a matter of the quadratic formula.

Feel free to try this, with systems of symmetric equations when they are asked in the forums. It usually works!

This post has been edited 2 times. Last edited by rchokler, Mar 8, 2018, 5:07 AM

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