Find all quadruples (r, s, t, u)

[anemone]

Find all quadruples $(r,\,s,\,t,\,u)$ of positive real numbers such that $rstu=1$, $r^{2012}+2012s=2012t+u^{2012}$ and $2012r+s^{2012}=t^{2012}+2012u$.

 

[Albert1]

Find all quadruples $(r,\,s,\,t,\,u)$ of positive real numbers such that $rstu=1---(1)$,
$r^{2012}+2012s=2012t+u^{2012}---(2)$ and
$2012r+s^{2012}=t^{2012}+2012u---(3)$.

Spoiler

by observation ,we may let:
$r=u=\dfrac {1}{s},$ and $s=t$

 

[anemone]

Spoiler

by observation ,we may let:
$r=u=\dfrac {1}{s},$ and $s=t$

Hi Albert,

Thanks for your reply...perhaps you want to prove that there are no other solutions other than the one you cited by using the AM-GM inequality formula?

 

[anemone]

Solution of other:

Spoiler

Rewrite the last two equations as

$r^{2012}-u^{2012}=2012(t-s)$---(1)

$t^{2012}-s^{2012}=2012(r-u)$---(2)

Next, observe that $r=u$ holds iff $t=s$ holds. In that case, the last two equations are satisfied and the condition $rstu=1$ leads to a set of valid quadruples of the form $(r,\,s,\,t,\,u)=(k,\,\dfrac{1}{k},\,\dfrac{1}{k},\,k)$ for any $k>0$.

We now show that there are no other solutions. Assume that $r\ne u$ and $t\ne s$. Multiply the equation (1) by (2) we have

$(r^{2012}-u^{2012})(t^{2012}-s^{2012})=2012^2(t-s)(r-u)$

and divide the LHS by the RHS to get

$\dfrac{r^{2011}+\cdots+r^{2011-i}u^i+\cdots+u^{2011}}{2012}\cdot \dfrac{t^{2011}+\cdots+t^{2011-i}s^i+\cdots+s^{2011}}{2012}=1$

Now, applying the AM-GM inequality to the first factor, we have

$\dfrac{r^{2011}+\cdots+r^{2011-i}u^i+\cdots+u^{2011}}{2012}>\sqrt[2012]{(ru)^{\small\dfrac{2011\times 2012}{2}}}=(ru)^{\small{\dfrac{2011}{2}}}$

The inequality is strict, since equality holds only if all terms in the mean are equal to each other, which happpens only if $r=u$.

Similarly, we find

$\dfrac{t^{2011}+\cdots+t^{2011-i}s^i+\cdots+s^{2011}}{2012}>\sqrt[2012]{(ts)^{\small\dfrac{2011\times 2012}{2}}}=(ts)^{\small{\dfrac{2011}{2}}}$

Multiply both inequalities we obtain $(ru)^{\small{\dfrac{2011}{2}}}(ts)^{\small{\dfrac{2011}{2}}}<1$ which is equivalent to $rstu<1$, a contradiction.

 

[CellCoree]

I would first recognize that this problem involves finding a set of four positive real numbers that satisfy three equations. This is a system of equations, and solving systems of equations is a common problem in mathematics and science. To find all possible solutions, I would approach this problem by first isolating one variable in each equation and substituting it into the other equations.

Starting with the first equation, $rstu=1$, we can isolate one variable by dividing both sides by the other three variables. This gives us $r=\frac{1}{stu}$. We can then substitute this value for $r$ into the other two equations to get:

$\frac{1}{stu^{2012}}+2012s=2012t+u^{2012}$ and $2012\left(\frac{1}{stu}\right)+s^{2012}=t^{2012}+2012u$

Next, we can rearrange these equations to get all the variables on one side and a constant on the other side:

$2012t-u^{2012}=\frac{1}{stu^{2012}}-2012s$ and $2012u-t^{2012}=s^{2012}-2012\left(\frac{1}{stu}\right)$

Now, we can use the substitution method to solve this system of equations. We can substitute the first equation into the second equation to get:

$2012u-\left(\frac{1}{stu^{2012}}-2012s\right)^{2012}=s^{2012}-2012\left(\frac{1}{stu}\right)$

This gives us a single equation with only three variables: $u$, $s$, and $t$. We can then use algebraic manipulation to solve for one of these variables in terms of the other two. For example, we can solve for $u$ by isolating it on one side of the equation:

$2012u=s^{2012}-2012\left(\frac{1}{stu}\right)+\left(\frac{1}{stu^{2012}}-2012s\right)^{2012}$

$u=\frac{s^{2012}+2012s-2012\left(\frac{1}{stu}\right)+\left(\frac{1}{stu^{2012}}\right)^{2012}}{2012}$

We