IMO LongList 1985 CYP2 - System of Simultaneous Equations

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IMO LongList 1985 CYP2 - System of Simultaneous Equations

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#1 h Sep 10, 2010, 10:57 PM • 2 Y

Y by Adventure10, cubres

Solve the system of simultaneous equations
$\sqrt x - \frac 1y - 2w + 3z = 1,$ $x + \frac{1}{y^2} - 4w^2 - 9z^2 = 3,$ $x \sqrt x - \frac{1}{y^3} - 8w^3 + 27z^3 = -5,$ $x^2 + \frac{1}{y^4} - 16w^4 - 81z^4 = 15.$

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Dr Sonnhard Graubner

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#2 h Oct 22, 2010, 9:33 PM • 3 Y

Y by Adventure10, Mango247, cubres

hello, i have found $x=1,y=1/2,z=1/3,w=-1/2$ .
Sonnhard.

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anonymouslonely

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#3 h Sep 20, 2011, 6:51 AM • 3 Y

Y by Adventure10, Mango247, cubres

can you post your proof, please?

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#4 h Sep 20, 2011, 7:12 AM • 3 Y

Y by Adventure10, Mango247, cubres

When I was posting IMO LongList 1985 problems, this problem really bothered me. I posted it here (with changed variables). Read that topic to find the solution.

Moderator says: updated link is https://artofproblemsolving.com/community/c4h366927

This post has been edited 1 time. Last edited by v_Enhance, Apr 29, 2023, 1:10 AM

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Dr Sonnhard Graubner

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#5 h Sep 20, 2011, 5:32 PM • 3 Y

Y by Adventure10, Mango247, cubres

hello, checking again my solution, i have found no mistake.
Sonnhard.

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#6 h Apr 29, 2023, 1:09 AM • 10 Y

Y by teomihai, HamstPan38825, Amir Hossein, FriIzi, bryanguo, peace09, EpicBird08, two_steps, Assassino9931, cubres

Solution from Twitch Solves ISL:

Let $a = \sqrt x$ , $b = -1/y$ , $c = 2w$ , $d = -3z$ . $\begin{align*} a + b - c - d &= 1 \\ a^2 + b^2 - c^2 - d^2 &= 3 \\ a^3 + b^3 - c^3 - d^3 &= -5 \\ a^4 + b^4 - c^4 - d^4 &= 15. \end{align*}$ This condition is the same as saying $a^n + b^n + (-1)^n + (-1)^n = c^n + d^n + (-2)^n + 1^n \qquad n = 1, 2, 3, 4.$ which is equivalent to saying the multiset $\{a,b,-1,-1\}$ is the same as the multiset $\{c,d,-2,1\}$ , (because Newton's formulas imply the polynomials with these roots have the same coefficients). Therefore, $\{a,b\} = \{-2,1\}$ while $c=d=-1$ .
Going back, with $a > 0$ this gives only one solution, which evidently works: $(x,y,w,z) = (1, 1/2, -1/2, 1/3).$

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#7 h May 1, 2023, 8:33 PM • 1 Y

Y by cubres

v_Enhance wrote:

Solution from Twitch Solves ISL:
Let $a = \sqrt x$ , $b = -1/y$ , $c = 2w$ , $d = -3z$ ....

This indeed deserves to be on the Kaywañan* Algebra Contest. Thanks Evan for the neat solution!

*A glimpse is attached

Attachments:

Kaywanian-MonsterTrigContest.pdf (176kb)

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#8 h Dec 2, 2023, 2:23 PM • 2 Y

Y by ehuseyinyigit, cubres

Set $a=\sqrt x$ , $b=-1/y$ , $c=2w$ and $d=-3z$ . The given equations can be rewritten as $a^n+b^n+2 \times(-1)^n =c^n+d^n+(-2)^n+1^n$ for $n\in \{1,2,3,4\}$ .

Hence, by Newton's formula, we deduce that $(a,b,-1,-1)$ is a permutation of $(c,d,-2,-1)$ .

Now since $a>0$ , it suffices to try all of the possible cases and find $(a,b,c,d)=(1,-2,-1,-1)$ which implies $(x,y,w,z) = (1, \frac{1}2, -\frac{1}2,\frac{1}3)$

$\mathbb{Q.E.D.}$

This post has been edited 1 time. Last edited by Cusofay, Dec 2, 2023, 2:23 PM
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#9 h Apr 16, 2024, 10:44 PM • 1 Y

Y by cubres

Solution

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#10 h May 5, 2024, 1:13 PM • 1 Y

Y by cubres

Denote $\sqrt x = a$ , $-\frac{1}{y} = b$ , 2w = c, -3z = d. Now we plug in these into the system and we get a + b - c - d = 1, $a^2 + b^2 - c^2 - d^2 = 3$ , $a^3 + b^3 - c^3 - d^3 = -5$ , $a^4 + b^4 - c^4 - d^4 = 15$ . But this is basically equivalent to $a^n + b^n + (-1)^n + (-1)^n = c^n + d^n + (-2)^n + 1^n$ for n = 1,2,3,4. By Newton's formulas the polynomials with these roots have the same coefficients which means that (a, b, -1, -1) is some permutation of (c, d, -2, 1). Since $a = \sqrt x$ , then $a \geq 0$ $\Rightarrow$ for a is left to equal 1 $\Rightarrow$ b = -2, c = d = -1 $\Rightarrow$ (a, b, c, d) = (1, -2, -1, -1) $\Rightarrow$ $(x, y, w, z) = (1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3})$ .

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combowomborhombo

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#11 h Aug 9, 2024, 1:08 AM • 1 Y

Y by cubres

Since the variables look pretty nasty, we can substitute stand-alone variables instead of contrived variables. The substitutions to be made are:

$a = \sqrt{x}, \quad b = -\frac{1}{y}, \quad c = 2w, \quad \text{and} \quad d = -3z$
This gives the following equations:

$\begin{align*} a + b - c - d &= 1, \\ a^2 + b^2 - c^2 - d^2 &= 3, \\ a^3 + b^3 - c^3 - d^3 &= -5, \\ a^4 + b^4 - c^4 - d^4 &= 15. \end{align*}$
These formulas can be rearranged to get:

$\begin{align*} a + b &= 1 + b + c, \\ a^2 + b^2 &= 3 + c^2 + d^2, \\ a^3 + b^3 &= -5 + c^3 + d^3, \\ a^4 + b^4 &= 15 + c^4 + d^4. \end{align*}$
We can replace the numbers $1, 3, -5, 15$ with $2^n - (-1)^n - (-1)^n + 1^n$ , where the equations now become:

$a^i + b^i + (-1)^i + (-1)^i = c^i + d^i + (-2)^i + 1^i \quad \text{for } 1 \le i \le 4.$
Since the first four Newton's sums are fixed in the degree 4 polynomial, we can conclude that the coefficients are the same. Therefore, the multisets $\{a, b, -1, -1\}$ and $\{c, d, -2, 1\}$ are permutations of each other.

Setting $c = d = -1$ , and since $a$ cannot be negative as it is under a square root, we have $a = 1$ and $b = -2$ .

Finally, computing the original variables, we get the solution as:

$(x, y, w, z) = \boxed{\left(1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}\right)}$

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eg4334

#12 h Aug 12, 2024, 4:21 AM • 1 Y

Y by cubres

The obvious substituiton to make is $a=\sqrt{x}$ , $b = -\frac{1}{y}$ , $c = 2w$ , and $d=-3z$ . This gives us the equations $a+b-c-d=1$ $a^2+b^2-c^2-d^2=3$ $a^3+b^3-c^3-d^3=-5$ $a^4 + b^4 - c^4 - d^4 = 15$ Now we can rearrange each of the equations as follows: $a^n + b^n + (-1)^n + (-1)^n = c^n + d^n + (-2)^n + 1^n, n \in \{ 1, 2, 3, 4\}$ Remark Now it follows that the polynomials $(x-a)(x-b)(x+1)(x+1)$ and $(x-c)(x-d)(x+2)(x-1)$ are equivalent. This is because of Newtons Sums and the fact that we have four Newton Sums and four roots. From here the finish is clear. $c=d=-1$ and $a=1$ while $b=-2$ (because $a > 0$ , we can discount $a=-2$ ). This translates to the solution $(x, y, w, z) = (1, \frac12, -\frac12, \frac13)$

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#13 h Aug 20, 2024, 1:38 AM • 1 Y

Y by cubres

Perform the substitution $a = \sqrt x$ , $b = -\frac 1y$ , $c = 2w$ , and $d=-3z$ . We obtain:
$\begin{align*} a + b - c - d &= 1, \\ a^2 + b^2 - c^2 - d^2 &= 3, \\ a^3 + b^3 - c^3 - d^3&= -5, \\ a^4 + b^4 - c^4 - d^4 &= 15. \end{align*}$ In fact, we actually have $a^k+b^k+(-1)^k + (-1)^k = c^k + d^k + 1^k + (-2)^k$ for $k=1, 2, 3, 4$ . Consider the polynomials $(x-a)(x-b)(x+1)^2$ and $(x-c)(x-d)(x-1)(x+2)$ . The first four Newton's sums of both of these polynomials are equal, so the polynomials must be identical. Therefore, the only solution is $a = 1, b=-2, c=-1, d=-1$ , which corresponds to $(w, x, y, z) = \boxed{\left(1, \frac 12, -\frac 12, \frac 13\right)}$ . $\blacksquare$

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#14 h Apr 16, 2025, 4:07 PM

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Solution

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#15 h Apr 25, 2025, 3:45 PM

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Define $a = \sqrt x, b = -\frac1y, c = 2w, d = -3z.$
$\begin{align*} a + b& = 1 + c + d \\ a^2 + b^2 & = 3 + c^2 + d^2 \\ a^3 + b^3 & = -5 + c^3 + d^3 \\ a^4 + b^4 & = 15 + c^4 + d^4 \\ \end{align*}$ Thus, $a^n + b^n + (-1)^n + (-1)^n = c^n + d^n + 1^n + (-2)^n \text{ for }n = 1, 2, 3, 4$ By choosing a polynomial of degree 4 with the roots $\{a, b, -1, -1\},$ we see that it must as be the same as $\{1, -2, c, d\}.$ Hence, $c = d = -1.$ By the definition of $a,$ we have that $a = 1,$ and $b = -2.$ Thus, our answer is $(a, b, c, d) = \left(1, \frac12, -\frac12, \frac13\right).$

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