Cyclic system of equations

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Source: OMEC Ecuador National Olympiad Final Round 2024 N3 P1 day 1

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KAME06

#1 h Feb 28, 2025, 12:14 AM

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Find all real solutions:
$\begin{cases}a^3=2024bc \\ b^3=2024cd \\ c^3=2024da \\ d^3=2024ab \end{cases}$

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#2 h Feb 28, 2025, 1:48 AM

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Multiply all together. Either $abcd=0$ or $abcd=2024^4$ .
If WLOG $a=0$ , then $c=0$ , then $d=0$ , then $b=0$ .
$\begin{cases}a^3=2024bc \\ c^3=2024da \end{cases} \implies a^3 c^3=2024^6=b^3 d^3 \implies ac=bd=2024^2$ $\begin{cases}a^3=2024bc \\ b^3=2024cd \end{cases} \implies \begin{cases}a^4=2024^3 b \\ a b^4=2024^5 \end{cases}$ Thus, $\left(\frac{a^4}{b}\right)^5= \left(ab^4\right)^3 \implies a^{17}=b^{17} \implies a=b=c=d=\sqrt{2024}$

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#3 h Apr 6, 2025, 11:56 PM

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Dear navier3072,

to me there seems to be a slight mistake in your last line which leads to a wrong result. I can't follow some of your implications as you do not explain them. Anyway, it is true that all four variables turn out to have the same value, but their common value is 2024 (in this case where neither of them is zero).

This post has been edited 2 times. Last edited by Rainbow1971, Apr 7, 2025, 12:02 AM

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#4 h Apr 7, 2025, 12:27 AM

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Rainbow1971 wrote:

...I can't follow some of your implications as you do not explain them. ...

I think the last line has quite a bit of steps missing.

Tag the equations: $a^3 = 2024bc \tag{1}$ $b^3 = 2024cd \tag{2}$ $c^3 = 2024da \tag{3}$ $d^3 = 2024ab \tag{4}$ Multiplying all equations together gives $a^3b^3c^3d^3 = 2024^4a^2b^2c^2d^2$ . If none of the variables are equal to 0, then we divide both sides by $a^2b^2c^2d^2$ to get $abcd = 2024^4 \tag{5}$ Multiplying (1) and (3) together gives $a^3c^3 = 2024^2abcd$ . Multiplying (2) and (4) together gives $b^3d^3 = 2024^2abcd$ as well. Substituting (5), these are equivalent to $a^3c^3 = b^3d^3 = 2024^6$ . Cube rooting transforms this into $ac = bd = 2024^2 \tag{6}$
Multiplying both sides of (1) by $a$ , we get $a^4 = 2024bac$ . Substituting (6), this is equivalent to $a^4 = 2024^3b$ . Divide both sides by $b$ to obtain $\frac{a^4}{b} = 2024^3 \tag{7}$ Multiplying both sides of (2) by $ab$ , we get $ab^4 = 2024abcd$ . Substitute (5) and we have $ab^4 = 2024^5 \tag{8}$ Raise both sides of (7) to the power of 3 to obtain $(\frac{a^4}{b})^5 = 2024^{15}$ . Raise both sides of (8) to the power of 5 to obtain $(ab^4)^3 = 2024^{15}$ . Therefore $(\frac{a^4}{b})^5 = (ab^4)^3$ . Expanding both of these results in $\frac{a^{20}}{b^5} = a^3b^{12}$ . Multiply both sides by $\frac{b^5}{a^3}$ to see that this is equivalent to $a^{17} = b^{17}$ . This is equivalent to: $a = b$ We can substitute this into (6) to see that $bc = bd$ . Divide both sides by $b$ to find that $c = d$ Substitution into (2) and (4) allows us to get $b^3 = 2024d^2$ and $d^3 = 2024b^2$ . Raise to the power of 2 and 3 respectively to get $b^6 = 2024^2d^4$ and $d^9 = 2024^3b^6$ . Substitute the former into the latter to get $d^9 = 2024^5d^4$ , from which we can conclude $d = 2024$ . Because $c=d$ we get $c = 2024$ . Plug these into (2), we get $b^3 = 2024^3$ and therefore $b = 2024$ . Because $a=b$ we have $a = 2024$ .

This post has been edited 2 times. Last edited by maromex, Apr 7, 2025, 12:29 AM
Reason: latex mistake

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#5 h Apr 7, 2025, 1:47 AM

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Thank you for your kind and helpful response.

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