How many triples $(a,b,c)$ with $a,b,c \in \mathbb{R}$ satisfy the following system? $$\begin{cases} a^4-b^4=c \\ b^4-c^4=a \\ c^4-a^4=b \end{cases}$$.
Problem
Source: Lusophon Mathematical Olympiad 2022 Problem 1
Tags: algebra, system of equations
no_room_for_error
04.11.2022 00:39
If one of the variables, say $c$, equals $0$, then it follows that $a^4-b^4=0,\ b^4=a$ and $-a^4=b$. $a^4=b^4$ implies that $a=\pm b$ and solving the two cases, we arrive at two possible solutions $(0, 0, 0)$ and $(1, -1, 0)$. Similarly, if $b=0$ or $a=0$, we find two more cyclic solutions $(0, 1, -1)$ and $(-1, 0, 1)$. Now, assume that $abc \neq 0$. Summing the three equations, we get $a+b+c=0$. Then: $$c=a^4-b^4=(a+b)(a-b)(a^2+b^2)=-c(a-b)(a^2+b^2) \implies (a-b)(a^2+b^2)=-1$$ Since $a^2+b^2 > 0$ it follows that $a-b < 0$. Similarly, from the other two equations, we get that $b-c<0$ and $c-a<0$. Summing, we arrive at $0 < 0$, a contradiction. Therefore, there are $4$ triples.
rchokler
04.11.2022 01:21
Indeed of $16$ total solutions, $4$ are real. Real: $(0,0,0)$ and cycles of $(1,-1,0)$ Complex: Cycles of $(e^{-2\pi i/3},e^{\pi i/3},0)$, $(e^{2\pi i/3},e^{-\pi i/3},0)$, $\left(\frac{e^{\pi i/6}}{\sqrt[6]{3}},-\frac{i}{\sqrt[6]{3}},\frac{e^{5\pi i/6}}{\sqrt[6]{3}}\right)$, $\left(\frac{e^{-\pi i/6}}{\sqrt[6]{3}},\frac{i}{\sqrt[6]{3}},\frac{e^{-5\pi i/6}}{\sqrt[6]{3}}\right)$
PennyLane_31
11.05.2023 21:38
Let $\begin{cases} a^4-b^4=c (1) \\ b^4-c^4=a (2) \\ c^4-a^4=b (3) \end{cases}$ First, notice that if $a=b$, by (1)$\implies c=0\implies a+b=0\implies 2a=0\implies a=0$. So, $(a,b,c)=(0, 0, 0)$ is a solution (check in (1), (2) and (3)). If only one of $a,b$ or $c$ is 0, WLOG, $a=0$, by (1), we have $0-b^4= c\implies c=-b^4$ and, by (2)$\implies b^4=c^4$. So, $c= -b^4= -c^4\implies 1= -c^3\Leftrightarrow c= -1\implies b= 1$. So, we will have the triple $(a,b,c)= (0, 1, -1)$, and so as the cyclic solutions $(a,b,c)= (-1, 0, 1); (1, -1, 0)$. Next, we will prove that there are no more solutions than those.
Let's assume now that $a\neq b\neq c\neq 0$. Now, the sum $(1)+(2)+(3)\implies a+b+c= 0$. We know that as $a,b, c\in \mathbb{R}\implies a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac)$. Since $a+b+c=0\implies a^3+b^3+c^3-3abc=0\implies a^3+b^3+c^3= 3abc$ $\implies a^3+b^3= 3abc-c^3\implies (a^3+b^3)(a-b)= (3abc-c^3)(a-b)\implies a^4-b^4-a^3b+b^3a= 3a^2bc-3ab^2c-c^3a+c^3b$.
By (1), we have $c-ab (a^2-b^2)= -c(-3a^2b+3ab^2+ac^2-bc^2)\implies c+c(-3a^2b+3ab^2+ac^2-bc^2)= ab(a^2-b^2)\implies c(-3a^2b+3ab^2+ac^2-bc^2+1)= ab(a+b)(a-b)$. Since $a+b+c=0\implies a+b=-c$. So, $\cancel{c}(-3a^2b+3ab^2+ac^2-bc^2+1)= ab(-\cancel{c})(a-b)\implies -3a^2b+3ab^2+ac^2-bc^2+1= -ab(a-b)\implies -3a^2b+3ab^2+ac^2-bc^2+1+a^2b-ab^2=0\implies -2a^2b+2ab^2+c^2(a-b)+1=0\implies -2ab(a-b)+c^2(a-b)=-1\implies (a-b)(c^2-2ab)=-1$. Similary, we have $(b-c)(a^2-2bc)=(c-a)(b^2-2ac)=-1$.
Now, in (1), we have: $(a^2)^2-(b^2)^2= (a^2+b^2)(a^2-b^2)= (a^2+b^2)(a+b)(a-b)=c$. Also, $a+b+c=0\implies a+b+(a^2+b^2)(a+b)(a-b)=0\implies (a+b)[(a^2+b^2)(a-b)+1]=0$. So, $(a+b)=0$ or $[(a^2+b^2)(a-b)+1]=0$, but $(a+b)=0\Leftrightarrow c=0$, and we already analyse this case. So, we must have $(a^2+b^2)(a-b)+1= 0\implies (a^2+b^2)(a-b)= -1= (a-b)(c^2-2ab)\implies (a^2+b^2)\cancel{(a-b)}= \cancel{(a-b)}(c^2-2ab)$. Notice that we can cancel $(a-b)$, because $a\neq b\implies a-b\neq 0$. Now, we have $a^2+b^2= c^2-2ab\implies a^2+2ab+b^2= c^2\implies (a+b)^2= c^2\implies |c|= |a+b|$. So, $c=a+b(i)$ or $c= a-b(ii)$ or $-c= a+b(iii)$ or $-c= a-b(iv)$. We can obviously see that $(iii)$ is a consequence from $a+b+c=0$, which give us nothing. So, let's see what happens in $(i), (ii)$ and $(iv)$. In $(i), a+b+c=0\implies a+b+a+b=0\implies 2(a+b)=0\implies a+b=0$, that we already see it before (case in c=0 is similar with a=0). In $(ii), a+b+c=0\implies a+b+a-b=0\implies 2a=0\implies a=0$ (we already see if a=0). And last but not least, in $(iv), -c=a-b\implies c= b-a$, and $a+b+c=0\implies a+b+b-a=0\implies 2b=0\implies b=0$(which is similar to a=0). So, there's no more cases if $a\neq b\neq c\neq 0$.
Therefore, there are only 4 solutions for the triples $(a,b,c)= (0, 0, 0); (0, 1, -1); (-1, 0, 1); (1, -1, 0)$