lol this is just 2023 aime ii p4 Adding the first two equations and subtracting double the third equation gives us cyclic permutations of \[ (a-c)(a^2 + b^2 + c^2 + 2ac + 2ba + 2bc) = 0\](or something along the lines of that. The motivation for doing this is subtracting each pair of equations to get another system of three equations, and then subtracting each pair of those equations, which is equivalent to adding two and subtracting doubling the third.) By analyzing the RHS of our given three equations, $a$, $b$ and $c$ are all positive, hence the second term in the expression above is positive. Hence, $a = c$, and applying cyclic reasoning gives us $a=b=c$; plugging this into the original equation gives us $a=b=c=2$.

ihatemath123 wrote: lol this is just 2023 aime ii p4 Adding the first two equations and subtracting double the third equation gives us cyclic permutations of \[ (a-c)(a^2 + b^2 + c^2 + 2ac + 2ba + 2bc) = 0\](or something along the lines of that. The motivation for doing this is subtracting each pair of equations to get another system of three equations, and then subtracting each pair of those equations, which is equivalent to adding two and subtracting doubling the third.) By analyzing the RHS of our given three equations, $a$, $b$ and $c$ are all positive, hence the second term in the expression above is positive. Hence, $a = c$, and applying cyclic reasoning gives us $a=b=c$; plugging this into the original equation gives us $a=b=c=2$. more case: $a+b+c=0$ result a=b>0 ,c<0 $ false!

Those values don't work; in general $a+b+c = 0$ is not a case, since we know $a$, $b$ and $c$ must all be positive.

ihatemath123 wrote: Those values don't work; in general $a+b+c = 0$ is not a case, since we know $a$, $b$ and $c$ must all be positive. you have right !

Pretty good question Let's say WLOG $a \geq b$ $\geq c$ 1.$a=b=c=2$ 2.$a>b=c$ $a^4+b^4=16a$(.... 1) $b^4+a^2b^2=16b$(.... 2) (.... 1)-(.... 2)=$a^2(a+b)=16$ $a^2(a+b) >2b^3$ Then $b<2$ Let's inspect (.... 2) $b^3+a^2b=16>8+2a^2$ Then $2>a$ but $2a^3>a^2(a+b)=16 \implies{ a>2}$ Contradicition. One case is remain which is 3.$a>b>c$ $a^4+b^2c^2=16a$ (... X) $b^4+c^2a^2=16b$(...Y) $c^4+a^2b^2=16c$(...Z) (...X)-2(...Y)-(...Z)=0 Which is a((a+b) ^2+ac)=c((b+c)^2+ac) $a>c$ Then we need to prove (a+b) ^2<(b+c)^2 But it is wrong then contradicition again. And we are done

first we have $16a-a^4 = b^2c^2$ $16b-b^4 = c^2a^2$ $16c-c^4 = a^2b^2$ $16a^3-a^6 = 16b^3-b^6=16c^3-c^6$ if 2 of equation equal thus we can pick $a^3 = b^3 \rightarrow a = b$ due to symetricity $a^4+a^2c^2 =16a$ $c^4+a^4 = 16c$ $c^2(a^2-c^2) = 16(a-c)$ $(a-c)(ac^2+c^3-16) = 0$ $c^3+\frac{a^2b^2}{c} = c^3+ac^2$ implies $a = b =c =2 $ then the other 3 are equal and 3 of them equal to two Other Cases are when $(a^3 \ne b^3 ) $ $a^3+b^3 = 16, b^3+c^3 = 16, c^3+a^3$ forcing $a = b = c$ also so contradiction only sols is $(a,b,c) = 2$