Redacted... cus I suck.

khan.academy wrote: $$a^2-b^2=-c(a-b) \implies a+b+c=0$$Also, $a,b,c$ are even. For the sake of contradiction, let $a>b$, then $$a^2>b^2 \implies bc>ca \implies b>a$$So, either $b=a$ or $b<0, |b|>a$. For $b=a$,we have, $a(a-c)=4$, so $a|4$. $$a=1,b=1 \implies c=-2, a=b=2 \implies c=-4 , a=b=4 \implies c=-8,$$None of these satisfy the equation above. For $b<0, b=c+a$, then $$a^2+c^2+ca=4, 2a^2+ac=4 \implies a=-c \implies a=2, c=-2, b=0$$which is the only solution. This solution is wrong. The main reason being that you cannot multiply/divide both sides of an inequality by an negative number. As a side note, check the official site (HBCSE) for the correct solution to the problem.

We will examine two cases: Case 1: $a=b$ Then, the two equations become identical and we have: $$a(a-c)=4$$We have the following cases: 1) $a=-4 \Longleftrightarrow c=-3, (a,b,c)=(-4,-4,-3)$ 2) $a=-2 \Longleftrightarrow c=0, (a,b,c)=(-2,-2,0)$ 3) $a=-1 \Longleftrightarrow c=3, (a,b,c)=(-1,-1,3)$ 4) $a=1 \Longleftrightarrow c=-3, (a,b,c)=(1,1,-3)$ 5) $a=2 \Longleftrightarrow c=0, (a,b,c)=(2,2,0)$ 6) $a=4 \Longleftrightarrow c=3, (a,b,c)=(4,4,3)$ Case 2: $a\neq{b}$ Substracting the two equations, we get: $$a^2-b^2=c(b-a) \Longleftrightarrow$$$$a=-(b+c)$$Replacing $a$ in the original equations, we have: $$b^2+bc+c^2=4$$If $b=0$ or $c=0$, we immediately find the triples $(a,b,c)=(2,0,-2)$, $(a,b,c)=(-2,0,2)$, $(a,b,c)=(2,-2,0)$, $(a,b,c)=(-2,2,0)$ Now, suppose that $b,c\neq{0}$ and that $b$ and $c$ have the same sign. Then: $$4=b^2+bc+c^2\ge{3bc}$$So, we only have to check $b=c=\pm{1}$, which give no solution. If $b$ and $c$ have opposite signs: $$4=b^2+bc+c^2\ge{-bc}$$So, we have that $b,c\in{[-4,4]}$ and by checking all cases, we get the triples $(a,b,c)=(0,2,-2)$, $(a,b,c)=(0,-2,2)$.