The problem I'm having trouble with is:
Let $a,b,c$ be nonzero real numbers and let $a^2 - b^2 = bc$ and $b^2 - c^2 = ca$. Prove that $a^2 - c^2 = ab$.
The solution strategy given in the course was to scale the two given equations by $s=\frac{1}{c}$, resulting in $a^2 - b^2 = bc$ becoming $a^2 - b^2 = b$ and $b^2 - c^2 = ca$ becoming $b^2 - 1 = a$. I see that $c$ is basically being set to 1, but I don't understand the justification. Doesn't scaling by $\frac{1}{c}$ by definition not change the equations, since
$(a/c)^2 - (b/c)^2 = (b/c)(c/c) \Longleftrightarrow \dfrac{a^2 - b^2}{c^2} = \dfrac{bc}{c^2} \Longleftrightarrow a^2 - b^2 = bc$
What am I missing?
You can think of this scaling as dividing all equalities by $c^2$, which is valid because $c\neq 0$. This division basically reduces a 3-variables problem into a 2-variables problem, which should be more tractable.
Let $A=a/c$, $B=b/c$ and $C=c/c=1$. Then, we have $$ a^2-b^2=bc\implies A^2-B^2=B;\\ b^2-c^2=ca\implies B^2-1=A. $$ It follows that $$ A^2-1=(A^2-B^2)+(B^2-1)=B+A. $$ Thus, it remains that we show $A+B=AB$. Note that $$ B=A^2-B^2=A^2-(A+1)=A^2-A-1 $$ so $$ AB=A+B\iff A^3-A^2-A=A^2-1\iff A^3-2A^2-A+1=0.\tag{i} $$ To show that (i) holds, we use $$ 1+A=B^2=(A^2-A-1)^2=A^4-2A^3-A^2+2A+1 $$ which implies $$ 0=A^4-2A^3-A^2+A=A(A^3-2A^2-A+1).\tag{ii} $$ Now we are done: (ii) implies that (i) is true, therefore we indeed have $AB=A+B$.