$a^2-b^2=bc$ and $b^2-c^2=ac \Rightarrow a^2-c^2=ab$
Some weeks ago our math teacher asked the following question and gave us a week to solve it:
If $a^2-b^2=bc$ and $b^2-c^2=ac ,$ Prove $a^2-c^2=ab$, Where $a,b,c$ are non-zero real numbers.
This seemed really easy at the first, but when i tried to prove it i just failed every time. After a week, I only came up with this idea: Assume our case is true. $a^2-c^2=ab\Rightarrow a^2-b^2+b^2-c^2=ab\Rightarrow bc+ac=ab\Rightarrow\frac{1}{abc}(bc+ac)=\frac{1}{abc}(ab) \Rightarrow$
$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Now if we Prove this, our case will be proved as well.
In my opinion this seemed like a really nice question so i wanted to share it with everyone.
2 Answers
$\begingroup$To begin with, note that by adding the two given equations together, we can immediately conclude that $$a^2-c^2=(a+b)c.\tag{$\star$}$$
Now, multiply both sides of $(\star)$ by $a-b,$ giving us $$(a^2-c^2)(a-b)=(a^2-b^2)c\\a^3-a^2b-ac^2+bc^2=bc^2\\a^3-a^2b-ac^2=0\\a(a^2-ab-c^2)=0,$$ so since $a\ne 0,$ we can conclude that $a^2-ab-c^2=0,$ so that $a^2-c^2=ab,$ as desired.
$\endgroup$ 5 $\begingroup$You can chug through this by eliminating the $a$ variable and following your nose. $$b(b + c) = b^2 + bc = a^2 = {(b^2 - c^2)^2 \over c^2}$$ We can assume that $b \neq -c$, since if $b = -c$ the first equation implies $a = 0$, contradicting that $a,b,c$ are all nonzero. So we can divide by $b+c$ to get $$b = {(b^2 - c^2)(b-c) \over c^2}$$ Multiplying by $c^2$ and expanding this becomes $$bc^2 = b^3 - c^2b - cb^2 + c^3 \tag 1$$ You want to show $a^2 - c^2 = ab$. Substituting your first equation into $a^2$ and the second into $ab$ this is equivalent to showing $$b^2 + bc - c^2= {b^2 - c^2 \over c}b$$ Multiplying by $c$ this is equivalent to $$b^2c + bc^2 - c^3= b^3 - c^2b \tag 2$$ Rearranging terms, $(2)$ is the same as $(1)$.
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