# A confusion regarding a specific high school algebric problem [duplicate]

In this equation, $$\frac{a^2 + b^2}{c^2} = \frac{b^2 + c^2}{a^2} = \frac{a^2 + c^2}{b^2} = \frac{1}{k}$$ If we want to find value of k we can calculate algebrically and we would get k=1/2. But the way we get to this we NEVER assume anything that a,b and c are all real numbers. So it should be applicable for every a, b and c no matter if it's real or not. But for the values like a^2=1, b^2=1 and c^2=-2 this relation is still satisfied but with these numbers we get k=-1. But how is this even possible? When we derived the solution for K fir random a,b and c we never assumed anything like they need to be real or complex. But for these numbers we get a different value of k . So. What's going on?

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Commented Jul 24 at 19:18
• To derive $k=1/2$ (hence $a^2=b^2=c^2$), you just need $a^2+b^2+c^2\ne0$. This is not the case in your example $1+1+(-2)$. Commented Jul 24 at 19:20
• @AnneBauval Oh.... that's exactly what I wanted to know....what a silly mistake I was making😅. Should I delete this question now ow is it okay if it's here?? Commented Jul 24 at 20:32

The equations in $$a,b,c$$ are equivalent to $$(a^2+b^2+c^2)(a+c)(a-c)=0,\quad (a^2+b^2+c^2)(a+b)(a-b)=0.$$ Either $$a=b=c$$ up to signs, so that the fractions are $$\frac{1}{2}$$, or $$a^2+b^2+c^2=0$$. Here, in the second case, we use that we have real numbers to exclude this, because by assumption $$a,b,c$$ are nonzero. Over the complex numbers there are indeed nonzero solutions, where the fractions can be equal to $$-1$$, because $$c^2=-(a^2+b^2)$$, and permutations of it.