Algebraic vs geometric approach different results confusion

Question

I consider the following system:

$$ x+y+z=A\\ x^2+y^2+z^2 = B $$ where $A,B ∈ ℝ; B ≥0$. Now, the geometric approach here is that these describe a sphere and a plane. So they can:

• Have no intersection: the system would then have no solutions. Since we have a very specific plane and a sphere here, this might be impossible, but mentioning it just in case
• Be tangent: the system would then have exactly one solution. An example of that happening is $A=15, B=75$
• Intersect via a circle: the system would then have infinitely many solutions and the set of solutions would form that circle.

So far so good. Now, let's try algebraic approach and consider this "trick": let's multiply 1st equation by $2w$ where $w ∈ ℝ, w ≠ 0$ and then add it to the first equation. This yields $x^2+2wx+y^2+2wy+z^2+2wz=2Aw+B$, or - after moving things around: $$(x+w)^2+(y+w)^2+(z+w)^2=3w^2+2Aw+B$$

And here's the thing: the $w$ we brought here was completely arbitrary. So we now try to make it so that the right hand side is 0, i.e. $3w^2+2Aw+B=0$. And here's what I don't understand:

• This equation could have no roots. Since the leading coefficient is positive, this would mean that the sum of squares on the left hand side is always positive. So as far as I see it, it corresponds to "infinitely many solutions" of the geometric approach.
• This equation could have exactly one root. Then it's clear that it corresponds to a "tangent case" from the geometric approach. For instance, if $A=15, B=75$ we have $ = 0$ meaning it's exactly such case.
• This equation could have two roots. This is the case I do not understand. What does it correspond to?

Furthermore, the algebraic approach doesn't seem to even work for the last case. For example, if we have $A=5, B=3$ then: $$ x+y+z=5\\ x^2+y^2+z^2 = 3 $$ And so $3w^2+10w+3$, so $ = 64$ and $w_1=-\frac{1}{3}, w_2=-3$. If we take $w=-3$ then it corresponds to equation $x^2-6x+y^2-6y+z^2-6z=3-(2∙3∙5)=-27 ⇒ (x-3)^2+(y-3)^2+(z-3)^2=0$ and there are several problems here:

• First, this implies $x=y=z=3$ because over reals the only way a sum of squares can add up to 0 is if all summands are 0 themselves. But $(3,3,3)$ doesn't satisfy the system
• Second, this isn't even unique because we could've used $w=-\frac{1}{3}$
• Third and most important - this was done with only simple linear operations so the resulting equation is equivalent to the system.

What do I miss with this last case when there are two solutions in $w$? And why does it fail algebraically whereas the case with only one solution in $w$ works correctly?

Answer 1

If you have the system$$\left\{\begin{array}{l}x+y+z=5\\x^2+y^2+z^2=3\end{array}\right.$$(which has no solutions) and then you multiply the first equation by $2w$ and then add it to the second one, you will get a new system:$$\left\{\begin{array}{l}x+y+z=5\\(x+w)^2+(y+w)^2+(z+w)^2=3w^2+10w+3.\end{array}\right.$$If now you put $w=-3$, this system becomes$$\left\{\begin{array}{l}x+y+z=5\\(x-3)^2+(y-3)^2+(z-3)^2=0.\end{array}\right.$$And, indeed, the only solution of the second equation is $(3,3,3)$. But $(3,3,3)$ is not a solution of the first equation, and therefore there is nothing new going on here: the conclusion is (again) that the system has no solution. The same thing occurs if you take $w=-\frac13$ instead.

Note the you are wrong when you claim that “the resulting equation is equivalent to the system”. It is not. The resulting equation together with the equation $x+y+z=5$ is what is equivalent to the original system.

Answer 2

@José Carlos Santos provided a good answer as to why the equation in the last case isn't equivalent to the whole system - but I was still left a bit wondering if just that fact definitely means the original system would have no solutions for any $A$ and $B$ for which two $w$ solutions occur. After some thoughts - yes, it is indeed the case:

We have: $3w^2+2Aw+B = 0$ with two solution, meaning $=4A^2-12B > 0$, thus $A^2>3B$. Let's use this fact on the original system. If we square first equation we get: $$ x^2+y^2+z^2+2(xy+yz+xz)=B+2(xy+yz+xz)=A^2\\ ⇒xy+yz+xz=\frac{A^2-B}{2} > \frac{3B-B}{2} = B $$

So we arrive at the following: $$ x^2+y^2+z^2=B\\ xy+yz+xz>B $$ Which concludes that $x^2+y^2+z^2<xy+yz+xz$, however that is impossible because it implies that: $$2x^2+2y^2+2z^2-2xy-2yz-2xz<0\\ ⇒(x-y)^2+(x-z)^2+(y-z)^2<0 $$ And that's impossible over $ℝ$. So indeed if we have two solutions in $w$ for this constructed equation then the system will have no real solutions.

Answer 3

For all $x,y,z,A,B ∈ ℝ$, the following statements are logically equivalent to each other:

1. $x + y + z = A$ and $x^2 + y^2 + z^2 = B,$
2. for all $w ∈ ℝ$: $$3\left((x + w)^2 + (y + w)^2 + (z + w)^2\right) = (3w + A)^2 + 3B - A^2\tag{1}\label{1},$$
3. there exists at least two $w ∈ ℝ$ for which ($\ref{1}$) holds.

I don't see any problem - except that you framed the problem wrong. This is a correct framing of it.