Prove that every pair of numbers that satisfies these equations is either (1,0) or (-1,0)

Question

Let $X$ be the set of pairs of real numbers $(x,y)$ that are solutions to both the equation $x^2+y^2=1$ and $x^2-y^2=1$. Prove that every element $(x,y)$ of $X$ is either $(1,0)$ or $(-1,0)$. Also, $(1,0) \in X$ and $(-1,0) \in X$.

So my idea is to let $(x_o,y_o)$ be an arbitrary element of $X$. So $$x_o^2+y_o^2=1$$ $$x_o^2-y_o^2=1$$

Then by adding the equations we get that $x_o^2= \pm 1$.

How do I go from here? My problem is I am not sure how to proof this in a way that shows both parts of the question.

Answer 1

You don't get $x_0^2 = \pm 1$. What you get is $x_0^2=1$ hence $x_0 = \pm 1$. This shows that if $(x_0,y_0) \in X$ then either $x_0=1$ or $x_0=-1$. In each case, the equation $$ x_0^2+y_0^2=1 $$ becomes, substituting $\pm 1$ in place of $x_0$, $$ y_0^2=0 $$ hence $y_0=0$. What this means is that if $(x_0,y_0) \in X$ then you have two options, two possible choices: either $x_0=1$ and $y_0=0$ or $x_0=-1$ and $y_0=0$. So $X \subseteq \{(1,0),(-1,0)\}$.

Now, in order to show that $X$ is actually equal to $\{(1,0),(-1,0)\}$, all you have to do is check that $(1,0)$ and $(-1,0)$ are indeed solutions to the problem. That is readily verified.

Note 1: You have to be careful about implications. For example, if I ask you to find the solution(s) of $$ a+1=2 \tag{$\star$} $$ you could proceed like this (just for the sake of illustrating what I mean).

Suppose that $a+1=2$. Then it must be the case that $(a+1)^2=2^2$, that is, $a^2+2a-3=0$. We can solve this quadratic equation: $$ a = \frac{-(2)\pm\sqrt{2^2-4(1)(-3)}}{2(1)} = 1 \text{ or }-3 $$ Hence you would have shown, by this argument, that if $a$ is a solution of $(\star)$ then it must either be $1$ or $-3$. But! You have not shown that both $1$ and $-3$ are solutions of $(\star)$ because you did not have equivalences throughout your argument (more precisely $a+1=2 \Rightarrow (a+1)^2=2^2$ but $(a+1)^2=2^2 \nRightarrow a+1=2$). In fact, $-3$ is not a solution of $(\star)$. But $1$ is one because $1+1=2$.

Note 2: In the case of your problem, you correctly argumented that if \begin{align} x_0^2+y_0^2=1 \tag{1}\\ x_0^2-y_0^2=1 \tag{2} \end{align} then by adding both equations it must be the case that $$ x_0^2=1 \tag{3} $$ In other words, you showed that if $(x_0,y_0)$ satisfies $(1)$ and $(2)$, then it satisfies $(3)$. But the converse is not true! If $(x_0,y_0)$ satisfies $(3)$ then it must not be the case that it satisfies $(1)$ and $(2)$. Indeed, $(1,1)$ is a counterexample. However it is the case that $(x_0,y_0)$ satisfies $(1)$ and $(2)$ if and only if it satisfies $$ x_0^2=1\\ y_0^2=0 $$

Answer 2

Add the equations and get $2x^2 = 2$ or $x^2 = 1$ so $x = \pm 1$. Since $x$ only occurs in $x^2$, both of these are solutions.

Subtract the equations and get $2y^2 = 0$ so $y = 0$.

Therefore the only solutions are $(x, y) = (1, 0), (-1, 0)$.