Yes, I'm here again, with another misunderstanding about the solution.
Question: Two positive integers $a$ and $b$ are chosen randomly, uniformly, and independently from the set of positive integers less than or equal to $1000$. What is the expected value of the number of quadrants through which the graph $x^a+y^b=1$ will pass?
Solution: Without loss of generality suppose that $a$ (or $b$, respectively) is even. Then $(x, y)$ is a solution if and only if $(−x, y)$ is a solution. Now we show that $y$ can be positive or negative. Note that $$y=\sqrt[b]{1-x^a}.$$ If $b$ is odd, then as long as $x$ is large enough, the expression will be negative. If $b$ is even, then symmetry means that $-\sqrt[b]{1-x^a}$ is a solution whenever $\sqrt[b]{1-x^a}$ is a solution.
Answer is $3.75$.
The solution shows that $y$ can be positive or negative, but there is nothing else. It also says $(x, y)$ is a solution if and only if $(-x, y)$ is, which doesn't have proof. It also doesn't explain how you get to $3.75$. Can someone please provide a better solution?
First let's establish the cases:
There are a number of ways to show this. Your algebraic argument can be strengthened and analyzed on a case-by-case basis. You can also use calculus. For any $a$ and $b$, $(1,0)$ and $(0,1)$ are on the curve $x^a+y^b = 1$. If we differentiate implicitly treating $y$ as a function of $x$: $$ ax^{a-1} + b y^{b-1} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = - \frac{ax^{a-1}}{by^{b-1}} $$ Apparently $\frac{dy}{dx} = 0$ at $(0,1)$, which means the tangent line to the curve is horizontal there. So the curve passes into both the first (on the right of $(0,1)$) and third (on the left) quadrants.
Similarly, $\frac{dx}{dy} = 0$ at $(1,0)$, which means the tangent line to the curve is vertical there, and the curve passes into the fourth quadrant as well.
If $b$ is even, then $(0,-1)$ is on the curve, and $\frac{dy}{dx} = 0$ there. So the curve has a horizontal tangent there, and because of this passes into the fourth quadrant.
If $a$ is even, then $(-1,0)$ is on the curve, and $\frac{dx}{dy} = 0$ there. So the curve has a vertical tangent there, and because of this also passes into the fourth quadrant.
As lulu points out, if $a$ and $b$ are both odd, then $x^a + y^b < 0$ for all $(x,y)$ in the third quadrant. So the curve will only pass through three quadrants.
Now let $X(a,b)$ be the random variable which assigns to $(a,b)$ the number of quadrants $x^a+y^b = 1$ passes through. We already know there are only two values of $X$: $3$ and $4$. So $$ E(X) = 3\cdot P(X=3) + 4\cdot P(X=4) $$ We also know that $X=3$ if $a$ and $b$ are both odd and $X=4$ otherwise. Let $\mathrm{OO}$ be the event that $a$ and $b$ are both odd. Then $$ E(X) = 3 \cdot P(\mathrm{OO}) + 4 \cdot (1-P(\mathrm{OO})) = 4 - P(\mathrm{OO}) $$ The last thing to determine is $P(\mathrm{OO})$. But this is pretty clear. The probability that $a$ is odd is $\frac{1}{2}$ since there are 1000 outcomes for $a$ and half of them are odd. The probability that $b$ is odd is also $\frac{1}{2}$. Since the two events are independent, the probability that they are both odd is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Thus $$ E(X) = 4 - \frac{1}{4} = 3.75 $$