Counting ordered pairs $(a,b)$ satisfying $a^2+b^2=(a+b)^2$, with $a$ and $b$ in the interval $[-100,100]$

Question

A lazy mathematician believes that $$a^2+b^2 = (a+b)^2.$$ If $a$ and $b$ are both integers from the interval $[-100, 100]$, find the number of ordered pairs $(a,b)$ that satisfy the equation above.

The equation implies that $2ab = 0$, hence either $a=0$ or $b=0$. This would result in the following pairs. When $a=0$ $$(0,-100), (0, -99), \dots, (0, 100).$$

When $b=0$ $$(-100, 0), (-99, 0), \dots, (100, 0)$$

Since we have twice the pair $(0, 0)$ we should subtract $1$ from the total counts. For me it seems that there's $200$ possible choices in each scenario. This would imply that the total would be $400-1 = 399$, but the correct answer was $401$. What am I missing in the countings?

Answer 1

The problem with your reasoning is that the integers from $-100$ to $100$ comprise $\color{green}{201}$ integers, not $\color{red}{200}$ integers. Thus, the answer is $2\cdot 201-1=\color{green}{401}$.

Answer 2

How many numbers are there in the sequence $-1, 0, 1$? What about $-10, -9, \ldots, 0, \ldots, 9, 10$? How about in $-100, \ldots, 100$?

Answer 3

$$a^2+b^2 = (a+b)^2\implies a^2+b^2 -(a+b)^2=0\implies \pm2ab=0$$ This means that one of $a$ or $b$ must be zero if the other is non-zero "or" both may be zero. This means we have $200$ non-zero values of $a$ when $b$ is zero plus $200$ values of $b$ when $a$ is zero plus the $1$ pair $(0,0)$ for a total of $401$ ordered pairs.