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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?

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    $\begingroup$ There's actually 201 scenarios in each one, but yeah the argument is correct $\endgroup$ – Something Jun 10 '20 at 12:45
  • $\begingroup$ Freshman's dream $\endgroup$ – RobPratt Jun 10 '20 at 17:41
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How many numbers are there in the sequence $-1, 0, 1$? What about $-10, -9, \ldots, 0, \ldots, 9, 10$? How about in $-100, \ldots, 100$?

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  • $\begingroup$ Absolutely spot on example $\endgroup$ – Omar S Jun 11 '20 at 9:46
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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 $2\cdot 201-1=\color{green}{401}$.

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$$a^2+b^2 = (a+b)^2\implies a^2+b^2 -(a+b)^2=0\implies \pm2ab=0$$ 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.

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