Mistakes give rise to problems- 5

First of all, note that $(a+b)^{2}=a^{2}+b^{2}$ implies that $a^{2}+2ab+b^{2}=a^{2}+b^{2} \Rightarrow 2ab=0 \Rightarrow \boxed{ab=0}$ .

Let us now find the number of pairs that satisfy that property in the given domain, which is $-10\leq a,b \leq 10$ .

Note that in order for any product to be equal to zero at least one of the multipliers in it must be equal to zero. Therefore, we get two possible cases:

$1^{\circ} a=0$

For $a=0$ we have a total of $21$ possible pairs, since $b$ can essentially be any integer so that $-10\leq b \leq 10$ .

$2^{\circ} b=0$

For $b=0$ we have a total of $21$ possible pairs, since $a$ can essentially be any integer so that $-10\leq a \leq 10$ .

Bear in mind that we now have to apply the inclusion-exclusion principle in order to count the total number of pairs, considering we counted the pair $(a,b)=(0,0)$ twice!

Finally, we get that the total number of such pairs is $21+21-1=\boxed{41}$ .

It's only possible when the value of 2ab is 0. so it can be in 20 ways for a=0, 20 ways for b=0, and 1 way for a & b both equals 0. Therefore it's 41.

This problem is similar to my problem Absolute Laziness .

By this equation

             ( a+b )^2 =a^2+b^2

And we know that

            ( a+b )^2 =a^2+b^2+2ab

So it means ab=0; this is only right for a=0 or b=0 and for both equal to zero. So from -10<=(a,b)<=10 Total solution is 41 Answer = 41