Absolute Laziness

We have $a^{2}+b^{2} = a^{2}+b^{2}+2ab$

hence $2ab = 0$

So we have either $a = 0$ or $b=0$ or both are equal to 0.

We can assign the $0$ value in 2 ways either to a or to b and the other variable has 201 choices so total number of ordered pairs = $2*201 = 402$ but clearly we have over counted the $(0,0)$ case hence the total number of possibilities should be $402-1 = \boxed{401}$ .

We HAVE a²+b²=a²+b²+2ab ==> ab=0 so a=0 or b=0 sow we have 400 pairs compose by 0 and a number between -100 &100 (0;a) or (a;0) and the pair (0;0) ,therfore the solution is 401

Did you mean If a and b are both integers or If a and b are both positive integers (as it states already)? With respect to both options, correct answer could be 401 or 0 ordered pairs.

2ab = 0 to satisfy the given equation and this could be possible if we consider one of the co-ordiate to be 0 and there are such 400 possible ways and one way if we put both co-ordinate to be i.e (0,0) so the ans.. 401