Radical mistake

First of all, remember that $\sqrt{a^2} = |a|$ and not $\sqrt{a^2} = a$

And also $|a|^2 = a^2$

Therefore, $\sqrt{a^2+b^2} = |a|+|b|$

Squaring both sides gives -

$a^2+b^2 = |a|^2 + |b|^2 + 2|ab| \Rightarrow |ab| = 0$

Keeping $a=0$ , we have 21 possibilities for $b$

Then keeping $b=0$ , we have 21 possibilities for $a$

But, now we have considered the case (0,0) twice so we need to exclude it.

Hence total solutions = $21+21-1 = \boxed{41}$

A poem:

Start by squaring this equation
On both sides; the same expression
Sans a term which is an exception
It's 2ab, such the rejection!

Subtract a^2 + b^2 on both sides
Getting 2ab as zero, quite nice
To make these two values equalize
a or b is zero, a surprise!

We count the ordered pairs, this manner:
We set a as zero, then count 'er:
We get 21 pairs, substituting b there
-10 to 10, 21 pairs!

Then we do the same for b
A value of zero, let it be
Then quite obviously we can see
21 pairs, such that values agree

Then we add the both cases together
We get 42, altogether
Counted (0, 0) twice, however:
Finally, 41, subtracting 1 there.

We get the final answer, 41 .

Squaring both sides of the equation we get that $a^{2}+b^{2}=a^{2}+2\sqrt{a^{2}b^{2}}+b^{2}$ Simplifying the equation we get that $2ab=0$ (technically it should be |ab|, but in this case it is not important) From the we see that either a or b is 0; We have 21 possible values for b if a is 0, and we have 21 possible values for a if b is 0, but the pair (0;0) is included in both answer sets, hence the answer is $21+21-1=\boxed{41}$

By squaring both sides of the equation, we get: $a^2+b^2=a^2+2\sqrt{a^2b^2}+b^2$ . If we subtract $a^2+b^2$ from both sides, we get $0=2\sqrt{a^2b^2}$ . When we divide 2, we get: $0=\sqrt{a^2b^2}$ . That means that either a or b has to be 0, or both. If only a is 0, then b can be and integer between $-10$ and $10$ excluding 0, meaning there are $\boxed{20}$ pairs. The same is true if only b is 0. If both a and b are 0, then we have $\boxed1$ more pair. That means that there are a total of $20+20+1=\boxed{41}$ different pairs that satisfy the above conditions.

Square both sides: $a^2+b^2=a^2+2\sqrt{a^2b^2}+b^2$ $0=2\sqrt{a^2b^2}$ $ab=0$ Now, we count when at least one of $(a,b)$ is 0. We could count this normally, or we could use complementary counting. If neither is 0, there are $20^2=400$ possibilities. The total number is $21^2=441$ . The answer is $441-400=\boxed{41}$ .

Desenvolvendo,

$\\ \sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2} \\\\ \sqrt{a^2 + b^2} = a + b \\\\ \left ( \sqrt{a^2 + b^2} \right )^2 = \left ( a + b \right )^2 \\\\ a^2 + b^2 = a^2 + 2ab + b^2 \\\\ 0 = 2ab \\\\ ab = 0$

Isto é, devemos encontrar a quantidade de pares ordenados cujo produto entre a abscissa e a ordenada é nulo.

São eles: $S = \left \{ \left ( - 10, 0 \right ),\left ( - 9, 0 \right ),...,\left ( 9, 0 \right ),\left ( 10, 0 \right ),\left ( 0, 0 \right ), \left (0, - 10 \right ),\left (0, - 9 \right ),...,\left ( 0, 9 \right ), \left ( 0, 10 \right ) \right \}$

Totalizando a quantidade de $\boxed{41 \text{ pares ordenados}}$

Either a or b must be zero for this equation to hold.

When any one of them is zero, the other one can have any value in [-10,10].

So total no. of pairs are 2*21=42.

But in this we have counted (0,0) twice.

So, the final answer is 41.

Squaring both sides gives $a^2 + b^2 = a^2 + b^2 + 2ab$ or $2ab = 0$ . Thus, at least one of $a$ and $b$ is $0$ . We consider $3$ cases:

Case $1$ : $a = b = 0$

This gives one solution.

Case $2$ : $a = 0, b \neq 0$

There are $20$ possible values of $b$ . Thus, this gives $20$ solutions.

Case $3$ : $a \neq 0, b = 0$

Similarily, this also gives $20$ solutions.

Now, we just note that all solutions described above satisfies the original equation. So, there are $\boxed{41}$ solutions.

We can think about the problem as a geometric problem in cartesian plane:

$a$ and $b$ are catheti of a right triangle so we have to find how many ordered pairs of catheti satisfy the condition that the sum of their lengths is equivalent to the length of the hypotenuse.

Thus, the triangle must be degenerate and we have to count all the pairs $(a,0)$ and $(0,b)$ without counting $(0,0)$ two times.

So, the answer is $\boxed{41}$

Squaring both sides of the given equation yields $a^2+b^2 = a^2 + b^2 + 2(\sqrt{a^2})(\sqrt{b^2})$ $0=2(\sqrt{a^2b^2})$ $0=|ab|$

This can only happen when either $a=0$ or $b=0$ (or both). So let us split this up by cases.

Case 1 : $a=0$ There are exactly 21 possibilities for b.

Case 2 : $b=0$ There are exactly 21 possibilities for a.

We might be tempted to think the final answer is 42. But we have counted $(a,b)=(0,0)$ twice, so the answer is actually $\boxed{41}$ .

• Square both sides: $a^2 + b^2 = a^2 + 2|a||b| + b^2$ => $|a||b| = 0$ .
• So either a is 0 => 21 options for b. Or b is 0 => 21 options for a. We counted a=b=0 twice, so exclude one of these. $21+21-1 = \boxed{41}$ .

possible only if a^2+b^2 is a perfect squre. that happens only when we take the pairs that contain atleast 1 (0). so...(0,0)(0,1)(0,-1)............(0,10)(0,-10)

:)

Sherry claims that $\sqrt{a^2+b^2}=\sqrt{a^2}+\sqrt{b^2}$ , but we know that $\sqrt{a^2+b^2}=a+b=\sqrt{a^2+2ab+b^2}$ . Therefore, if both hold true, then $\sqrt{a^2+b^2}=\sqrt{a^2+2ab+b^2}$ . After squaring both sides and further simplification, we find that $2ab=0$ . This is only possible if: 1. $a$ is equal to 0 2. $b$ is equal to 0

Let us take the first case. Since $a$ (which is equal to $0$ ) will always make $2ab$ equal to $0$ , $b$ can be any integer that satisfies $-10 \leq b \leq 10$ . There are $21$ possibilities for the first case. Similarly, there are $21$ possibilities for the second case, when $b$ is equal to $0$ . However, we are over-counting the situation when both $a$ and $b$ are equal to $0$ , so we must subtract $1$ . Therefore, our answer is $21+21-1=\boxed{41}$ .

We manipulate the equation by squaring both sides and simplifying: $\begin{aligned} \sqrt {a^2 + b^2} &=& \sqrt{a^2} + \sqrt{b^2} \\ a^2 + b^2 &=& a^2 + b^2 + 2ab \\ ab &=& 0. \end{aligned}$ So, we want at least one of $a$ and $b$ to be $0$ . We use complementary counting. There are $21$ choices for $a$ and $21$ choices for $b$ , so $21^2 = 441$ possible ordered pairs $(a, b)$ .

The number of these ordered pairs for which neither $a$ nor $b$ is $0$ is $20^2 = 400$ . Thus, our answer is $441 - 400 = \boxed{41}.$

This reminds me very strongly of the triangle inequality, and I feel like there should be some relationship here to either the triangle inequality or the Cauchy-Schwarz inequality if we consider $a$ and $b$ as vectors in $R$ with norms $|a|$ and $|b|$ . But for equality to hold in either of those cases the vectors must be scalar multiples of each other, whereas here the solution requires orthogonal vectors.

Maybe someone more familiar with the topic can shed some wisdom and develop an abstract argument involving orthogonality for why at least one of $a$ or $b$ must be zero? Am I looking for a connection where none exists?

$\sqrt{a^2 + b^2} = \sqrt{a^2}+ \sqrt{b^2}$

$a^2 + b^2 = a^2 + \sqrt{a^2b^2} + b^2$

$0 = \sqrt{a^2b^2}$

$0 =ab$

On $-10 \leq a \leq 10$ , and $b=0$ , there are 21 solutions, and there also 21 on $-10 \leq b \leq 10$ and $a=0$ . Total, subtracting the one solution that appears twice, there are $\boxed{41}$ solutions.

\sqrt{a^2+b^2}=\sqrt{a^2} + \sqrt{b^2} = a + b Squaring both the sides, one sees this is possible only if either a = 0 or b = 0 or both are zeros.

Note that to satisfy the above relation at least one of a and b must be 0 and $a^2+b^2$ must be a perfect square .The first ordered pair is $(0,0)$ .

Taking at least one of a and b 0 maximum value of $a^2+b^2$ is 100 and minimum is 0.There are 10 squares in this range and each square can be obtained in four ways.For example 25 can be obtained by $(5,0),(0,5),(-5,0),(0,-5)$ .

This gives us $10 \times 4 =40$ ordered pairs.Total ordered pairs = $40+1=\boxed{41}$

Simplifying the equation and squaring both sides, $a^2+b^2=a^2+2ab+b^2$ .

Therefore, $2ab=0$ . For this to occur, at least $1$ of $(a,b)$ must be equals to $0$ .

When $a=0$ , there are $21$ choices for $b$ , since $-10\leq b \leq 10$ and $b$ is an integer. This applies to the case when $b=0$ as well. However, we have doubled counted the case when $a=0$ and $b=0$ . Thus, the total number of cases is $21\times 2 - 1 = \boxed{41}$ .

$\sqrt{a^2 + b^2} = a + b$ $a^2 + b^2 = a^2 + 2ab + b^2$ $0 = 2ab \Rightarrow ab =0$ .

If $a=0$ , we have $10 - -10 + 1 = 21$ possibilities (including (0,0)). So, with $b = 0$ , we get $21 - 1 = 20$ possibilities. Finally, the answer is $21 + 20 = \boxed{41}$ .

If √(a^2+b^2 )=√(a^2 )+√(b^2 ) Then √(a^2+b^2 )=a+b 〖√(a^2+b^2 )〗^2=(a+b)^2 a^2+b^2=a^2+2ab+b^2 The squared terms cancel on both sides so we have 2ab=0 This is only possible when either a or b is 0.
For -10≤a≤10 and -10≤b≤10 we count 21 ordered pairs for each giving us 42, but because the ordered pair (0,0) duplicates in each set we have 41 possible pairs.

Let $a^2 + b^2 = c^2$

Then $\sqrt{a^2 + b^2} = |c|$

Also $\sqrt{a^2} + \sqrt{b^2} = |a| + |b|$

So $\sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2}$

Is equivalent to $|c| = |a| + |b|$

This is analogous to the triangle inequality , $c \leq a + b$ .

Where $a, b,c$ are the sides of a triangle with $a<c$ and $b<c$ .

The triangle inequality becomes the equality $c=a+b$ exactly when the triangle with sides $a, b, c$ has $0$ area, or in other words when $a = 0$ or $b=0$ or both.

Therefore, the original equation will be satisfied by any ordered pair in the set $\{(0,b),(a,0),(0,0)\}$ with $a$ and $b$ in the set $\{\pm1,\pm2,\pm3,\ldots,\pm9,\pm10\}$ .

So the amount of possibilities for $a$ is 20, the amount of possibilities for $b$ is 20, and there is the case that both $a=0$ and $b=0$

Number of solutions, $n = 20+20+1 = \boxed{41}$

Consider the problem geometrically. The LHS is the Euclidean distance of a point to the origin, and the RHS is the Manhattan Distance. These two distances are equal only on the axes. So there is the origin (1point) and 10 points on both axes on either side. So the total number of points is 41.

Haha, I was thinking of $(0,b), (a,0)$ and $(0,0)$ and thus 3...

$\sqrt{a^2+b^2} = \sqrt{a^2} + \sqrt{b^2} = |a| + |b|$

Squaring both sides;

$a^2+b^2 = |a|^2 + |b|^2 + 2|a||b| = a^2 + b^2 + 2|a||b|$

$2|a||b| = 0$

$|a| = 0$ or $|b| = 0$

Assume $a=0$ and $b$ is non-zero we have $10+10 = 20$ solutions.

For $a$ to be non-zero and $b=0$ we have another $20$ solutions.

Not forgetting $a=0$ and $b=0$ , we have $20+20+1 = \boxed{41}$ solutions in all.

Squaring and simplifying gave ab=0. So either both equal to zero or one of them. So I got 21. To start with the problem I had in mind that except 0, all the rest in the domain will be -tive or +tive. I forgot it for the first answer. Corrected for the second remembering well to have (0,0) only once..

The only possibility that the above equation holds true is when either of them is '0' or both of them are 0.. Case 1:Both a and b are zeros. (0,0) ---> 1 possiblity Case 2:If a = 0 , then b can take any values between -10 to 10 (leaving 0 cause it is already taken care in case 1) ----> 20 possibities Case 3: if b = 0 , then it is again another 20 possibilities.. therefore total 41 ways

$\sqrt{a^2 +b^2} = \sqrt{a^2}+\sqrt{b^2}=a+b$

Squaring both sides, we get...

$\left(\sqrt{a^2+b^2}\right)^2=\left(a+b\right)^2$

$\Longrightarrow a^2+b^2=a^2+b^2+2ab$

$\Longrightarrow 2ab=0$

$\Longrightarrow ab = 0$

Thus, either $a$ or $b$ or both of them are zero...

Fixing $a=0$ , we get $-10\leq b\leq10$ , hence $21$ values of $b$ ...

Fixing $b=0$ , we get $-10\leq a\leq10$ , hence $21$ values of $a$ ...

Note that the pair $(a,b)=(0,0)$ occurs in both cases, hence twice...

Therefore, the required result is: $21+21-1=\fbox{41}$ ...

from given relation,after simplifying we get 2ab=0,now take a or b as 0,and consider the cases

Square both sides and cancel equal terms. Then we have 2ab=0. So, if a=0, there are 21 options for b. The same happens when b=0. This totals 42 ordered pairs. However, the case a=b=0 was counted twice, so the answer is 41.

$a^{2}$ + $b^{2}$ = $|a|^{2}$ + $|b|^{2}$ , and $\sqrt{a^{2}}$ + $\sqrt{b^{2}}$ = |a|+|b|Substitute into the equation given in the question, we get

$\sqrt{|a|^{2}+|b|^{2}}$ =|a|+|b|

Squaring both side of the equation, we get $|a|^{2}$ + $|b|^{2}$ = $|a|^{2}$ +2|a||b|+ $|b|^{2}$ . Therefore 2|a||b|=0.

Consider either a or b is 0. If a =0, b =-10,-9,...,9,10, but b not equal to 0. We have 20 choices for b (from -10 to -1, and from 1 to 10). Similarly,If b =0, a also have 20 choices.There is also a pair which both a and b are 0. So total is 20+20+1= $\boxed{41}$

$\sqrt{a^2 + b^2}$

$\implies \sqrt{a^2 + b^2 + 2ab}$ [Let $2ab = 0$ ]

$\implies \sqrt{(a + b)^2}$

$\implies (a+b)$

And $\sqrt{a^2} + \sqrt{b^2}$

$\implies (a+b)$

So, it is only possible when $2ab = 0$

$\implies 2ab =0$

$\implies ab = 0$

We can put value of $b = -10 to 10$ and $a = 0$

And $a = -10 to 10$ and $b = 0$

There is 21 integers from -10 to 10

So, there is total ${(21 \times 2) - 1}$ ordered pairs of integers.

We counted $0,0$ twice so there is -1

And the answer is $\boxed{41}$