Common silly mistake

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

Since we know $a,b,c >= 0$ we can simplify the RHS to:

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

Squaring on both sides,

$a^2 + b^2 - 2\sqrt{(a^2+b^2)(a^2+c^2)} + a^2 + c^2 = b^2 - 2bc + c^2$

Collecting like terms,

$2a^2 - 2\sqrt{(a^2+b^2)(a^2+c^2)} = -2bc$

Isolating the square root,

$-a^2 + \sqrt{(a^2+b^2)(a^2+c^2)} = bc$

$\sqrt{(a^2+b^2)(a^2+c^2)} = a^2 + bc$

Squaring on both sides again and expanding the product of polynomials,

$a^4 + a^2b^2 + a^2c^2 + b^2c^2 = a^4 + 2a^2bc + b^2c^2$

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

$a^2 ( b^2 + c^2 - 2bc ) = 0$

Factoring the polynomial,

$a^2 (b-c)^2 = 0$

So either $a = 0$ , giving us $100$ possible solutions, or $a \not= 0$ and $b = c$ , giving us $9 \times 10 = 90$ more solutions.

So, 190.

First of all since $b,c > 0$ we get $\sqrt{b^2} = b$ , $\sqrt{c^2} = c$ , Hence $\sqrt{a^2 + b^2} - \sqrt{a^2 + c^2} = \sqrt{b^2} - \sqrt{c^2}$ $\Updownarrow$ $\sqrt{a^2 + b^2} - \sqrt{a^2 + c^2} = b - c$ $\Downarrow$ $2a^2 + b^2 + c^2 - 2\sqrt{(a^2 + b^2)(a^2 + c^2)} = b^2 + c^2 - 2bc$ $\Updownarrow$ $a^2 + bc = \sqrt{(a^2 + b^2)(a^2 + c^2)}$ $\Updownarrow$ $(a^2 + bc)^2 = (a^2 + b^2)(a^2 + c^2)$ $\Updownarrow$ $2a^2bc - a^2b^2 - a^2c^2 = 0$ Now it is important that one step was mere implication, hence we need to also check our solutions, now if $a = 0$ , then $b,c \in \{0,1,2,\dots,9\}$ . If $a \neq 0$ , then $2bc - b^2 - c^2 = 0 \Leftrightarrow (b - c)^2 = 0 \Leftrightarrow b = c$ Hence if $a \in \{1,2,\dots,9\}$ , then $(b,c) \in \Big\{(0,0),(1,1),\dots,(9,9)\Big\}$ Also obviously the original equation holds if $b = c$ or $a = 0$ , hence by combinatorial rule of product there are $10 \cdot 10 + 10 \cdot 9 = \boxed{190}$ solutions.

let $\sqrt{a^2 + b^2} - \sqrt{a^2 + c^2} = \sqrt{b^2} - \sqrt{c^2}$

if $a = 0$ then:

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

$\sqrt{0^2 + b^2} - \sqrt{0^2 + c^2} = \sqrt{b^2} - \sqrt{c^2}$

$\sqrt{b^2} - \sqrt{c^2} = \sqrt{b^2} - \sqrt{c^2}$

and we've number of ordered triples $(0,b,c)$ is $10 \times 10 = 100$

if $a > 0$ then must $b = c$ so number of ordered triples $(a,b,b) = (a,c,c) = (a,b) = (a,c)$ is $10 \times 10 = 100$

but the triples $(0,b,c)$ is repeated twice, when $a = 0$ and $b = c$ so we subtract $10$ :

$(0,0,0)$ $(0,1,1)$ $(0,2,2)$ ... $(0,9,9)$

then number of ordered triples is $100 + 100 - 10 = 190$

$\begin{aligned} \sqrt{a^2+b^2} - \sqrt{a^2+c^2} &= \sqrt{b^2} - \sqrt{c^2} = b-c \\ 2a^2+b^2+c^2-2\sqrt{\left( a^2+b^2 \right)\left( a^2+c^2 \right)} &= b^2+c^2-2bc \\ 2\sqrt{\left( a^2+b^2 \right)\left( a^2+c^2 \right)} &= 2 \left( a^2+2bc \right) \\ 4\left( a^2+b^2 \right)\left( a^2+c^2 \right) &= 4 \left( a^2+2bc \right)^2 \\ \left( a^2+b^2 \right)\left( a^2+c^2 \right) &= \left( a^2+2bc \right)^2 \\ a^4+a^2b^2 + a^2c^2 + b^2c^2 &= a^4 + b^2c^2 + 2a^2bc \\ a^2 \left( b^2+c^2 \right) &= a^2 ( 2bc ) \\ \end{aligned}$

So $a^2=0 \implies a=0$ or $b^2+c^2=2bc \implies b=c$ (by AM-GM).

Hence, by the inclusion-exclusion principle, the number of ordered triples satisfying the constraints is

$100 + 100 - 10 = \boxed{190}.$

to make this equation valid, a must be equal to zero or b must be equal to c.

There are 100 situations where a=0 and 100 situations where b=c. There are 10 situations that overlap so there 100+100-10=190 possible solutions.

So there are 190 triples of integers that satisfy the equation.

Simplifying the given equation, we get $2a^{2}bc=a^{2}b^{2}+a^{2}c^{2}$ So, either $a=0$ , or $(b-c)^{2}=0$ $=>b=c$ When a=0, we get 10 possible values of b and c each. So total possible (a,b,c) combinations are $1*10*10=100$ When b=c, we get 9 possible values of a (1 through 9) and 10 possible values of (b,c) (0 through 10). So total possible (a,b,c) combinations are $9*10*1=90$ Sum total = 100+90=190

The statement will be true if b=c or if a=0.

There are 100 cases where b=c and 100 cases where a=0.

And 10 cases are overlapping, therefore are counted twice.

Therefore the number of cases is 100+100-10=190.

The reasons for there being 100 cases where b=c and 100 cases where a=0 are left to the reader as an exercise.

Squaring on both sides and then arranging gives us - $2a^2 + 2\sqrt{b^2c^2}=2\sqrt{(a^2+b^2)(a^2+c^2)}$ Now again square on both sides and arranging gives us - $2\sqrt{b^2c^2} = b^2+c^2$ We know that $|x|=\sqrt{x^2}$ So, $2|bc|=b^2+c^2 \Rightarrow 2|bc| - 2bc = (b-c)^2$

Since bc is always positive (because questions says a,b,c are integers from (0 - 9) So, $2|bc| - 2bc = 0 = (b-c)^2 \Rightarrow b=c$

Case 1 : when $b=c$ , and $a$ ranges from 1-9, we have $10\times9 = 90$ possiblities

Case 2 : now taking $a=0$ , the equality becomes $\sqrt{0+b^2}-\sqrt{0+c^2} = \sqrt{b^2}-\sqrt{c^2}$ , which is true for all $b$ and $c$ . Hence total of $10\times10=100$ possibilities.

Therefore, we get total $190$ ordered triples of integers $(a,b,c)$ which satisfy the above condition