Common silly mistake

Rearranging the equation we get $\sqrt{a^2+b^2}+\sqrt{c^2}=\sqrt{a^2+c^2}+\sqrt{b^2}$ and squaring we obtain $a^2+b^2+c^2+2\sqrt{a^2c^2+b^2c^2}=a^2+c^2+b^2+2\sqrt{a^2b^2+c^2b^2}$ Simplifying and squaring again we are left with $a^2c^2+b^2c^2=a^2b^2+c^2b^2$ which actually is $a^2b^2=a^2c^2$ Now, we have two cases:

• $a=0$ Then substituting in the original equation we obtain an identity, which means $b,c$ can take any value. So we have $10$ choices for each one, that is $10\cdot10=100$ couples.
• $a\neq0$ Then we simplify $a^2$ and get $b^2=c^2$ since $b,c$ are non-negative we can take the square root and get $b=c$ . So for every choice of $a$ we have $10$ choices for $b,c$ .

In conclusion the total number of the triples is $10\cdot10+9\cdot10=190$

if $\sqrt{a^2+b^2} -\sqrt{a^2 +c^2} = \sqrt{b^2} - \sqrt{c^2} = b - c$ , we can easily see that

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

So,we can see that $2a^2bc = a^2(b^2 + c^2)$

a2+b2−−−−−−√−a2+c2−−−−−−√=b2−−√−c2−−√ Since we know a,b,c>=0 we can simplify the RHS to:

a2+b2−−−−−−√−a2+c2−−−−−−√=b−c Squaring on both sides,

a2+b2−2(a2+b2)(a2+c2)−−−−−−−−−−−−−−√+a2+c2=b2−2bc+c2 Collecting like terms,

2a2−2(a2+b2)(a2+c2)−−−−−−−−−−−−−−√=−2bc Isolating the square root,

−a2+(a2+b2)(a2+c2)−−−−−−−−−−−−−−√=bc (a2+b2)(a2+c2)−−−−−−−−−−−−−−√=a2+bc Squaring on both sides again and expanding the product of polynomials,

a4+a2b2+a2c2+b2c2=a4+2a2bc+b2c2 a2b2+a2c2=2a2bc a2(b2+c2−2bc)=0 Factoring the polynomial,

a2(b−c)2=0 So either a=0, giving us 100 possible solutions, or a≠0 and b=c, giving us 9×10=90 more solutions.

So, 190.