Solve for integer solutions

[anemone]

Determine all integer pairs of solutions for $(a,\,b)$ such that $\sqrt{a-\sqrt{b}}+ \sqrt{a+\sqrt{b}}=\sqrt{ab}$.

 

[kaliprasad]

Determine all integer pairs of solutions for $(a,\,b)$ such that $\sqrt{a-\sqrt{b}}+ \sqrt{a+\sqrt{b}}=\sqrt{ab}$.

Spoiler

square both sides to get$2a+2\sqrt{a^2-b} = ab$
or
$2\sqrt{a^2-b} = ab-2a$
square both sides to get
$4(a^2-b) = a^2b^2-4a^2b+4a^2$
or $a^2b^2= 4a^2b - 4b$
so $b=0$ or $b= 4- \frac{4}{a^2}$
b= 0 => a = 0
$b= 4- \frac{4}{a^2}$
gives a = 1 or 2
a = 1 gives b=0 this gives rise to contradiction
a= 2 gives b = 3 and this is the solution
so $(a,b) = (0,0)$ or $(2,3)$
it can be checked that $(a,b) = (2,3)$ is a solution by taking square root and adding. I have done the same

 

[anemone]

Spoiler

square both sides to get$2a+2\sqrt{a^2-b} = ab$
or
$2\sqrt{a^2-b} = ab-2a$
square both sides to get
$4(a^2-b) = a^2b^2-4a^2b+4a^2$
or $a^2b^2= 4a^2b - 4b$
so $b=0$ or $b= 4- \frac{4}{a^2}$
b= 0 => a = 0
$b= 4- \frac{4}{a^2}$
gives a = 1 or 2
a = 1 gives b=0 this gives rise to contradiction
a= 2 gives b = 3 and this is the solution
so $(a,b) = (0,0)$ or $(2,3)$
it can be checked that $(a,b) = (2,3)$ is a solution by taking square root and adding. I have done the same

Thanks for participating, kaliprasad and for your solution!

My solution:

Spoiler

Squaring two times the given equality, we get $a^2((b-2)^2-4)+4b=0$, since $b≥0$, this implies $((b-2)^2-4)≤0$, solving it for $b$ we get $0≤b≤4$, and we get $(a,\,b)=(0,\,0)$ and $(2,\,3)$.