2023/027) Find Positve integers x and m such that $\sqrt{x} + \sqrt{x+60} = \sqrt{m}$

 For the above to hold we must have $x= na^2$ and $x+60= nb^2$ where n,a,b are integers

So we get $n(b^2 - a^2) = 60$

or$n(b+a)(b-a) = 60$

now for a anb b to be integer we must have b+a and b-a to be both even or odd 

we need to find 60 as product of 3 numbers with b+ a and b- a to be different

this gives us following cases

$n=1, b + a= 30, b- a = 2$ giving $n=1,b= 16, a = 14$ giving $x= 196, m = 900$

$n=3, b + a= 10, b- a = 2$ giving $n=3,b= 6, a = 4$ giving $x= 48, m =300$

$n=4, b + a= 5, b- a = 3$ giving $n=4,b= 4, a = 1$ giving $x= 4, m = 100$

$n=4, b + a= 15, b- a = 1$ giving $n=4,b= 8, a = 7$ giving $x= 196, m = 900$ (solution repeats)

$n=5, b + a= 6, b- a = 2$ giving $n=5,b= 4, a = 2$ giving $x= 20, m = 180$

n= 15 gives a+ b and b-a to be same so no further solution

So solution set $(196,900),(48,300), (4,100),(20,180)$