Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares. What is the smallest possible value of $b$? Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares.
What is the smallest possible value of $b$?
Let, $2a-b=k^2, a-2b=p^2, a+b=q^2$.
$k^2=p^2+q^2$ after adding any of the two equations. 
How to proceed further?

If you subtract the last two you get $q^2-p^2=3b$.  If you add the first and last you get $k^2+q^2=3a$.  No primitive Pythagorean triangle has legs that differ by a multiple of $3$, so we need a triangle that has a common factor of $3$.  The smallest such is $9-12-15$ and we find
$$3b=144-81=63\\b=21\\3a=225+144=369\\a=123$$
This is the smallest $b$ because the difference of the two legs must be at least $3$.  If the shorter leg is $c$ we have $(c+3)^2-c^2=6c+9$ and $b$ will grow with the shorter leg.  
