Answer: Yes.
You can re-arrange the equation as:
3(a+b+c+d)(a+b-c-d) = (a-b+c-d)(-a+b+c-d)
So if a+b+c+d is prime then either a-b+c-d or -a+b+c-d is a multiple of a+b+c+ d.
But since they are both smaller than a+b+c+d in
magnitude, then one of them must be zero. Since the right hand side
of the equation is zero, so is the left hand side. That means a+b-c-d=0,
so a+b=c+d, so a+b+c+d=2(a+b), which is an even number bigger than 2,
proving the statement.
How the heck did I ever manage to rearrange
a²+ab+b²-c²-cd-d²=0
to get
3(a+b+c+d)(a+b-c-d) = (a-b+c-d)(-a+b+c-d) ?
The trick was to write:
a² + ab +
b² = (3(a+b)² + (a-b)²)/4
and find a similar expression for
c²+cd+d², equate these appropriately, regroup then use difference of two squares.
Note, this equation is in effect a
diagonalization of the quadratic form a²+ab+b². That is if we define new variables X=a+b, Y=a-b then the quadratic form becomes
3X²/4 +
Y²/4 which is diagonal because there are no cross terms -- that is, the coefficient of XY is zero. | 