I'm trying to prove the following equation has no solutions, with the following conditions on a, b, c and d:
These 4 variables are integers. They can be negative, but each of them must have a unique absolute value. That is, if one of the variables is equal to x, then none of the others may be -x. Note that this rules out the trivial solution a = b = c = d = 0.
The equation is
3a^2 + a + 3b^2 + b = 3c^2 + c + 3d^2 + d
What techniques could I employ to show this has no solutions?
Both sides of the equation must also be equal to 2e^2, which means they must be equal to a number that is 2 times a perfect square.
in other words,
3a^2 + a + 3b^2 + b = 3c^2 + c + 3d^2 + d = 2e^2.
My first instinct is to rearrange to
3a^2 + 3b^2 - 3c^2 - 3d^2 = c + d - a - b
Now the right hand side must both be even (from the 2e^2 part) and divisible by 3 (from the left hand side), which implies that
c + d - a - b = 0 (mod 6).
But that's going to give a lot of cases to consider if I just start considering a, b, c and d mod 6. Is there a better way to proceed?
Finding integer solutions to the equation
isn't difficult. ( is a solution). A reasonable technique is to complete the squares on each of the four expressions, (you find that the numbers introduced cancel out), and then switching a pair from one side of the equation to the other. That leads to a 'difference of two squares' situation and hence an equation of the form . Then find sets of values for that produce integer values for
The real problem is to find (at least two ) different solutions of the equation
for some value(s) of .