Thread: Is it possible to show this equation has no solutions

1. Is it possible to show this equation has no solutions

Hi,

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.

2. Re: Is it possible to show this equation has no solutions

Originally Posted by kiwijoey

What techniques could I employ to show this has no solutions?
You can try to rearrange the equation such that it has the sum of squared numbers on one side and 0 on the other.

Originally Posted by kiwijoey

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.
So $2e^2 = 2e^2$ ?

3. Re: Is it possible to show this equation has no solutions

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?

4. Re: Is it possible to show this equation has no solutions

Finding integer solutions to the equation

$3a^2+a+3b^2+b=3c^2+c+3d^2+d$

isn't difficult. ( $a=41,b=8,c=40,d=12$ 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 $w.x=y.z$. Then find sets of values for $w,x,y,z$ that produce integer values for $a,b,c,d.$

The real problem is to find (at least two ) different solutions of the equation

$3a^2+a+3b^2+b=2n^2$

for some value(s) of $n$.