# Is it possible to show this equation has no solutions

• Dec 27th 2011, 09:28 PM
kiwijoey
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.
• Dec 27th 2011, 09:50 PM
pickslides
Re: Is it possible to show this equation has no solutions
Quote:

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.

Quote:

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 \$\displaystyle 2e^2 = 2e^2\$ ?
• Dec 27th 2011, 10:12 PM
kiwijoey
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?
• Dec 29th 2011, 01:40 AM
BobP
Re: Is it possible to show this equation has no solutions
Finding integer solutions to the equation

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

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

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

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

for some value(s) of \$\displaystyle n\$.