Show that there does not exist any positive integer pair (a, b) satisfying the equation a² + 3ab – 2b² = 156

When I took number theory in college, I remember my professor telling us that sometimes it would be easier to solve problems like this one by using modular arithmetic.  Those of you who have studied number theory should be able to follow my argument; however, for those of you who are unfamiliar with this topic, I will give a brief overview of the ideas involved.  Hopefully it will make sense!

The method I will use here involves 'reading the equation modulo 3'.  This means that I am only really considering what the equation would look like after I threw away all the multiples of 3 and just kept the remainders.  This is how arithmetic mod 3 works; the only 3 numbers we consider are 0, 1 and 2.  All other numbers are replaced by their remainders after they have been divided by 3.  So for example, the number 28 is equivalent to 1 in mod 3 arithmetic, because

28 = 27+1 = 3(9)+1 = 0+1 = 1

Now, let me get back to the given problem:

a^2 + 3ab - 2b^2 = 156

a^2 = 2b^2 - 3ab + 156

a^2 + b^2 = 3b^2 - 3ab + 156

a^2 + b^2 = 3[b^2 - ab + 52]

Now, notice that the right side of the equation is an integral multiple of 3.  Therefore, the left hand side must be a multiple of 3 also.  So, we need to find integers a and b such that a^2+b^2 is a multiple of 3.

Now, let's consider how the squaring operation works in mod 3.  Here are the squares:

0^2 = 0

1^2 = 1

2^2 = 4 = 3+1 = 0+1 = 1

So, the only squares in mod 3 arithmetic are 0 and 1.  So, in mod 3, a^2 + b^2 would have to be one of the sums 0+0, 0+1, 1+0, or 1+1.  However, the only one of these sums which is a multiple of 3 is the first one!  This means that a and b must both be 0 in mod 3 arithmetic.  In layman's terms, this means that a and b both have to be multiples of 3.  This is the idea that I'll use to show that the equation has no positive integral solutions.

Now, let a=3m, b=3n, where m and n are some unknown positive integers.  Then we can write the original problem as:

a^2 + 3ab - 2b^2 = 156

(3m)^2 + 3(3m)(3n) - 2(3n)^2 = 156

9m^2 + 27mn - 18n^2 = 156

9[m^2 + 3mn - 2n^2] = 156

This is far enough!  The left hand side of this equation is a multiple of 9, but the right is not.  Therefore, we can find no integral solutions to the given equation.

Thank you K Sengupta for this interesting problem.  I enjoy the puzzles that you post here and they always give me inspiration to dust off the old textbooks and try to improve my mathematics.  I look forward to seeing the solution you post to this one.

-John

Posted by John Reid on 2007-07-01 00:34:26