looks like we're talking
Spoiler:
(a^2-1)b^2 - a^2:
if a= 1 or 2 (mod 3), (a^2-1)b^2 - a^2 = 0*b^2 - 1 (mod 3) = -1 (mod 3), and no perfect squares can do that;
if a=0 (mod 3), (a^2-1)b^2 - a^2 = (-1)b^2 (mod 3), it actually seems feasible if b=0 (mod 3). In that case, we need to test modulus 9.
If a=0 (mod 9), (a^2-1)b^2 - a^2 = (-1)b^2 (mod 9), so b=3 or 6 (mod 9). Let's try some actual numbers now:
a=0, we have -1b^2: b=0
a=3, we have 8b^2-9: not gonna work
a=6, we have 35b^2-36, first glance b=6 (mod 9). However, let b=6k (k E Z) then 35b^2-36 = (35k^2-1)*36. Clearly, no matter what k is, 35k^2-1 just won't be a perfect square.
What do you think, it seems that delta will be perfect square only when a=b=0?
Thanks going to romsek and dannydengler for their efforts. Here's my solution:
The correct and only answer is indeed a = b = c = 0. So we are going to start with the assumption that some of a, b, c are not equal to 0.
First, let's modify the problem a bit. I'm going to take a, b, and c and divide out all the possible factors of 2 such that we still have integers. In other words we're looking for a maximal value of n such that we have the integers and that one of must be an odd number, else n is not maximal.
The problem to solve is now
We can clean this up a bit and obtain
Now, must be divisible by 4 as is divisible by 4. We already know that one of is odd, so if the sum must be even we have to have another of to be odd as well.
So we want to look at in terms of modulo 4. We have two odd numbers and one even one. Odd numbers are of the form 1, 3 (mod 4). When we square these we find that all odd numbers squared are equivalent to 1 (mod 4). Even numbers are of the form 0, 2 (mod 4) and when we square them we find that all even numbers squared are 0 (mod 4).
So are two odd numbers and one even one. Squaring each individually: . But we need to it be 0 mod 4 because . As these two facts do not agree there is no answer to the problem besides a = b = c = 0.
-Dan