# Thread: Problem #17: Diophantine Equation

1. ## Problem #17: Diophantine Equation

Find all integers a, b, c such that $a^2 + b^2 + c^2 + 2abc = 0$

(Source: Wisconsin Mathematics, Science, Engineering Talent search)

-Dan

2. ## Re: Problem #17: Diophantine Equation

looks like we're talking

Spoiler:
$a=b=c=0$

3. ## Re: Problem #17: Diophantine Equation

Originally Posted by romsek
looks like we're talking

Spoiler:
$a=b=c=0$
Correct! Are there any other solutions? How could you show if there are/aren't?

-Dan

4. ## Re: Problem #17: Diophantine Equation

a^2+b^2+c^2+2abc=0
c^2+(2ab)c+(a^b+b^2) =0

delta = (2ab)^2 - 4*1*(a^2+b^2) = 4a^2*b^2 - 4a^2 - 4b^2 and you want that to be a perfect square.
Or just look at a^2*b^2 - a^2 - b^2...

5. ## Re: Problem #17: Diophantine Equation

(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?

6. ## Re: Problem #17: Diophantine Equation

Originally Posted by dennydengler
(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?
Nice work, but a little too much on the "guess and check" for my blood.

Hint: Try another base on the original problem. What does mod 2 give you?

-Dan

Edit: Sorry, I meant try mod 4.

7. ## Re: Problem #17: Diophantine Equation

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 $a = 2^n \cdot a_0 \text{ and } b = 2^n \cdot b_0 \text{ and } c = 2^n \cdot c_0$ and that one of $a_0, ~ b_0,~c_0$ must be an odd number, else n is not maximal.

The problem to solve is now $(2^n a_0)^2 + (2^n b_0)^2 + (2^n c_0)^2 + 2 \cdot (2^n a_0 ) (2^n b_0) (2^n c_0) = 0$

We can clean this up a bit and obtain
$a_0^2 + b_0^2 + c_0^2 + 2^{n + 1}a_0 b_0 c_0 = 0$

Now, $a_0^2 + b_0^2 + c_0^2$ must be divisible by 4 as $2^{n + 1}a_0 b_0 c_0$ is divisible by 4. We already know that one of $a_0,~b_0,~c_0$ is odd, so if the sum must be even we have to have another of $a_0,~b_0,~c_0$ to be odd as well.

So we want to look at $a_0^2 + b_0^2 + c_0^2$ 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 $a_0,~b_0,~c_0$ are two odd numbers and one even one. Squaring each individually: $a_0^2 + b_0^2 + c_0^2 \equiv 1 + 1 + 0 \equiv 2 \text{ mod(4) }$. But we need to it be 0 mod 4 because $2^{n + 1}a_0 b_0 c_0 \equiv 0 \text{ mod(4)}$. As these two facts do not agree there is no answer to the problem besides a = b = c = 0.

-Dan