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Math Help - Problem #17: Diophantine Equation

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    Forum Admin topsquark's Avatar
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    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
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    Re: Problem #17: Diophantine Equation

    looks like we're talking

    Spoiler:
    $a=b=c=0$
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    Re: Problem #17: Diophantine Equation

    Quote Originally Posted by romsek View Post
    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
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    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...
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    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?
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    Re: Problem #17: Diophantine Equation

    Quote Originally Posted by dennydengler View Post
    (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.
    Last edited by topsquark; August 4th 2014 at 06:44 PM.
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    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
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