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Does $a^2 + b^2 = 6 c^2$ have any integer solution?

My thought: (0,0,0) is obviously a solution but I don't think there are any others.

Because if I take equation of modulo 6, it gives $a^2 + b^2 = 0 \pmod 6$ a and b can only be 3 mod 6 or 0 mod 6. However, I am not sure what is the next step here. Any help is appreciated

poetasis
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user13851309
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  • Please don't remove what you tried from the question text since, as you changed it, note that it was then not an appropriate question to ask here (i.e., it had basically no context so it could be closed as such). If you're concerned about something like the formatting, note that your a^2 + b^2 = 0 mod 6, showing as $a^2 + b^2 = 0 mod 6$, can be written instead as a^2 + b^2 \equiv 0 \pmod{6}, with this being displayed as something like $a^2 + b^2 \equiv 0 \pmod{6}$. – John Omielan Feb 10 '22 at 04:08
  • Wolfram Alpha thinks $\space (0,0,0)\space$ is the only solution. – poetasis Feb 12 '22 at 04:01
  • https://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Feb 12 '22 at 05:16

3 Answers3

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Hint: You've shown that $a$ and $b$ are both multiples of $3$. So $a = 3j$ and $b = 3k$ for some integers $j,k$. What happens when you put this information into the equation $a^2+b^2=6c^2$ (assuming $c\ne 0$, as you've already separated out that case).

paw88789
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Consider a $\pmod{3}$ argument.

Since $6c^2$ is a multiple of $3$, you must have that

$a^2 + b^2$ is a multiple of $3$.

However, for any integer $n$, either

$n^2 \equiv 0 \pmod{3}$ or $n^2 \equiv (+1) \pmod{3}$.

Therefore, $a$ and $b$ must each be a multiple of $3$.

Let $r$ denote the largest positive integer exponent such that $3^r$ divides $a$.

Let $s$ denote the largest positive integer exponent such that $3^s$ divides $b$.

Since the constraint $a^2 + b^2 = 6c^2$ is symmetrical around $a$ and $b$, you can assume, without loss of generality, that $r \leq s$.

Since $3^r$ divides $a$ and $3^r$ divides $b$, you must have that $3^{(2r)}$ divides $6c^2.$

This implies that $3^{(2r - 1)}$ divides $c^2$.

This implies that $3^r$ divides $c$.

Therefore, you can let

  • $d = \frac{a}{3^r}$.
  • $e = \frac{b}{3^r}$.
  • $f = \frac{c}{3^r}$.

Then, $~\displaystyle d^2 + e^2 = 6f^2$
where $d$ is not a multiple of $3$.

This implies that $d^2 \equiv 1\pmod{3}.$

Then, you will have that $e^2 \equiv 0\pmod{3}$ or

$e^2 \equiv 1\pmod{3}$.

In either case, you will not have that $d^2 + e^2$ is a multiple of $3$.

This yields a contradiction, since $d^2 + e^2 = 6f^2$.

user2661923
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A simple calculation shows that both $a$ and $b$ should be divisible by $3$. Put $a=3a_1$ and $b=3b_1$. Then we have $9a_1^2+9b_1^2=6c^2$. Consequently, $c$ has to be divisible by $3$. Again, let's suppose $c=3c_1$. Therefore, we will get: $$a_1^2+b_1^2=6c_1^2.$$ Repeating this taken step implies that $a,b$ and $c$ are infinitely many times divisible by $3$ which is a contradiction unless all the three variables are zero.

Reza Rajaei
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