Find all solutions in set of integers for $a^2 + 5b^2 = 3c^2$

Question

I haven't learnt Diophantine equations to solve this equation (just know Modular Arithmetic). So given the equation we are to solve for all solutions in $\Bbb Z$. So what I've done so far:

Case 1: $\Bbb Z$ $\{0\}$

When $a=b=c=0$ the solution exists (since $0^2 + 5(0^2) = 3(0^2) = 0$)

Case 2: $\Bbb Z \setminus \{0\}$

For $a^2+5b^2=3c^2$ to hold, if $3c^2$ is even (that is, when c is also even), then a and b has to be even (to make $a^2+5b^2$ even).

If $3c^2$ is odd (that is, when c is also odd), then either one of a or b has to be odd (to make $a^2+5b^2$ even).

And attempting in every other way I know I just rearranged the equations in terms of a, b and c to see what I get:

$a =\pm \sqrt{3c^2-5b^2}, b=\pm \sqrt{\frac{3c^2-a^2}5}, c= \pm \sqrt{\frac{a^2+5b^2}3}$

And from this I know that $3c^2\geq5b^2$ but I am not sure how to show this mathematically and where to go from here without using Diophantine equations. I've also looked at Find all solutions: $x^2 + 2y^2 = z^2$ and attempted as suggested in the solutions but making no progress. Any suggestions would be much appreciated.

Answer 1

For all integers $n$,

$n^2$ has to be $0$, $1$ or $-1$ mod$(5)$.

Suppose $a^2 + 5b^2 = 3c^2$.

Since $a^2\equiv 3c^2$ mod$(5)$,

$c^2\not\equiv \pm1$ mod$(5)$ because $a^2\not\equiv\pm3$ mod$(5)$.

Therefore, $a\equiv c\equiv 0$ mod$(5)$. So $a$ and $c$ are both multiples of $5$.

Let $a=5d$ and $c=5e$.

We get $(5d)^2+5b^2=3(5c)^2$.

This simplifies to $5d^2+b^2=15c^2$, or equivalently $b^2=15c^2-5d^2=5(3c^2-d^2)$.

Therefore, $b$ is a multiple of $5$.

So $a$, $b$, and $c$ are all multiples of $5$.

$\\$

EDIT to help the OP:

The above proof shows that if $a,b,c$ are integers and $a^2 + 5b^2 = 3c^2$, then $\frac{a}{5},\frac{b}{5},\frac{c}{5}$ are also integers.

It is also the case that if $a^2 + 5b^2 = 3c^2$, then $(\frac{a}{5})^2 + 5(\frac{b}{5})^2 = 3(\frac{c}{5})^2$. This you can verify.

So you can apply the same logic to $\frac{a}{5},\frac{b}{5},\frac{c}{5}$. You will get that $\frac{a}{25},\frac{b}{25},\frac{c}{25}$ are integers, and $(\frac{a}{25})^2 + 5(\frac{b}{25})^2 = 3(\frac{c}{25})^2$.

You can repeat this procedure and find out that $\frac{a}{5^n},\frac{b}{5^n},\frac{c}{5^n}$ are integers for all $n\geq 1$. If one of these $a,b,c$ is a non-zero integer, we would get an immediate contradiction. Therefore, we have $a=b=c=0$.