Find all solutions in set of integers for $a^2 + 5b^2 = 3c^2$ 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.
 A: 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 $a=b=c=0$.
