Take the equation $3x^2-5y^2+7z^2 = 0$.
If we take this $mod \: 4$ we get:
$3x^2+3y^2+3z^2 \equiv 0 \: mod \: 4$
All of the squares modulo $4$ are either $0$ or $1$. $3x^2+3y^2+3z^2$ will never be equivalent to 0, then.
Is that the way one goes about these problems?
Thanks.
Almost. If there is a solution in integers $(x,y,z)$ that are not all zero, then there exists a solution $(x,y,z)$ with $\gcd(x,y,z) = 1.$ This would be accomplished by finding $g = \gcd(x,y,z)$ and replacing $(x,y,z)$ by $(x/g,y/g,z/g).$ NOTE: if we write $$ h(x,y,z) = 3x^2 - 5 y^2 + 7 z^2, $$ how does $$ h(x/g, \; y/g, \; z/g) $$ compare with $h(x,y,z)?$
Your calculation shows that if $$ 3x^2 - 5 y^2 + 7 z^2 \equiv 0 \pmod 4, $$ then $x,y,z$ are all even, hence their gcd cannot be one.
This technique is often called "infinite descent," but this version is pretty specific and comes out fairly simple in quadratic forms. There is even a special word for this, that the form $3x^2 - 5 y^2 + 7 z^2$ is said to be anisotropic in $\mathbb Q_2.$