In one paper about namber theory author stated 2 lemmas

Lemma 1. If $p$ is a prime $\equiv3(mod $ $4)$ then $x^2+y^2-pz^2$ represents a non-zero rational number $m$ if and only if $m$ is not of the form $kps^2$ with $\left(\frac{k}{p}\right)=1$ or $ks^2$ with $k\equiv p(mod$ $8)$.

Lemma 2. If $p$ and $q$ are odd primes with $p = 1(mod$ $4)$ and $\left(\frac{q}{p}\right)=-1$ then $x^2+qy^2-pz^2$ reperesents a non-zero rational number if and only if $m$ is not of the form $pks^2$ with $\left(\frac{k}{p}\right)=-1$ or $qks^2$ with $\left(\frac{k}{q}\right)=-1$.

To prove these lemmas author refers to general theorem of Hasse-Minkowski. Can they be proved without using such strong results?

In one paper about number theory author stated 2 lemmas

Lemma 1. If $p$ is a prime $\equiv3(mod $ $4)$ then $x^2+y^2-pz^2$ represents a non-zero rational number $m$ if and only if $m$ is not of the form $kps^2$ with $\left(\frac{k}{p}\right)=1$ or $ks^2$ with $k\equiv p(mod$ $8)$.

Lemma 2. If $p$ and $q$ are odd primes with $p = 1(mod$ $4)$ and $\left(\frac{q}{p}\right)=-1$ then $x^2+qy^2-pz^2$ reperesents a non-zero rational number if and only if $m$ is not of the form $pks^2$ with $\left(\frac{k}{p}\right)=-1$ or $qks^2$ with $\left(\frac{k}{q}\right)=-1$.

To prove these lemmas author refers to general theorem of Hasse-Minkowski. Can they be proved without using such strong results?