Integer solutions of : $u+pv=(x+py)(z+pt)$

Question

Let $p$ be an odd prime and $u,v$ be integers. Then, there exist integers $x,y,z,t$ such that $$u+pv=(x+py)(z+pt)$$ How can I find all integers $x,y,z,t$ knowing $u,v$? What I have done: $$p^2(yt)+p(yz+xt-v)+xz-u=0$$ If $yt=0$ then it's trivial. So I suppose that $yt\neq 0$ We have a quadratic equation in $p$ hence $$(yz+xt-v)^2-4yt(xz-u)=r^2$$ where $r $ is an integer. $$(yz+xt)^2-2v(yz+xt)+y^2-4yt(xz-u)=r^2$$ $$(yz-xt)^2-2v(yz+xt)+y^2+4ytu=r^2$$ Which did not get me too far. Any hints?

Answer 1

I think I have found it. The above equation yields $2$ odd primes $p,q$. Then, it is easy to show that: $$zx-u=pqyt$$ $$v-xt-yz=(p+q)yt$$ Then $$u=zx-pqyt$$ $$v=(p+q)yt+xt+yz$$ It follows that: $$(zx-pqyt)+p[yt(p+q)+xt+yz]=(x+py)(z+pt)$$