In another word we can solve the problem if we find three distinct representation of $pt^2$ as difference of two squares. for odd $t$ Note that we have $pt^2=\boxed {1}. $ $\left\{\frac{pt^2+1}{2}\right\}^2-\left\{\frac{pt^2-1}{2}\right\}^2$ $\boxed {2}.$ $\left\{\frac{(p+1).t}{2}\right\}^2-\left\{\frac{(p-1).t}{2}\right\}^2$ For any $t$, now let us find another representation. note that we have this identity : $(a^2-b^2).(c^2-d^2)=(ac+bd)^2-(bc+ad)^2$............$(\star)$ so we choose $t$ so that we have $t^2=c^2-d^2$ and now plug $a=\left\{\frac{p+1}{2}\right\}$ and $b=\left\{\frac{p-1}{2}\right\}$ in $\star$ to get another representation of $pt^2$. (this last identity can be identical with first two if we choose $t$ carelessly, however it is easy to show the existence of such $t$ so that last identity will be totally distinct from previous two.)

Note that the equation $1 + pa^2 = a^2$ has infinitely many solutions. Let $x_1, x_2, x_3>1$ be three values which correspond to values of $a$ that result in a solution. Let $t = cx_1x_2x_3, x = cx_2x_3, y = cx_1x_3, z = cx_1x_2$ for any $c \in \mathbb{N}$. $\square$