For the equation:

$$a^2+pb^2+(p+k^2)z^2=2c^2+2kcz$$

If the number $k$ is the problem any, and $p$ is such as this: $p=\frac{t^2}{2}-1$

Then the solution can be written:

$$a=\pm{t}n^2+2(tpr\mp(p+1)kj)ns-(2p(p+1)kjr\pm{t}((p+1)(p+k^2)j^2+pr^2))s^2$$

$$b=\pm{t}n^2-2(tr\pm(p+1)kj)ns+(2(p+1)kjr\mp{t}((p+1)(p+k^2)j^2+pr^2))s^2$$

$$z=2(p+1)j((p+1)kjs-tn)s$$

$$c=(p+1)(n^2+((p+1)(p+k^2)j^2+pr^2)s^2)$$

$n,s,j,r$ - integers which we are set.