solution of  the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ i am wondering if there is a complete solution for the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ in which $a,b,c,d,k$ are integer(not all zero) and $p$ is odd prime. 

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.
If you can represent numbers as:  $p=3k^2-t^2$
This decision when the coefficients are related through the equation of Pell.  $t^2-3k^2=-p$
To simplify calculations we will make this change.
$$x=(\pm{t}-2k)n^2+2j(t\mp3k)ns-(2kj^2+2kpe^2\pm{t}(pe^2-2j^2))s^2$$
$$y=(\pm{t}-2k)n^2+2j(2t\mp3k)ns-(8kj^2+2kpe^2\pm{t}(pe^2-2j^2))s^2$$
$$r=2e(tn-3kjs)s$$
$$f=n^2+(pe^2-2j^2)s^2$$
Then the solution can be written:
$$a=pr^2+(p+k^2)f^2-xy$$
$$b=r(x+y)$$
$$z=f(x+y)$$
$$c=pr^2+(p+k^2)f^2+x^2$$
$n,s,e,j$ - integers which we ask.
