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. 
 A: I'm not getting much 2-adic information for this one, but it should be easy enough to check all solutions mod 8 and mod 16 and see what happens.
To restrict anything, one property requires $p \equiv \pm 3 \pmod 8$ and the other requires   $p \equiv  3 \pmod 4.$ Put them together, when $$p \equiv  3 \pmod 8  $$  and
$$  p | k, $$
then all four of your letters $$ a,b,c,d = 0.$$
The proof uses two flavors of anisotropy for binaries. The assumption is that at least one of
$ a,b,c,d $ is nonzero and  $\gcd(a,b,c,d) = 1.$
 First we have forced 
$a^2 - 2 c^2 \equiv 0 \pmod p,$ so $a,c \equiv 0 \pmod p$ as $(2 | p) = -1.$ But then
$ p b^2 + p d^2 \equiv 0 \pmod {p^2},$ or
 $  b^2 +  d^2 \equiv 0 \pmod p,$  so $b,d \equiv 0 \pmod p$ as $(-1 | p) = -1.$ So 
 $ p | \gcd(a,b,c,d)$ contrary to assumption.
Otherwise, given a fixed $(p,k)$ once you have a nontrivial solution you get infinitely many
using automorphs of the indefinite part in variables $(c,d).$ That is, there may be many parametrized families of solutions of one type or another. But you can figure some of those out with a computer algebra system more easily than I can by hand.
The next interesting case is when $12 k^2 + 8p$ is a square, which means that the binary form 
$T(c,d)=2c^2+2kcd-(p+k^2)d^2$ factors. So $3 k^2 + 2p$ is a square, which is not possible for even $k,$ so $k$ is odd and $2p \equiv 6 \pmod 8,$ or   $p \equiv -1 \pmod 4.$ Unless $p=3$ we also need 
  $p \equiv -1 \pmod 3,$ or   $p \equiv -1  \equiv 11  \pmod {12}.$
For example, with $p=11, k=1,3 k^2 + 2p = 25, p + k^2 = 12,$ we have
$$ a^2+11b^2-2c^2-2cd+12d^2 = a^2+11b^2-2(c-2d)(c+3d).$$
The value of the factorization is that we can take, for instance, $c = 2 d + 1, c + 3 d = 5 d + 1,$ and
$$ a^2+11b^2-2(5d+1) = 0.$$ Now $a^2 + 11 b^2$ is not even unless it is also 
divisible by $4.$ We also need $ a^2 \equiv b^2 \equiv 1 \pmod 5.$ Put them together, we have a parametrized solution of sorts, with $$  a \equiv 1,4 \pmod 5, \; \;   b \equiv 1,4 \pmod 5, \; \;  a \equiv  b \pmod 2$$ take $c = 2 d + 1$ and
$$ d = \frac{ a^2+11b^2-2}{10}.$$  
A: Perhaps it would help if we knew where the question is coming from. 
For what it's worth, you can write the equation in the form
$$ a^2 + (c-kd)^2 + pb^2 + pd^2 = 3c^2, $$
so you are looking at a parametrized subset of the equation
$$ A^2 + B^2 + pC^2 + pD^2 = 3c^2. $$
If $p \equiv 1 \bmod 4$, then $p(B^2+D^2) = R^2 + S^2$ is a sum of two squares, and your solutions must occur among those of
$$ R^2 + S^2 + T^2 + U^2 = 3c^2. $$
Both quadrics can be parametrized by the standard method of sweeping lines if you know
one solution. For arbitrary primes $p$ such a solution seems to be difficult to find. And even armed with such a parametrization you then would have to figure out which of them satisfy the additional conditions coming from the original equation. 
A: There are lots!  Here is a selection:
$(a,b,c,d,k,p)=(3,3,4,1,1,3)$
$(a,b,c,d,k,p)=(3,3,4,1,7,3)$
$(a,b,c,d,k,p)=(3,3,4,2,2,3)$
A: 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.
