Solving two diophantine equations. Find at least one 5-tuple of positive integers which satisfy the following two equations
$$a^2-d^2=3(b^2-c^2)$$ 
$$e^2-b^2=3(d^2-c^2)$$
such that no three of the 5 positive integers $a, b, c, d, e$ form an Arithmetic progression and $a<b<c<d<e$.
Please help.
 A: Here is a nice context to this seemingly random system. If there is a $5$-tuple $a,b,c,d,e$ such that,
$$\begin{aligned}
a^2-d^2&=3(b^2-c^2)\\[1.5mm]
e^2-b^2&=3(d^2-c^2)
\end{aligned}\tag1$$
then for $k = 2,4$,
$$(a + c)^k + (-a + c)^k + (2 d)^k = (c + e)^k + (c - e)^k + (2 b)^k\tag2$$
and for $k = 1,2,3,5$,
$$(a + 2 c)^k + (-a + 2 c)^k + (-2 c + e)^k + (-2 c - e)^k =\\
 (2 b + c)^k + (-2 b + c)^k + (-c + 2 d)^k + (-c - 2 d)^k\tag3$$
There is an infinite number of co-prime solutions to $(1)$. An example is,
$$a,b,c,d,e = x - 3 y,\; x + 2 y,\; x + 3 y,\; z,\; -x + 7 y$$
where $x,y,z$ satisfy,
$$x^2+24y^2 = z^2\tag4$$
The condition $(4)$ can be easily solved as,
$$x,y,z = 2mn,\; m^2-24n^2,\; m^2+24n^2$$ 
and if we wish that $a<b<c<d<e$, then $m \approx 6n$ for $n$ above a bound. For example, if $m,n = 38,\,6$, then after removing common factors,
$$a,b,c,d,e = -321,\; 404,\; 549,\;577,\;901$$
and since $(1)$ involves squares, then signs don't matter. 
P.S. Using Lucian's small solution $a,b,c,d,e = 4,\,41,\,54,\,61,\,64$ on $(2),(3)$, one also gets,
$$58^k+50^k+122^k = 118^k+10^k+82^k,\;\;k=2,4$$
$$112^k+104^k+(-44)^k+(-172)^k = 136^k+(-28)^k+68^k+(-176)^k,\;\; k=1,2,3,5$$
A: To solve the system of equations:
$$\left\{\begin{aligned}&a^2-d^2=3(b^2-c^2)\\&e^2-b^2=3(d^2-c^2)\end{aligned}\right.$$
It is better to use such solutions.
$$a=p+(5k-3t)s$$
$$d=p-(k-3t)s$$
$$b=p+(3k-t)s$$
$$c=p+(k+t)s$$
$$e=p+(5t-3k)s$$
$p,s,k,t - $ integers asked us.
A: From given above, the below mentioned equations have parametric solution,
$a^2-d^2=3(b^2-c^2)$
$e^2-b^2=3(d^2-c^2)$
$(a,b,c,d,e)=(2,k-1,2k,3k+1,4k+2)$

