Diophantine D(12)-quadruple By Diophantine $D(12)$-quadruple I mean the set $\{a,b,c,d\}$ of positive integers in which the product of any 2 distinct elements plus 12 is a perfect square in $\mathbb{N}.$
I am trying to prove that every element of the Diophantine $D(12)$-quadruple is even.
I was trying to look at the equations $xy+12=z^2$, where $x,y\in\{a,b,c,d\}$ modulo $4, 8, 16$, but all I managed to prove is that at most $2$ elements are odd, which means nothing for me.
 A: The answer can be found in this paper 
https://web.math.pmf.unizg.hr/~duje/pdf/acta1.pdf
Let $\{a_1,a_2,a_3,a_4\}$ be a $D(12)$-quadruple. 
Assume that one if its elements, say $a_1$, is odd. 
It is easy to see that the square of an integer
is congruent to $0, 1, 4$ or $9 \pmod{16}$. 
Therefore, $a_i a_j \equiv 4, 5, 8$ or $13 \pmod{16}$, for
$i, j = 1, 2, 3, 4$, $i \neq j$. This implies that if some $a_i$ is even, then it is
divisible by $4$. Since $a_ia_j$ is not divisible by $16$, we conclude that there
must be at most one even number, i.e. at least three odd numbers, among
$a_i$, $i = 1, 2, 3, 4$. 
Let $a_1, a_2, a_3 $ be odd. We have $a_1a_2 \equiv 5 \pmod{8}$, $a_1a_3 \equiv 5 \pmod{8}$, $a_2a_3 \equiv 5 \pmod{8}$. 
By multiplying these congruences, we obtain
$(a_1a_2a_3)^2 \equiv 5^3 = 125 \equiv 5 \pmod{8}$,
which is impossible, since the square of an integer is congruent to $0, 1$ or $4$ mod $8$.
A: Idea:
Say $D(12)=\{1,4,13,d\}$ then $$13d+12= n^2\implies n = 13k\pm 5$$
so $$d= 13k^2\pm10k +1$$ 
Now try to find if for some $k$ you can have also $$4d+12 = m^2\;\;\;\wedge \;\;\; d+12=s^2$$
