Finding the magic number as following Let $s$ and $t$ be distinct positive integers with $s+t$ and $s-t$ are a square numbers.
A pair $(s,t)$ called magic if there is exist positive integer $u$, such that
$12s^2 + t^2 = 4t^2u^3$. Does it exist a magic number?
I try that $s+t = m^2$ and $s-t = n^2$ for some positive integer $m, n$, such that
$2t = (m-n)(m+n)$. LHS is even, so RHS must be even. There are 2 cases, when both $m$ and $n$ are odd, and, when both $m$ and $n$ are even.
And then, what next? I stuck at here. Any idea?
 A: Too long for a comment. As Ross Molikan points out, the problem boils down to solve $3k^2+1=4u^3$.
We work in the ring $R=\mathbb{Z}[j]$, where $j=e^{2i\pi/3}$. The ring $R$ is a PID (it is even euclidean), hence a UFD.
Set $z=1+k\sqrt{-3}=k+1+2k j$. The equation may be rewritten $zz^*=4u^3$, where $*$ denotes complex conjugation (which induces an automorphism of $R$).
Since $2$ is known to be irreducible in $R$, $2$ divides $z$ or $z^*$ in $R$, but then $2$ divides $z$ in both cases (apply complex conjugation). Since $z=(k+1)+2kj$, this implies that $k+1$ is even, and that $k$ is odd. We then have $z=2y$ with y=$\frac{k+1}{2}+kj$, with $k$ odd. In particular, $2\nmid y$ in $R$.
Now the equation is equivalent to $yy^*=u^3$.
We claim that $y$ and $y^*$ are coprime in $R$. Indeed , if $t\in R$ is a common divisor of $y$ and $y^*$, it divide $y+y^*=\frac{z+z^*}{2}=1$, and so $t$ is a unit.
Since $y, y^*$ are coprime and $yy^*$ is a cube, $y=\alpha w^3$, where $\alpha$ is a unit of $R$ and $w\in R$. Notice now that the units of $R$ are $\pm 1,\pm j,\pm j^2$
Assume first that $\alpha=\pm 1.$ Changing signs (since $-1$ ) , one may assume that $\alpha=1$.
Hence $y=w^3$, so $z=2w^3$. We now use the fact that an element $w$ of $R$ may be written under the form $w=\frac{a+b\sqrt{-3}}{2}$, where $a,b$ have same parity.
We then get $z=2w^3=\dfrac{a^3-9 ab^2+(3a^2b-3b^3)\sqrt{-3}}{4}=1+k\sqrt{-3}$.
In particular, $4=a(a^2-9b^2)$. Note that if $a$ and $b$ are even, then $a^2-9b^2$ must be divisible by $4$, and then $a (a^2-9b^2)$ is divisible by $8$, contradiction. Hence $a$ and $b$ are odd, so $a=\pm 1$. If $a=1$, then $3=-9b^2\leq 0$, contradiction. Hence $a=-1$, so $9b^2=5$, another contradiction.
It remains to examine the case $\alpha=\pm j, \pm j^2$.  Since $-1$ is a cube, one may assume that $\alpha=j$ or $j^2$. If $\alpha=j^2$, conjugating yields that $z^*=2j (w^*)^3$.
So replacing $k$ by $-k$, one may assume that $z=2jw^3$. This seems to be the difficult case. Still thinking about it...Maybe somebody will be able to continue further.
A: First note that $t$ must be even as the two terms in the equation except $t^2$ are even.  Let $t=2v$ and now we are looking for solutions to $3s^2+v^2=4v^2u^3$. $s$ and $v$ must have the same parity.  If they are both even we can divide both by $2$ and the equation will still be satisfied, so the minimal solution will have both odd.  Now $s$ must be a multiple of $v$, so let $s=kv$ and we have $3k^2+1=4u^3$.  This is an elliptic curve and there are those who can find integer solutions on them, but I am not one.  I only find $k=1,u=1$ by a quick search up to $k=458$.  This becomes $2s=t, u=1$ but then $s-t=-s \lt 0$ and it cannot be a square.  If there is not another integer point on the curve, there is no solution.
A: Comment:
An experimental approach:
To make sure (s-t) and (s+t) are squares we may consider following Pythagorean triple:
$a=2i+1$, $b=2i(i+1)$ and $c=2i(i+1)+1$
Where :
$2i(i+1)+1-2i(i+1)=1=1^2$
$2i(i+1)+1+2i(i+1)=4i^2+4i+1=(2i+1)^2$
And:
$$(2i+1)^2+[2i(i+1)]^2=[2i(i+1)+1]^2$$
So we must have:
$$u^3=\frac{12[2i(i+1)+1]^2+[2i(i+1)]^2}{4[2i(i+1)]^2}$$
Or:
$$u^3=\frac{12[(b+1)^2+b^2}{4b^2}$$
I could find no integral solution for u for i up to $10^6$.
