<s>[INST] 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?
 [/INST] 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.
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