# 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?

• Try first the simpler case of when $u=1$. – Memes Sep 27 at 2:46
• It means that, $12s^2+t^2 = 4t^2 \Leftrightarrow 12s^2 = 3t^2 \Leftrightarrow t^2 = 4s^2$. Then, $(s,t) = (1,4)$. Right? – user795084 Sep 27 at 2:48
• But, that's not a square number. – user795084 Sep 27 at 2:50
• I'm sorry about that, I didn't read the question fully! – Memes Sep 27 at 3:02
• Does it help to write $12s^2=t^2(4u^3-1)$? For what it's worth, $4u^3-1$ is always odd, so $t$ must be a multiple of $2$. – Memes Sep 27 at 3:05

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.

• So, there is no solution? – user795084 Sep 27 at 3:27
• @user795084: there is no solution unless there is a solution for larger $k$ than I tried. – Ross Millikan Sep 27 at 3:47

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 $$10^6$$.

• But, $s+t=18$ and $s-t = 8$ not a square number. – user795084 Sep 27 at 4:07
• Both $s+t$ and $s-t$ are a square number, Sir. – user795084 Sep 27 at 4:11
• Yes you are right . I will edit my answer. – sirous Sep 27 at 4:43

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.