# Diophantine question

This came up when I did a brand-new (or maybe it's just "birdtracks" in disguise :-) graph-based construction of the E8 family. x,y,z are dimensions and thus integer. x<0 actually doesn't hurt. (For $y=x,z=(3 (-2 + o) o)/(10 + o)$ the standard E8 setup results, o must now divide 360 etc. pp.) So here is the equation:

$3 x (2 + x) (-2 + x + x^2 - 2 y) y (-x + x^2 - 2 z) z=q^2$

With rational y or z I wouldn't pester MO - I'd solve it on the spot. But here some nasty division properties are involved, and I'm lousy in number theory. Is there a finite, easy describable solution list?

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Numerical search suggests there's lots of solutions: any of the LHS factors can be set to zero, or $x=1$ and $q = \pm 6yz$, and quite a bit besides that. – Noam D. Elkies Feb 6 '13 at 20:56
@Peter: No, not necessary. @Noam: Obviously (although: zero dimension...not sure if want)...the "besides" is the interesting part. (I hoped for a general applicable method.) Note that negative y or z is fishy - I'm still pondering if negative dimensions on the RHS of a Clebsch-Gordon expansion make any sense at all. – Hauke Reddmann Feb 7 '13 at 10:00
Well the $x=1$ family wasn't that obvious... (Though maybe it's obvious in the "standard E8 setup".) The usual heuristics suggest that there should be a sparse but infinite set of "random" solutions (if $\max(x,y,z)\in[H,2H)$ there are about $H^3$ numbers of size $H^6$ so we expect about $H^0$ of them to be squares), and a list of solutions up to say $10^3$ might also turn up new some parametric families. – Noam D. Elkies Feb 7 '13 at 16:37
There are still quite a lot besides those. For example, take $x=2$, and any $z$. Then the equation simplifies to $n_z((y-1)^2 -1)=q^2$, which is Pell's equation if $n_z$ is not a square. Choose $z$ so that $6(z^2-z)$ is not a square (another Pell equation, but this time we want non-solutions) and you'll get infinitely many more solutions. I think the same tricks will work with any $x$ value: complete squares, get a Pell-type equation for $z$ in terms of $x$ and choose a non-solution $n_z$, then write the Pell equation for $y$ in terms of that, and solve. – Zack Wolske Feb 7 '13 at 16:52

Here is an algorithm which will generate many families of solutions, provided you are willing to compute solutions to $a^2 - nb^2 = 1$, and will generate all of them if you instead solve $a^2 - nb^2 = k^2$.

Define the functions $F=\frac{x^2-x}{2}$, $G=\frac{x^2+x-2}{2}$ and $H=3x^2+6x$. Then your equation becomes $$H(2y^2 - 2Fy)(2z^2-2Gz)=q^2$$ Multiplying both sides by 4, completing the square, and substituting $Y=2y-F$ and $Z=2z-G$, we get
$$H(Y^2 - F^2)(Z^2 - G^2) = (2q)^2$$ Choose $n=n_{x,z}=H(Z^2 - G^2)$ so that it is not a square. Then $n$ divides $2q$, so factoring it out and writing $Q=\frac{2q}{n}$, you now want to solve $Y^2 - nQ^2 = F^2$. If you're willing to take just the solutions where $F$ divides $y$ - or the ones where $G$ divides $z$, since their roles are interchangeable - then you are just solving the classical Pell's equation. You can choose $n_{x,z}$ either by listing values for $x$ and $z$, or by taking the squarefree part of $H$ and solving another Pell type equation, but now you want $n$ to not be a solution to the equation (which will happen most of the time).

Noam Elkies's solutions have $F=G=0$, so you don't need to worry about making $n$ non-square, because there are just two terms in the final equation, and you only need to equate the squares.

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Let $P(x,y,z)=3 x (2 + x) (-2 + x + x^2 - 2 y) y (-x + x^2 - 2 z) z$ be the product which we wish to be a square. As noted,

• $P(1,y,z)=\left(6yz\right)^2.$

A nice parametric solution is

• $P(2r,3r,r+1)=\left(12r(r+1)(2r^2-2r-1)\right)^2$

Also, for fixed $x,y$ or $x,z$ we are left with a Pell Equation.

The parametric solution arises from first setting $y=x+z-1$ to make the two quadratic factors equal $P(x,x+z-1,z)=(x^2-x-2z)^23x(x+2)z(x+z-1).$ There are many other parametric solutions such as

• $P(r^2,r^2+2,3)=\left(3r(r^2+2)^2(r^2-3)\right)^2$ and
• $P(3r^2-2,3r^2-2,1)=\left(3r(3r^2-4)(3r^2-2)(3r^2-1)\right)^2.$

Although none that I know of with the right-hand side the square of a fourth degree polynomial in $r$.

Most solutions to $P=\Box$ do not have $y=x+z-1$. There are $1442$ choices of $2 \le,x \le 2000,1 \le z \le 2000$ which make $P(x,x+z-1,z)$ a square. Of them $1000$ have $x=2(z-1).$

For fixed $x$ and $z \gt \frac{x^2-x}2$ we have $P(x,y,z)=Ay(y-B)$ for some constants. If $A \gt 0$ has square-free part $\alpha \gt 1$ then we are left with a potentially solvable Pell equations. We see that $\gcd(y,y-B)$ divides $B$ so we seek solutions $$\{y,y-B\}=\{m\alpha_1u^2,m\alpha_2v^2\}$$ where $m\cdot \gcd(\alpha_1,\alpha_2)$ divides $B$ and $\alpha_1\alpha_2=\alpha.$

For example $x,z=3,5$ reduces to having $P(3,y,5)=2\cdot30^2\cdot y\cdot(y-5)$ a square.

There are no solutions to $\{y-5,y\}=\{u^2,2v^2\}$ (in either order) but $\{y-5,y\}=\{5u^2,10v^2\}$ means considering $P(3,5w,5)=2\cdot150^2 \cdot (w-1)\cdot w.$ We need $(w-1)w=2\Box$ with familiar solutions $(w-1)w=1\cdot2,8\cdot 9,49 \cdot 50,288 \cdot 289\cdots.$

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$Y^2-5$ may never be twice a square, but $y(y-5)$ certainly can be. Which one do you mean? – Gerry Myerson Feb 8 '13 at 4:21
Badly said, but I simply meant that $y,y-5=Y^2,2T^2$ (in one order or the other) has no solutions however $y,y-5=5Y^2,10T^2$ does. – Aaron Meyerowitz Feb 8 '13 at 4:32