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Fibonacci apparently found some solutions to this problem:

Find rational solutions of:

$$x+y+z+x^2=u^2$$

$$x+y+z+x^2+y^2=v^2$$

$$x+y+z+x^2+y^2+z^2=w^2$$

How would you find solutions to this using the mathematics available in Fibonaccis's time? (of course by this I mostly mean without using calculus, series, and modern maths. Also please exclude modular arithmetic notation if possible.) I was able to find little bits of information by adding and subtracting equations, such as $z^2=w^2-v^2$, $y^2=v^2-u^2$, and $y^2+z^2=w^2-u^2$, but I really do not know what to do. Thanks.

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Is your goal to find all solutions or some solutions ? –  Ewan Delanoy Jun 21 '13 at 6:07
    
@EwanDelanoy I don't know if there are a finite number of solutions, but if there were an infinite number, a proof of that would be nice. –  Ovi Jun 21 '13 at 6:13
    
A preliminary analysis: you are searching for rational solutions $(x,y,z)$ s.t. $y^2+z^2=w^2-u^2$. If $w^2-u^2<0$ there are none; if $w^2-u^2=0$ one has the solutions $(x,0,0)$ with rational $x$ s.t. $x+x^2=u^2=w^2$. Existence of rational solutions of the 2nd degree polynomial in $x$ depends on $w$. One has 2 rational solutions if $w=\frac{q^2-1}{4}$ for some real $q$, otherwise there are none. It remains to study the case $w^2-u^2>0$ –  Avitus Jun 21 '13 at 6:46
    
@Avitus From OP's last derived equation (essentially a Pythagorean quadruple), then $w^2-u^2$ will be greater than $0$, unless $y$ and $z$ are trivially $0$. –  alex.jordan Jun 21 '13 at 7:01
    
@alex.jordan I am not sure about this because I know nothing about $w$ and $u$, which I presume just to be fixed. If $y=z=0$, then there exists still space for non trivial rational solutions $(x,0,0)$. –  Avitus Jun 21 '13 at 7:04

1 Answer 1

up vote 2 down vote accepted

This is not a full answer in that not all solutions are described. But the discussion yields two infinite parametrized families of solutions. And the methods could possibly be studied longer to find more families, and possibly parametrize all solutions. As proof that this works before you invest in studying it, check that the solution it predicts at the end is valid.

There is a known trick for parametrizing rational points on quadratic surfaces, that I think extends to hypersurfaces.

Take the first equation. $(x,y,z,u)=(0,0,0,0)$ is a rational solution. Suppose $(X,Y,Z,U)$ is a different rational solution. Then the line connecting these two points in $4$-space is parametrized by $(x,y,z,u)=t(X,Y,Z,U)$. This line intersects the surface $x+y+z+x^2=u^2$ in precisely two places, since the intersection is found by solving for $t$ in $tX+tY+t^2Z^2=t^2U^2$. One solution is clearly given by $t=0$, and the other is given by $t=\frac{X+Y}{U^2-Z^2}$. Now since the line is parametrized by rational numbers, the intersection of this line with the plane $u=1$ has all rational coordinates: $(a,b,c,1)$. We can solve for $t$ to bring the fourth coordinate to $1$, and have $t=1/U$. So $$\begin{align}a&=X/U\\b&=Y/U\\c&=Z/U\end{align}$$

This establishes a map from rational points on $x+y+z+x^2=u^2$ to rational points on $u=1$. But this map is reversible. Take any rational triple $(a,b,c,1)$ and consider the line connecting this point to $(0,0,0,0)$. This line is parametrized by $(x,y,z,u)=s(a,b,c,1)$, and intersects $x+y+z+x^2=u^2$ in two places. To find both, we substitute: $as+bs+cs+a^2s^2=s^2$, and along with $s=0$, the other solution is with $s=\frac{a+b+c}{1-a^2}$.

So rational solutions to your first equation are given by $$\begin{align}x&=a\frac{a+b+c}{1-a^2}\\y&=b\frac{a+b+c}{1-a^2}\\z&=c\frac{a+b+c}{1-a^2}\\u&=\frac{a+b+c}{1-a^2}\end{align}$$ where $a,b,c$ are any triple of rationals excluding $a=\pm1$.

One infinite family of solutions to the system arises out of this if we take $b=c=0$: $(x,y,z,u,v,w)=\left(\frac{a^2}{1-a^2},0,0,\frac{a}{1-a^2},\pm\frac{a}{1-a^2},\pm\frac{a}{1-a^2}\right)$.


We can see what happens if we throw these into the next equation.

$$\frac{(a+b+c)^2}{1-a^2}+(a^2+b^2)\left(\frac{a+b+c}{1-a^2}\right)^2=v^2$$

Unfortunately this equation is degree 6:

$$(1+b^2)(a+b+c)^2=v^2(1-a^2)^2$$

So trying to proceed as before but this time in $(a,b,c,v)$-space won't work. Lines will not be guaranteed to intersect the surface at two points, which is a crucial element of what we did above.

If we are merely hunting families of solutions, and give up (for now) on finding all solutions, then it would help to have $1+b^2$ be a square. That is, to have $1+b^2=d^2$. We can do this by finding any primitive Pythagorean triple $(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$ and dividing by one of the left terms. Say we choose the second term, so that for integers $m$ and $n$, we have $$\begin{align}b&=\frac{m^2-n^2}{2mn}\\d&=\frac{m^2+n^2}{2mn}\end{align}$$ Now the earlier equation reduces to $$d(a+b+c)=v(1-a^2)$$

If we take $c=0$ (implying $z=0$) then we have another family of solutions to the system that arises out of this. Taking $m,n$ to be free nonzero integers, $a$ a free rational not equal to $1$, we have $$(x,y,z,u,v,w)=\left(a\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},\frac{m^2-n^2}{2mn}\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},0,\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},\frac{m^2+n^2}{2mn}\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},\pm\frac{m^2+n^2}{2mn}\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2}\right)$$

For example, $m=1$, $n=2$, $a=3/5$ yields $(-9/64, 45/256,0,-15/64, -75/256,75/256)$.


It seems reasonable that some other family could be worked out this way that does not demand $z=0$.

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