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Are there integral solutions to the equation $x^2+y^2-z^2=1$?

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The advanced approach is to rewrite this as $x^2+y^2= z^2+1$ and use unique factorization in $\mathbb Z[i]$. Assuming you disallow the obvious answers where $x$ or $y$ is $\pm 1$, you get that $x+yi=(a+bi)(c+di)$ and $z+i=(a+bi)(c-di)$ for some $a,b,c,d$. That yields a condition on $a,b,c,d$ and an explicit formulas form $x,y,z$ in terms of $a,b,c,d.$

Specifically, if $bc-da=1$ then $x=ac-bd, y=bc+ad, z=ac+bd$ is a solution to your equation. I think this gives all of them, but there is some concern about units.

This doesn't give us all ordered triples, since $y=bc+ad=bc-ad+2ad=1+2ad$, and hence this only gives us $y$ odd. However, it is clearly true that one of $x,y$ has to be odd, so this might sill give all un-ordered pairs $x,y$.

For example, $(a,b,c,d)=(7,2,4,1)$ then $x=26, y=15, z=30$ is the solution.

A more elementary solution is to assume $y=2n+1$ is odd (one of $x,y$ must be odd.) Then $x$ and $z$ have to have the same parity, and we get: $$4n(n+1)=y^2-1 = z^2-x^2=(z-x)(z+x)$$

Dividing by $4$: $$n(n+1)=\frac{z-x}{2}\frac{z+x}{2}$$

So we just need a factorization, $n(n+1)=UV$ with $U<V$. Then $z=V+U$ and $x=V-U$.

The most basic answer is $V=n(n+1)$ and $U=1$. Then $x=n(n+1)-1, y=2n+1, z=n(n+1)+1$.

The "obvious" answer, $U=n$, $V=n+1$ just gets you $x=1, y=2n+1, z=2n+1$.

Since $n(n+1)$ is even, we can write $U=2$, $V=\frac{n(n+1)}{2}=T_n$. Then $$x=T_n-2, y=2n+1, z=T_n+2$$

When $n>0$ we see that the number of distinct positive solutions $(x,2n+1,z)$ is $$\frac{\tau(n(n+1))}{2}=\frac{\tau(n)\tau(n+1)}2$$

If $UV=n(n+1)$ we can get $a,b,c,d$ from the first solution by defining: $$a=(V,n), b=(U,n+1), c=(V,n+1), d=(U,n)$$

then $$bc=n+1, ad=n, ac=V, bd=U$$

So $bc-ad=1$, $bc+ad=2n+1=y$, $V-U=ac-bd=x$ and $V+U=ac+bd=z$.

So any result from the second method is a result from the first method, which means the first method gets all positive solutions, too.

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  • $\begingroup$ Ha ha +1. Had the almost same realization :) $\endgroup$ – user17762 Apr 4 '13 at 21:26
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Recall the Brahmagupta formula $$(ad-bc)^2 + (ac+bd)^2 = (a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$$ Now find $a,b,c,d$ such that say $ad-bc = 1$. Then set $z = ac+bd$, $x=ac-bd$ and $y = ad+bc$ to get a solution.

For instance, one such one parameter family solution following the above procedure is $$x = 2t, y = 2t^2-1, z = 2t^2$$ where $a = d = t, b = (t+1), c = (t-1)$ since $t^2 - (t^2-1) = 1$.

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Yes: $(1,1,1)$. Are you after all of them?

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  • $\begingroup$ (7,4,8) is another solution I found. Is there a parametric equation that can generate all solutions? $\endgroup$ – fermats Apr 4 '13 at 21:02
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the equation:

$X^2+Y^2=Z^2-1$

Solutions can be written using the solutions of Pell's equation: $p^2-2k(k-1)s^2=1$

$k$ - given by us.

Solutions hav e form:

$X=2kps-2(k-1)s^2$

$Y=2(k-1)ps+2ks^2$

$Z=p^2+2(k^2-k+1)s^2$

And more:

$X=2p^2-2(3k-2)ps+2(2k-1)(k-1)s^2$

$Y=2p^2-2(3k-1)ps+2k(2k-1)s^2$

$Z=3p^2-4(2k-1)ps+2(3k^2-3k+1)s^2$

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We can take the equation to $x^2 + y^2 = 1 + z^2$ so if we pick a $z$ then we just need to find all possible ways of expressing $z^2 + 1$ as a sum of two squares (as noted in the comments we always have one way: $z^2 + 1$). This is a relatively well known problem and there will be multiple possible solutions for $x$ and $y$, there is another question on this site about efficiently finding the solutions of this should you wish to do so.

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  • $\begingroup$ Um, $z^2+1$ is expressed as the sum of two squares. What you want is whether there is more than one way to express $z^2+1$ as the sum of two squares, which is essentially whether $z^2+1$ has more than one odd prime factor. $\endgroup$ – Thomas Andrews May 25 '14 at 2:35
  • $\begingroup$ @ThomasAndrews yes, that was unclear. Thanks! $\endgroup$ – Alex J Best May 25 '14 at 10:12
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Of course for the equation $X^2+Y^2=Z^2+t$

There is a particular solution:

$X=1\pm{b}$

$Y=\frac{(b^2-t\pm{2b})}{2}$

$Z=\frac{(b^2+2-t\pm{2b})}{2}$

But interessuet is another solution:

$X^2+Y^2=Z^2+1$

If you use the solution of Pell's equation: $p^2-2s^2=\pm1$

Making formula has the form:

$X=2s(p+s)L+p^2+2ps+2s^2=aL+c$

$Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$

$Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$

number $L$ and any given us.

The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple. $a^2+b^2=c^2$

This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$

Allows you to find Pythagorean triples with a given difference.

$a=2s(p+s)$

$b=p(p+2s)$

$c=p^2+2ps+2s^2$

$b-a=\pm{k}$

Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.

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You can use the isomorphism T between SO(2,1) and SL(2) which is heavily used in physics. What it boils down to is that for every unimodular 2x2 matrix there is a 3x3 matrix of hyperbolic motions that has 3 unit (in Lorentzian sense) vectors for its columns.

So in those columns you not only get 2 solutions to your problem but also a solution to

$x^2+y^2-z^2=-1$

Here it is in Mathematica code:

T[{{a_, b_}, {c_, d_}}] := {
{(a^2 + b^2 - c^2 - d^2)/2, (a^2 - b^2 - c^2 + d^2)/2, -ab + cd},
{-ac - bd, bd - ac,ad + bc},
{(a^2 + b^2 + c^2 + d^2)/2, (a^2 - b^2 + c^2 - d^2)/2, -ab - cd}
}

So when you take for instance modular matrix m={{1,4},{2,9}}, you get the following columns:

u={-34,-38,51} v={31,34,-46} w={14,17,-22}

The first solves the "-1" equation, the second and third solve yours.

(https://arxiv.org/pdf/1508.00920.pdf)

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If $s$ and $t$ are any integers, then $x=2st+t+1$, $y=2s(st+t+1)+1$, $z=2s(st+t+1)+t+1$ satisfy $x^2+y^2-z^2=-1$. This is a simple way to get a 2-parameter family of solutions (without having to solve auxilary equations such as Pell's equation). I stumbled across this solution by considering a problem in finite geometry.

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