User john r ramsden - MathOverflow

Answer by John R Ramsden for Reference request: an elementary proof of Brouwer fixed-point theorem.

2012-09-25T09:42:43Z

There is an interesting essay on Brouwer's Fixed Point theorem, including a contructive proof, at Kevin Brown's MathPages site http://www.mathpages.com/home/kmath262/kmath262.htm

Looking at the home page, I see he has now written a string of books. If his articles are any indication, these books are doubtless excellent and well worth buying.

Answer by John R Ramsden for Diophantine Equation with Polynomial Coefficients

2012-09-23T20:21:52Z

Denoting $s = 12 n^2 - 1$ and using $a^2 + b^2 + c^2 = r$ your final equation can be rearranged as $ (\frac{y}{z})^2 + r (\frac{x}{z})^2 + s c^2 = r s$.

Denoting $d, e = (\frac{x}{z})^2, (\frac{y}{z})^2 $ respectively, this becomes $(r - c^2) (s - d^2) = c^2 d^2 + e^2$, and a homogenized version of the latter can have integer solutions only if the primes dividing the LHS factors satisfy the usual mod 4 conditions that determine the existence of sum-of-two-squares representations.

edit: That gives necessary conditions but is presumably only a starting point, in view of the extra constraint that $c$ and $d$ are present on both sides of the equation.

Rational solutions of $x (y - z) y (z - x) z (x - y) = t^2$

2012-07-28T22:46:57Z

I am interested in finding every rational solution of $x (y - z) y (z - x) z (x - y) = t^2$ (expressed in homogenous form, to show its symmetry).

Among other approaches I am pursuing, it is clear that letting $t = x y z u$ (or a similar transform that flips all of x, y, z to the other side, each to the power 1, such as $t = x (y - z) u$) results in an equation which is quadratic in each of, x, y, z. So, holding two of these fixed, say x, y, and with u also fixed, one can in general leapfrog from one solution (x, y, z) to another (x, y, z') where z, z' are the two roots of the quadratic in z.

But naturally the next question is how many disjoint "3D lattices" of solutions formed in this way are needed to include every rational solution. Since u is fixed for every solution in any given lattice, I would imagine there must be an infinite number of these lattices, but that one can hop between them via the transformations that flip x, y, z to the right hand side and back again, if that makes sense.

Answer by John R Ramsden for Fiction books about mathematicians?

2012-07-15T10:17:38Z

According to Wikipedia, http://en.wikipedia.org/wiki/Giordano_Bruno, Giordano Bruno was a mathematician, among other things, and S J Parris has written three excellent novels (and counting ..) that feature Bruno as the main character as a detective and secret agent combined:

http://www.amazon.com/Heresy-S-J-Parris/dp/0385531281

http://www.amazon.co.uk/Prophecy-Giordano-Bruno-S-Parris/dp/0007317735

http://www.amazon.co.uk/Sacrilege-Giordano-Bruno-S-Parris/dp/000731776X

Numerical method for finding characteristics of a square wave

2012-03-24T09:21:09Z

I have a test rig which at short intervals samples the current drawn by a WiFi chip beaconing and sleeping between each beacon. The result is a good approximation to a square wave, with slight spiky irregularities or "whiskers" at the boundaries, but broadly speaking a series of plateaus (plateaux?!) of equal height.

I would like a simple numeric method to extract from the data the best fit "idealized" characteristics, mainly the average plateau height and width and sleeping gap between steps. I suppose deducing the period would be most of what is needed for a start.

This problem is probably only borderline on-topic here, if that, being near the interface of maths and computer algorithms. Also it is obvious up to a point how one might proceed, by dexterous use of averages. But there may be other situations where the wave form is not so simple, for example a sloping line instead of a plateau.

So I wondered if there might be a simple but more effective and robust approach than fiddling around with averages. although I probably wouldn't want anything too sophisticated such as Fast Fourier Transforms (if those might be appropriate). But I am open to and interested in any suggestions. So now's your chance to solve a real industrial problem ;-)

Answer by John R Ramsden for Not especially famous, long-open problems which anyone can understand

2012-06-23T09:04:04Z

Ramanujan's conjecture [*] If $2^x$ and $3^x$ are both rational (hereafter assumed) integers for some non-zero $x$ then $x$ is an integer.

[*] I think that is the accepted name for this problem. He certainly proved the weaker corresponding result with $2^x$, $3^x$, and $5^x$ all assumed to be integers.

Unlike some of the other fascinating conjectures already listed here, this one seems "obviously" true. Yet I gather little progress has been made on it. It must be hard to find a foothold, so to speak, or know where to start.

Another easily understood example is the Erdos-Straus Conjecture [ http://planetmath.org/ErdHosStrausConjecture.html ], that for every integer n > 1 there is at least one set of positive integers $x, y, z$ with $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{4}{n}$ The result is trivially true if negative integers are also allowed.

In this case, by contrast, it's easy(ish) to "almost" prove it, and with patience and ingenuity one can proceed (apparently) ever closer to a solution. But a few annoying special cases always seem to slip through the net!

One more example - I think a high school kid would have little difficulty understanding the ABC conjecture [ http://en.wikipedia.org/wiki/Abc_conjecture ], or following the simple proof of the corresponding result for polynomials [ http://en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem ]

Rank of $x (x^2 - 1) = c (c^2 - 1) y^2 $ over $\mathbb{Q}$ for given rational values of $c$

2012-06-17T18:11:26Z

Can anything be said in general about the rank etc over $\mathbb{Q}$ of the family of Weierstrass equations (in slightly non-standard form) $x (x^2 - 1) = c (c^2 - 1) y^2$ for various given rational values of $c$ ? I have a good reason for asking, so this isn't idle curiosity.

Naturally, it would be simplest if the Weierstrass equation has the same behaviour for each of these values of $c$, or least with a manageable (finite) amount of variation. Obviously there is always a solution $x, |y| = c, 1$; but that might be a trivial solution of a rank 0 case.

Dually automorphic varieties

2012-04-21T17:13:05Z

Is it possible to have a pair of varieties or schemes (well surfaces are what interest me), say $V$, $W$ and a pair of rational maps $F: V \rightarrow W$ and $G: W \rightarrow V$ such that $F \circ G$ and $G \circ F$ are automorphisms of $V$ and $W$ respectively but the latter are not birationally equivalent (all over $\mathbb{Q}$)?

Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c?

2012-05-19T08:14:41Z

I need this result for something else. It seems fairly hard, but I may be missing something obvious.

Just one non-trivial solution for any given $c$ would be fine (for my application).

Answer by John R Ramsden for Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c?

2012-05-19T16:04:04Z

Many thanks for your replies. I thought I'd cracked it, but after posting noticed a mistake at the end. However, I'll leave the following, as far as it goes, in case it suggests any alternative angles to others.

Firstly, note that the equation can be expressed as:

$\dfrac{x - 1}{z} \dfrac{x + 1}{z} \dfrac{y - 1}{z} \dfrac{y + 1}{z} = c$

So taking:

$X, Y, Z, T = \dfrac{x - 1}{z}, \dfrac{y + 1}{z}, \dfrac{x + 1}{z}, \dfrac{y - 1}{z}$

(which is obviously unirational, i.e. "reversible") we can express it as:

$X + Y = Z + T$

$X Y Z T = c$

Now (reusing the original x, y, z for convenience) take:

$X, Y, Z, T = \dfrac{c x}{d}, \dfrac{y}{d}, \dfrac{z}{d}, \dfrac{t}{d}$

Then the preceding pair becomes:

$c x + y = z + t$

$x y z t = d^4$

Now as a special case assume t = c, so the first of this pair gives:

$t = c = \dfrac{z - y}{x - 1}$

and the second then becomes:

$x y z (z - y) = (x - 1) d^4$

Finally, letting:

$x, y, z = p d, q d, r d$

we obtain:

$c = \dfrac{(r - q) d}{p d - 1} = \dfrac{d}{p q r}$

But this, although it looks tantalizingly simple, is the end of the line for the present attempt!

Answer by John R Ramsden for Omniscient bots gathering on $\mathbb{Z}^2$

2012-01-21T08:42:55Z

Looks easy if you aren't worried about the efficiency of the algorithm. If the bots can agree about which direction is bottom-left, they can start by drawing the smallest bounding rectangle containing them all, each assigning themselves a unique row-major order in this rectangle and mapping that to the row-major order in a target square disjoint from the bounding rectangle but in some predictable position relative to it.

Then the algorithm for i = 1, 2, n is simply for the i-th bot to shuffle towards its target position one step at a time while the remaining bots do nothing until this position has been reached.

Choosing the bots in row-major order, in other words imagining that the rectangle and the target square are each cut into rows stitched end-to-end as one long strip guarantees that no bot whose turn has started is enclosed by other bots and therefore unable to move to its target position.

Answer by John R Ramsden for Which Diophantine equations can be solved using continued fractions?

2012-02-25T11:58:47Z

I was about to mention H J S Smith's algorithm for finding integer solutions to $x^2 + y^2 = p$ for $p \equiv 1$ mod 4; but this is referred to in a related thread at http://mathoverflow.net/questions/49866/applications-of-finite-continued-fractions

(Apologies if that thread is easily found from this one; but I wouldn't have noticed it without doing a Google search, and perhaps some other readers are equally inexperienced in StackOverflow ways or unobservant!)

Also, what about higher-dimensional continued fractions, expressed as matrix recurrence relations? I seem to recall that these can be used to find rational solutions of equations involving some kinds of cubic forms.

Answer by John R Ramsden for Impossible Heronian Triangles (Ratio of 2 Sides)

2012-01-22T19:34:44Z

Starting with $Z(Z - m^2) = (Z - n^2) b^2$, to which Allan MacLeod's elliptic curve can be reduced by taking $W = (Z - n^2)b$, one can find a general parametrization of m, n as follows.

Letting $a, x = Z/b, m/b$ gives $a^2 - a b x^2 = a b - n^2$, in which then letting $n = a y$ gives $b (x^2 + 1) = a (y^2 + 1)$. This implies $x^2 + 1, y^2 + 1 = a z, b z$ for some rational $z$, and multiplying these gives after composition $(\frac{x y + 1}{z})^2 + (\frac{x - y}{z})^2 = a b$.

Letting $a, b = k A, k B$, where $A, B$ are coprime integers, the factor $k^2$ in the preceding equation can be absorbed into each square, and we can conclude that $A, B$ are each a sum of two squares, say $p^2 + q^2, r^2 + s^2$ resp, so that $B (x^2 + 1) = A (y^2 + 1)$ becomes $(p x + q)^2 + (p - q x)^2 = (r y + s)^2 + (r - s y)^2$

The latter has general solution as follows, for rational $u, v$ with $u^2 + v^2 = 1$ :

$p x + q, p - q x = u (r y + s) + v (r - s y), v (r y + s) - u (r - s y) w$

So that:

$ x = \frac{u (r y + s) + v (r - s y) - q}{p} = \frac{p - v (r y + s) - u (r - s y)}{q}$

which expresses $y$ and then $x$ rationally in terms of $u, v$ and $p, q, r, s$ (and the latter appear homogenously, so one of them is disposable i.e. can be assumed equal to 1).

edit: I should clarify that this isn't a rational parametrization of one elliptic curve, which of course is impossible if $m^2, n^2$ are non-zero and distinct. What it does is start with a supposed "symbolic" solution and express the roots parametrically in a way consistent with that solution. In other words it constructs a multi-dimensional pencil of all elliptic curves having the required form.

The variety $x_1 + x_2 + .. + x_n = 0$, $x_1 x_2 .. x_n = 1$ for n > 4

2012-01-14T07:35:49Z

For $n = 3$ the variety $x_1 + x_2 + .. + x_n = 0$, $x_1 x_2 .. x_n = 1$ is an elliptic curve, and for $n = 4$ it is rational [edit: or so I thought, before seeing the other replies].

What can be said, for example regarding rationality, for larger $n$ (for values of $x_i$ where the variety has no components of the same form with smaller $n$)? Is this variety of a standard type, such as toric or Calabi-Yau?

One might assume that it remains rational for $n > 4$; but, given that the degree of the product increases, that might well not be so.

Answer by John R Ramsden for The variety $x_1 + x_2 + .. + x_n = 0$, $x_1 x_2 .. x_n = 1$ for n > 4

2012-01-14T09:35:12Z

(expanded comment)

For $n = 3$ in $X Y (X + Y) + 1 = 0$ letting $Y = t X$ gives $t (t + 1) = - 1 / X^3 = x^3$ say. Then letting $y = 2 t + 1$ gives $4 x^3 + 1 = y^2$, which is a non-degenerate elliptic curve.

For $n = 4$, in $X + Y + Z + T = 0, X Y Z T = 1$ Letting $X Y, Z T = -u, -1/u$ resp satisfies the second and in the first gives $X Z (X + Z) = X/u + u Z$. Then letting $t = X/Z$ this becomes $t u (t + 1) Z^2 = t + u^2$, in which letting $t = u v, w = 1/Z$ gives $u v (u v + 1) = (u + v) w^2$.

I thought I had a parametrization of the latter; but having found a slip in my algebra, now I'm not so sure. Maybe it is K3.

Answer by John R Ramsden for Does this surface contain all perfect cuboids?

2011-08-27T20:28:46Z

Looks OK to me (the approach anyway - I didn't check the numerator calculation). Ruslan Sharipov also found an explicit equation for the perfect cuboid surface, in a recent ArXiv paper at http://arxiv.org/abs/1104.1716. His derivation was much more intricate than yours, but the result looks very similar!

This surface is known to be a so-called surface of general type [ http://en.wikipedia.org/wiki/Surface_of_general_type ] and thus has only a finite number of rational points.

Most reckon it has no non-trivial (i.e. with all non-zero) rational points, and with equations like this it tends to be "small or nothing". Various people over the last century or so have claimed proofs of this; but I think the problem is still generally agreed to be open.

It would be interesting to look at congruence conditions on a homogenized version of your equation or Sharipov's. Maybe you would find the high degree strongly limited the number of solutions modulo smallish primes such as 17 and 23, although with four variables kicking around (in the homogenized equation) there are a lot of combinations!

Answer by John R Ramsden for Are nontrivial integer solutions known for $x^3+y^3+z^3=3$?

2011-06-02T18:09:10Z

As $x^3 + y^3 = c$ for any given (suitable) c is an elliptic curve, perhaps a reasonable strategy would be to try various integers $f$, $g$ for which $c := f^3 - 3 g^3$ is small and establish the Mordell-Weil rank of the curve.

If this is ever positive (for values other than the known solutions the OP mentioned) then one would establish that there were other non-trivial rational solutions, even if these had still not been found.

Edit: Rereading the OP's post, I notice they are asking for integer solutions rather than rational solutions, and I recall now that there are rational parametrizations anyway. So perhaps this approach isn't very useful after all.

Answer by John R Ramsden for Parametrization of the intersection of an ellipsoid with a sphere

2011-05-30T17:09:54Z

You'll almost certainly find that sampling the parameters uniformly will not sample points uniformly on the intersection.

But a general parametrization, involving Jacobian elliptic functions, can be found inductively as follows (where for convenience I denote $ a_i = 1 / \lambda_i $).

Let $x_1 = 1 + p$ and, for $i > 1$, $x_i = p q_i $. Plugging these into the n-sphere gives either $p = 0$, which may or may not lead to a point in the intersection, or $p = -2 / ( 1 + q_2^2 + .. + q_n^2 )$.

Plugging the second equation for p into the $x_i$, and these into the ellipsoid, gives the remaining condition as follows, denoting $r^2 := q_2^2 + .. + q_n^2 $ :

$ a_1^2 (r^2 - 1)^2 + 4 a_2^2 q_2^2 + .. + 4 a_n^2 q_n^2 = (r^2 + 1)^2 $

Denoting $ s^2 := ((r^2 + 1)/2)^2 - a_1^2((r^2 - 1)/2)^2 $ and $ y_i := q_i / r$, these become:

$ y_2^2 + .. + y_n^2 = 1$

$ (a_2 \frac{r}{s} y_2)^2 + .. + (a_n \frac{r}{s} y_n)^2 = 1$

Noting that $r^2 = (\frac{r^2 + 1}{2})^2 - (\frac{r^2 - 1}{2})^2 $ identically, we see that $r$ and $s$ must satisfy:

$ \frac{r^2 - 1}{r^2 + 1} = sn(a_1, u) $

$ \frac{2 r}{r^2 + 1} = cn(a_1, u) $

$ \frac{2 s}{r^2 + 1} = dn(a_1, u) $

giving:

$ r = \frac{dn (1 + cn + sn)}{cn (1 + cn - sn)} $

$ s = \frac{1 + cn + sn}{1 + cn - sn} $

Finally, when you get down to:

$ z_{n-1}^2 + z_n^2 = 1$

$ (a_{n-1} z_{n-1})^2 + (a_n z_n)^2 = 1 $

you can treat this as a pair of linear equations in $z_{n-1}^2$ and $z_n^2$.

Regards

John R Ramsden

Looking up the Mordell-Weil rank and generator(s) of a Weierstrass Equation

2011-03-05T20:28:57Z

Is there a web site where one can look up a Weierstrass equation, by discriminant say, or coefficients of some readily derivable "standard" form, and find the rank of its solutions over Q neatly listed, along with a set of generators and torsion group?

After several web searches, the nearest I've come is the Cremona Tables. But they list by conductor, whatever that is, and there seems no obvious way for an amateur such as myself to translate the data on that site into a form usable for the above mentioned purpose.

Failing that, I'd be content with an eay to use Mathematica or Sage package to achieve this, with idiot-proof instructions.

Right now, I'm especially interested in the equation $ x^2 + x (x + 1)^2 = y^2 $, which obviously has some rational solutions, and I'm pretty sure has positive rank over Q.

But the answer for that specific equation, although useful at present, obviously won't help with others that may interest me in future - "Teach a Chinaman to fish" and all that ..

P.S. If the Cremona Tables can easily be used to look up equations such as the example above, I'd very much appreciate a simple walkthrough, using the above as an example, and I think others would also find this useful.

The surface $ x^2 y^2 + 1 = (x^2 + y^2) z^2 $

2011-01-02T13:49:29Z

Hi, I'm trying to find all rational points on the surface of the title, in connection with the Euler Brick (AKA Rational Box) problem.

This surface is equivalent to $ x^2 z^2 - 1 = (x^2 - z^2) y^2 $, and it is easy to find an addition formula for rational points by jogging the second into a form conformable with the equations defining Jacobian elliptic functions:

A rational point on the second form clearly implies rational $a, b, .. e$ (and conversely) satisfying the pair $ (a b)^2 - 1 = e (a b c)^2 $ and $ (a b)^2 - b^4 = e (a b d)^2 $ and dividing each by $ (a b)^2 $, one can express the variables as follows with modulus $ k = b^2 $ :

$ sn(k,u) = 1 / (a b) $

$ cn(k,u) = \sqrt e . c $

$ dn(k,u) = \sqrt e . d $

Because in the addition formulae for Jacobian elliptic functions, $cn$ and $dn$ arise only in products of pairs, and "cross products" $cn . dn$ occur only in the resulting $sn$, this means that a pair of rational solutions with $e = e_1$ and $e = e_2$ implies a rational solution with $E = e_1 e_2$ to the pair:

$ C^2 + E S^2 = 1 $

$ D^2 + b^4 E S^2 = 1 $

Treating this as a Diophantine pair in its own right, without reference to Elliptic functions, one can absorb $ S^2 $ into E, and plug the expression for $E$ given by the first into the second to obtain:

$ D^2 - b^4 C^2 = 1 - b^4 $

For any given $b$ this is a conic curve, with a rational point $ C, D = 1, 1 $ and hence has a rational parametrization over Q, in b^4 and some parameter t say.

Now I'm sure the original surface is not rational (I gather it is a K3 surface). But I wonder if the above can be reversed, to some extent, to at least give a procedure for finding all rational points on the surface from those of the pencil of conics parametrized by $b$.

Answer by John R Ramsden for Examples of common false beliefs in mathematics.

2010-12-18T19:50:37Z

Here's another howler some people commit: If m, n are integers such that m divides n^2 then m divides n.

It's true sometimes, for example if m is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers p, q, r with p squarefree such that $ m = p q^2 $ and $ n = p q r $

The usual counterexample is that 8 divides 4^2 but not 4 ;-)

Answer by John R Ramsden for Examples of common false beliefs in mathematics.

2010-12-18T12:34:58Z

The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$ (x + y) (x^2 + y^2) = z^2 $

On defining $ u = x/y $ and $ v = z/y $ one obtains $ y (u + 1) (u^2 + 1) = v^2 $, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $ x y (x y + 1) (x + y) = z^2 $ rational? I'm almost sure it isn't; but how can one be sure?

Answer by John R Ramsden for Analysis of a quadratic diophantine equation

2010-10-30T19:57:10Z

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

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

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$, we now investigate how to specialize this to integer solutions.

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible (given the other constraints of the problem).

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$ [*]

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd or both even..

Both odd leads to the same contradiction as Case 1 as it requires $A$ and $2 m / A$ both odd.

So they must be both even, which is the case iff $A \equiv m \mod(2)$ (provided that when $m$ is odd, $(m - 1) / B$ is even, in other words $B$ does not divide out the power of 2 dividing $m - 1$).

Furthermore from the form of $z$, $t$, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

Note that the above isn't an explicit integer solution. All I have done is reduce the problem to the pair [*], to which I have a draft solution that needs checking. But if anyone else wishes to nip in first with a solution to these then obviously feel free!

Comment by John R Ramsden

2013-05-11T20:07:37Z

That argument works, but with mod $p$, where $p$ is the smallest prime dividing $n$

Comment by John R Ramsden

2013-03-09T21:47:21Z

The fact that $x_3 + x_4$ is even doesn't mean it can be replaced by $2(x_3 + x_4)$, because $x_3$ and $x_4$ can both be odd. I suggest you have a loop for $x_3 + x_4$ equal n = 2, 4, .., for each iteration of which there is a set of possible pairs $x_3$, $x_4$, and for each such $n$ take the Cartesian product of this set of pairs with the set of pairs $x_1$, $x_2$ whose sum equals $(100 - n) / 2$.

Comment by John R Ramsden

2013-02-03T18:58:32Z

Can someone briefly explain why this was marked down? (N.B. I'm not disputing the reason one way or another, just curious)

Comment by John R Ramsden

2012-08-04T08:12:52Z

That link in JSE's comment is now math.u-bordeaux1.fr/~jcouveig/publi/volk.pdf

Comment by John R Ramsden

2012-07-29T09:02:24Z

Many thanks for the interesting and informative comments

Comment by John R Ramsden

2012-06-09T09:23:15Z

I was about to point out a paper on the Odd Perfect Number problem which appeared only yesterday on the ArXiV, but then noticed that Arnie is the author - DOH! In case anyone else is interested, the paper is at arxiv.org/abs/1206.1548.

Comment by John R Ramsden

2012-05-20T14:41:08Z

@Noam - That's fantastic, very many thanks! I remembered your xyz(x+y+z)=c paper, but didn't immediately make the connection because hitherto I've dealt (or tried to) with only individual c's and in that case xyz(x+y-z)=c is a different beast (although I imagine birationally equivalent to the former).

Comment by John R Ramsden

2012-05-20T14:40:46Z

@Daniel - I'll add an explanatory comment in a few days, or a reference to a paper, after I've checked the workings. Also, a short summary would sound preposterously implausible. But let's just say it relates to string theory, and Google returns no results on the phrase "Diophantine symmetry breaking" ;-)

Comment by John R Ramsden

2012-02-26T08:41:47Z

@Drike, because $n \mathbb Q = \mathbb Q$ for any non-zero rational $n$ then a bijection $f$ can be assumed to have all its coefficients in $\mathbb Z$. But I don't see how one can conclude they can be assumed to be in $ mathbb \N$. For example, what if there are two terms of even segree in $x$ and $y$ with coefficients of opposit sign?

Comment by John R Ramsden

2012-01-15T08:31:41Z

Very interesting Noam, and useful as an extended worked example, in what can otherwise seem a bewilderingly abstract topic! One reason I am interested in this variety is that for $N = 4$, with the extra condition $x_1 x_2 = a^2$ with rational $a$, it is birationally equivalent to a rational box with two face diagonals and body diagonal rational.

Comment by John R Ramsden

2012-01-14T13:31:11Z

DOH! It occurred to me that the code you mentioned must be the program given in your reply. I thought you were talking about Magma web calculator itself!

Comment by John R Ramsden

2012-01-14T13:27:18Z

Wow, yes the code would be very useful - I have plenty of other surfaces that need checking! Is there a link to a web page the code can be downloaded from? My email address, in case you need it, is jhnrmsdn@yahoo.co.uk

Comment by John R Ramsden

2011-05-14T11:01:08Z

Perhaps my comment wasn't quite as ridiculous as I/we suspected. See Definition 1, on page 8, of the recent ArXiv paper arxiv.org/abs/1105.2052 titled "Topological recursion and mirror curves". Their second "multiplicative" equation is slightly different to the one I quoted, involving as it does the exponents which they say represent charges. But aside from that, and a scaling which introduces a constant in the "additive" equation, their pair closely resembles mine!

Comment by John R Ramsden

2011-03-05T21:39:02Z

That's fantastic! Very many thanks for your prompt replies, Guys (William and GH so far). Just a quick related question. I couldn't find anywhere the "compact" coefficient notation, ($a_1$, $a_2$, .. , $a_5$) was defined, i.e. what term goes with what coefficients. I could probably guess, or infer from examples or by trial and error; but it would be handy to see this mentioned somewhere. Same goes for the "reduced" form ($a_1$, $a_2"). I imagine that must represent $x^3 + a_1 x + a_2 = y^2$; but it could be $4 x^3 + .. $

Comment by John R Ramsden

2010-10-31T20:46:35Z

Many thanks Gerry. It looks much clearer in TeX, and I'll use that in future replies. On checking the working I found that although broadly OK, it needed couple of minor corrections and an elaboration towards the end. (I hasten to add, this was unrelated to Gerry's cosmetic changes, but solely my sloppiness!) That's one of the great things about this Wiki system compared with usenet - One can go back and fix things.