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MaplePrimes Announcement

Maple Conference 2025: It's almost here!

by: KaskaKowalska 170 Products: Maple , Maple Learn

Yesterday at 1:53 PM

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There is still time to register for Maple Conference 2025, which takes place November 5-7, 2025.

The free registration includes access to three full days of presentations from Maplesoft product directors and developers, two distinguished keynote speakers, contributed talks by Maple users, and opportunities to network with fellow users, researchers, and Maplesoft staff.

The final day of the conference will feature three in-depth workshops presented by the R&D team. You'll get hands-on experience with creating professional documents in Maple, learn how to solve various differential equations more effectively using Maple's numerical solvers, and explore the power of the Maple programming language while solving interesting puzzles.

Access to the workshops is included with the free conference registration.

We hope to see you there!

Kaska Kowalska
Contributed Program Co-chair

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Featured Post

Rouben Rostamian

8581

Rupert's cube

It's possible to carve a hole through a unit cube, without splitting it into pieces, so that another unit cube can pass through that hole. This is know as the Prince Rupert problem and was first analyzed by John Wallis, a contemporary of Isaac Newton. Here's what the result looks like:

If your computer can play audio, have a look at Ruperts Cube with music!

Here is the worksheet that produced the cube and the animation: ruperts-cube.mw

October 22

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Featured Post

dharr

8612

Solving Diophantine quadratics in two variables

Maple's isolve can solve some but not all of the quadratic equations in two variables describing the conic sections. Here I show that by transforming the equations, Maple can then solve these missing cases.

I was recently surprised that isolve cannot solve the following simple Diophantine equation

which has the obvious (to a human) solution . This led me to think about the case of conic sections, which have the following general equation (= 0 implied), where I assume are integers.

The above equation has discriminant positive, indicating that is is a hyperbola.

>	$h_params := {a = 2, b = 4, c = 1, d = 0, e = 0, f = -7}; P__h := eval(P, h_params); disc := -4ac+b^2; eval(disc, h_params); plot__h := implicitplot(P__h, x = -10 .. 10, y = -10 .. 10, colour = red)$

Here's a parabola case (discriminant = 0) that isolve also has trouble with

>	$p_params := {a = 2, b = 4, c = 2, d = 1, e = 2, f = 7}; eval(disc, p_params); P__p := eval(P, p_params); {isolve(P__p)}; plot__h := implicitplot(P__p, x = -10 .. 10, y = -10 .. 10, colour = red)$

But this has at least one solution

>	$eval(P__p, {x = 7, y = -7})$

Maple seems to do better in the elliptic case (discriminant negative) and finds two solutions. Examination of the plot suggests there are no other solutions.

>	$e_params := {a = 2, b = 4, c = 3, d = 1, e = 2, f = -7}; eval(disc, e_params); P__e := eval(P, e_params); {isolve(P__e)}; plot__e := implicitplot(P__e, x = -10 .. 10, y = -10 .. 10, colour = red)$

I show that transformation of the general equation to the form , where D is the discriminant, allows Maple to solve the hyperbolic case, as well as the elliptic case it already knows how to solve; another transformation works for the parabolic case. Maple appears to be able to solve all (solvable) cases of the transformed equations, though this is not clear from the help page. The transformation is discussed in Bogdan Grechuk, Polynomial Diophantine Equations: A Systematic Approach, Springer, 2024, Sec. 3.1.7. doi: 10.1007/978-3-031-62949-5

A complete classification of the conics, including degenerate cases, is given in https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections. If the determinant, , of the following matrix is zero, we have a degenerate case.

A__Q := Matrix(3, 3, [a, (1/2)*b, (1/2)*d, (1/2)*b, c, (1/2)*e, (1/2)*d, (1/2)*e, f]); delta := LinearAlgebra:-Determinant(A__Q)

The case of is just the linear case, which Maple can solve, and is one case where . Other degenerate parabola cases are two coincident lines or two parallel lines. Degenerate hyperbolas are two intersecting lines, and degenerate ellipses are a single point. Maple can solve all these cases, e.g.,

>	$expand((2x+3y+1)(2x+3y)); {isolve(%)}; expand((2x+3y)(2x+3y)); {isolve(%)}; expand((x-2)*(y-3)); {isolve(%)}; x^2+y^2 = 0; {isolve(%)}$

${{x = -3*_Z1, y = 2*_Z1}, {x = -2-3*_Z1, y = 1+2*_Z1}}$

${{x = -3*_Z1, y = 2*_Z1}}$

${{x = _Z1, y = 3}}$

(The intersecting lines case above only finds one of the lines.) The transformation will consider only the case where at least one of or is non-zero. This misses hyperbolas with ; Maple seems to handle these bilinear equations, e.g.,

>	$bl_params := {a = 0, b = 2, c = 0, d = 1, e = 1, f = 2}; P__bl := eval(P, bl_params); {isolve(P__bl)}$

Transformation for non-zero discriminant
At least one of or must be non-zero; if necessary exchange and to ensure is non-zero.

We multiply by and change to new variables and .

>	$itr := {X = (-4ac+b^2)y+bd-2ae, Y = 2ax+b*y+d}; tr := solve(itr, {x, y})$

${X = (-4*a*c+b^2)*y+b*d-2*a*e, Y = 2*a*x+b*y+d}$

${x = (1/2)*(4*Y*a*c-Y*b^2+2*a*b*e-4*a*c*d+X*b)/((4*a*c-b^2)*a), y = -(2*a*e-b*d+X)/(4*a*c-b^2)}$

The transformed equation has the form or , where D is the discriminant.

>	$Q := collect(normal(eval(-4Pa(-4a*c+b^2), tr)), {X, Y}); m := -coeff(coeff(Q, X, 0), Y, 0)$

For positive discriminant D, this is a general Pell's equation, , which Maple knows how to solve. (The definition of Pell's equation requires that D is not a square, but Maple can also solve the simpler case where D is a square.) For the hyperbola above, we have an infinite number of solutions, parameterized by an arbitrary integer _Z1.

>	$Q__h := eval(Q, h_params); solXY__h := {isolve(Q__h)}$

${{X = -12*(1+2^(1/2))^(1+2*_Z1)-12*(1-2^(1/2))^(1+2*_Z1)-4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = -3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))-2*(1+2^(1/2))^(1+2*_Z1)-2*(1-2^(1/2))^(1+2*_Z1)}, {X = -12*(1+2^(1/2))^(1+2*_Z1)-12*(1-2^(1/2))^(1+2*_Z1)-4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = 3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))+2*(1+2^(1/2))^(1+2*_Z1)+2*(1-2^(1/2))^(1+2*_Z1)}, {X = -12*(1+2^(1/2))^(1+2*_Z1)-12*(1-2^(1/2))^(1+2*_Z1)+4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = -3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))+2*(1+2^(1/2))^(1+2*_Z1)+2*(1-2^(1/2))^(1+2*_Z1)}, {X = -12*(1+2^(1/2))^(1+2*_Z1)-12*(1-2^(1/2))^(1+2*_Z1)+4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = 3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))-2*(1+2^(1/2))^(1+2*_Z1)-2*(1-2^(1/2))^(1+2*_Z1)}, {X = 12*(1+2^(1/2))^(1+2*_Z1)+12*(1-2^(1/2))^(1+2*_Z1)-4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = -3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))+2*(1+2^(1/2))^(1+2*_Z1)+2*(1-2^(1/2))^(1+2*_Z1)}, {X = 12*(1+2^(1/2))^(1+2*_Z1)+12*(1-2^(1/2))^(1+2*_Z1)-4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = 3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))-2*(1+2^(1/2))^(1+2*_Z1)-2*(1-2^(1/2))^(1+2*_Z1)}, {X = 12*(1+2^(1/2))^(1+2*_Z1)+12*(1-2^(1/2))^(1+2*_Z1)+4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = -3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))-2*(1+2^(1/2))^(1+2*_Z1)-2*(1-2^(1/2))^(1+2*_Z1)}, {X = 12*(1+2^(1/2))^(1+2*_Z1)+12*(1-2^(1/2))^(1+2*_Z1)+4*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), Y = 3*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))+2*(1+2^(1/2))^(1+2*_Z1)+2*(1-2^(1/2))^(1+2*_Z1)}}$

Transforming back to the original coordinates

>	$sol__h := {seq(eval(eval(tr, h_params), solXY), `in`(solXY, solXY__h))}$

${{x = -2*(1+2^(1/2))^(1+2*_Z1)-2*(1-2^(1/2))^(1+2*_Z1)-(5/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = (3/2)*(1+2^(1/2))^(1+2*_Z1)+(3/2)*(1-2^(1/2))^(1+2*_Z1)+(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = -2*(1+2^(1/2))^(1+2*_Z1)-2*(1-2^(1/2))^(1+2*_Z1)+(5/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = (3/2)*(1+2^(1/2))^(1+2*_Z1)+(3/2)*(1-2^(1/2))^(1+2*_Z1)-(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = -(1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)-(1/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = (3/2)*(1+2^(1/2))^(1+2*_Z1)+(3/2)*(1-2^(1/2))^(1+2*_Z1)-(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = -(1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)+(1/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = (3/2)*(1+2^(1/2))^(1+2*_Z1)+(3/2)*(1-2^(1/2))^(1+2*_Z1)+(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = (1+2^(1/2))^(1+2*_Z1)+(1-2^(1/2))^(1+2*_Z1)-(1/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = -(3/2)*(1+2^(1/2))^(1+2*_Z1)-(3/2)*(1-2^(1/2))^(1+2*_Z1)-(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = (1+2^(1/2))^(1+2*_Z1)+(1-2^(1/2))^(1+2*_Z1)+(1/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = -(3/2)*(1+2^(1/2))^(1+2*_Z1)-(3/2)*(1-2^(1/2))^(1+2*_Z1)+(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = 2*(1+2^(1/2))^(1+2*_Z1)+2*(1-2^(1/2))^(1+2*_Z1)-(5/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = -(3/2)*(1+2^(1/2))^(1+2*_Z1)-(3/2)*(1-2^(1/2))^(1+2*_Z1)+(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}, {x = 2*(1+2^(1/2))^(1+2*_Z1)+2*(1-2^(1/2))^(1+2*_Z1)+(5/4)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1)), y = -(3/2)*(1+2^(1/2))^(1+2*_Z1)-(3/2)*(1-2^(1/2))^(1+2*_Z1)-(1/2)*2^(1/2)*((1+2^(1/2))^(1+2*_Z1)-(1-2^(1/2))^(1+2*_Z1))}}$

Evaluate for some _Z1 values, say _Z1=0 and _Z1=5.
It is evident from above that integer solutions in may transform to non-integer solutions in but that doesn't occur for these two cases

sol__h0 := evala(eval(sol__h, _Z1 = 0)); sol__h5 := evala(eval(sol__h, _Z1 = 5))

Check they are solutions to

For negative discriminant and negative , the equation has no solutions. In the classification scheme for the conics, the "imaginary ellipse" case (no real solutions) occurs when . For negative discriminant, we must have and the same sign, and this is the case of negative .

. In this case Maple returns NULL for both the untransformed and transformed case, which can mean no solutions or just that isolve couldn't find any.

>	$ie_params := {a = 3, b = 0, c = 5, d = 0, e = 0, f = 8}; P__ie := eval(P, ie_params); {isolve(P__ie)}; Q__ie := eval(Q, ie_params); {isolve(Q__ie)}$

For negative discriminant and positive or , the ellipse is real, and there are a finite number of solutions. Maple solves the untransformed and transformed equations.
Here we need to filter out non-integer solutions

>	$P__e; {isolve(P__e)}; Q__e := eval(Q, e_params); solXY__e := {isolve(Q__e)}; {seq(eval(eval(tr, e_params), solXY), `in`(solXY, solXY__e))}; sol__e := select(proc (q) options operator, arrow; eval(x::integer, q) and eval(y::integer, q) end proc, %)$

Transformation for zero discriminant
For the case of zero discriminant (parabola), we need a different transformation.

The transformed equation is of the form

We consider the parabolic example above, for which Maple finds no solutions without transformation.

>	$P__p; {isolve(P__p)}$

For the transformed problem, Maple finds an infinite number of solutions

>	$Q__p := eval(Q0, p_params); solXY__p := {isolve(Q__p)}; sol__p := {seq(eval(eval(tr0, p_params), solXY), `in`(solXY, solXY__p))}$

${{X = 1+8*_Z1, Y = -8*_Z1^2-2*_Z1-7}, {X = 3+8*_Z1, Y = -8*_Z1^2-6*_Z1-8}, {X = 5+8*_Z1, Y = -8*_Z1^2-10*_Z1-10}, {X = 7+8*_Z1, Y = -8*_Z1^2-14*_Z1-13}}$

${{x = 17/2+8*_Z1+8*_Z1^2, y = -8*_Z1^2-6*_Z1-8}, {x = 29/2+16*_Z1+8*_Z1^2, y = -8*_Z1^2-14*_Z1-13}, {x = 8*_Z1^2+4*_Z1+7, y = -8*_Z1^2-2*_Z1-7}, {x = 8*_Z1^2+12*_Z1+11, y = -8*_Z1^2-10*_Z1-10}}$

Two of the general solutions will not give integer solutions, so could be filtered out, but it is easier to filter after choosing some specific _Z1 values.

eval(sol__p, _Z1 = 0); sol__p0 := select(proc (q) options operator, arrow; eval(x::integer, q) and eval(y::integer, q) end proc, %); eval(sol__p, _Z1 = 5); sol__p5 := select(proc (q) options operator, arrow; eval(x::integer, q) and eval(y::integer, q) end proc, %)