42 is the new 33

A most interesting article appeared in Quanta magazine recently titled: Sum-of-Three-Cubes Problem Solved for ‘Stubborn’ Number 33. The article begins:

Mathematicians long wondered whether it’s possible to express the number 33 as the sum of three cubes — that is, whether the equation 33 = x³+ y³+ z³ has a solution. They knew that 29 could be written as 3³ + 1³ + 1³, for instance, whereas 32 is not expressible as the sum of three integers each raised to the third power. But the case of 33 went unsolved for 64 years. Now, Andrew Booker, a mathematician at the University of Bristol, has finally cracked it: 

He discovered that 

(8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³ = 33.

Apparently he used a very efficient search algorithm but he still required the use of a supercomputer running for three weeks to come up with the solution. Reading the article I also learned that there are no integer solutions to the equation x³+ y³+ z³ = n if \(n\equiv 4 \) mod 9 or \(n \equiv 5\) mod 9. Thus 31 and 32 cannot be expressed as the sums of three cubes. Of course, I've no idea why this is so and I may investigate the reason at some point in the future.

Interestingly, the only number below 100, for which a representation as a sum of three cubes has not been found, is 42. This is a number that featured in my blog post about magic cubes. Between 101 and 1000, there are 11 other "stubborn" numbers for which a representation has not been found. These numbers are 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975. All of these numbers have the property in common that \(n\equiv 3 \) mod 9 or \(n \equiv 6\) mod 9.

The equation x³+ y³+ z³ = n is an example of a Diophantine equation (a polynomial equation whose unknown variables must take integer values). For mathematicians, "A major result would be to prove the conjecture that n = x³ + y³ + z³ has infinitely many solutions for every whole number \(n\), except those \(n\) that have a remainder of 4 or 5 after being divided by 9."

Figure 1: solutions for values of m
between 1 and 10

My source for the following information comes from here. For some such Diophantine equations, there are infinitely many solutions. For example, x³+ y³+ z³ = 1 has infinitely many solutions because of the identities:$$(1 + 9m^3)^3 + (9m^4)^3 + (-9m^4 - 3m)^3 = 1$$ $$(1 - 9m^3)^3 + (9m^4)^3 + (-9m^4 + 3m)^3 = 1$$By assigning various integer values to \(m\), the solutions unfold. Figure 1 shows an example using values of m between 1 and 10.

If we write the equation in the form:$$x^3 + y^3 + z^3 = t^3$$ and set \(t\) to be an integer then Ramanujan found that:$$x = 3n^2 + 5nm - 5m^2$$ $$y = 4n^2 - 4nm + 6m^2$$ $$z = 5n^2 - 5nm - 3m^2$$ $$t = 6n^2 - 4nm + 4m^2$$If we set \(m=1\) and \(n=2\), then \(t=20\) and \(t^3=8000\). Thus for the equation:$$x^3+y^3+z^3=8000$$there is the solution \(x=17\), \(y=14\) and \(z=7\) or $$17^3+14^3+7^3=8000$$While these results are interesting, they clearly only work for certain numbers and provide no help at all in solving a still outstanding problem like:$$x^3+y^3+z^3=42$$A specific curiosity that should be mentioned is:$$ 3^3 + 4^3 + 5^3 = 6^3$$Another is the smallest cube number that is the sum of three different positive cubes. This turns out to be 216 or 6^3 and itself the sum of \( -3^3+6^3+3^3 \). 729 or \( 9^3 \) is the next such number since \(8^3 + 6^3 + 1^1 = 729\). 729 is a perfect cube and a perfect square since \(27^2=729\).

Here is the excellent Numberphile video on YouTube in which Andrew Booker talks about his discovery:

UPDATE (7th September 2019): a solution for 42 has been found:

Solving Frobenius Equations and Computing Frobenius Numbers

A Frobenius equation is an equation of the form:$$a_1x_n + \ldots + a_nx_n=m$$where \( a_1, \dots, a_n \) are positive integers, \( m \) is an integer and the coordinates \(x_1, \dots, x_n \) of solutions are required to be non-negative integers.

The Frobenius number of \( a_1, \ldots, a_n \) is the largest \( m \) for which the Frobenius equation \(a_1x_n + \ldots + a_nx_n=m \) has no solutions.

These definitions were taken from the Wolfram Language & System Documentation Center Tutorial. A Frobenius equation and number are both seemingly abstract concepts but there are practical, concrete applications. In the tutorial, the following example is given: in how many ways can a total of 42 cents be created using 1, 5, 10, and 25 cent coins? The problem can be expressed as a Frobenius equation:$$x_1+5x_2+10x_3+25x_4=42$$WolframAlpha solves this using the command: FrobeniusSolve[{1, 5, 10, 25}, 42] and the following output is produced:

{{2, 0, 4, 0}, {2, 1, 1, 1}, {2, 2, 3, 0}, {2, 3, 0, 1}, {2, 4, 2, 0}, {2, 6, 1, 0}, {2, 8, 0, 0}, {7, 0, 1, 1}, {7, 1, 3, 0}, {7, 2, 0, 1}, {7, 3, 2, 0}, {7, 5, 1, 0}, {7, 7, 0, 0}, {12, 0, 3, 0}, {12, 1, 0, 1}, {12, 2, 2, 0}, {12, 4, 1, 0}, {12, 6, 0, 0}, {17, 0, 0, 1}, {17, 1, 2, 0}, {17, 3, 1, 0}, {17, 5, 0, 0}, {22, 0, 2, 0}, {22, 2, 1, 0}, {22, 4, 0, 0}, {27, 1, 1, 0}, {27, 3, 0, 0}, {32, 0, 1, 0}, {32, 2, 0, 0}, {37, 1, 0, 0}, {42, 0, 0, 0}}

The Frobenius number is found using FrobeniusNumber[{1, 5, 10, 25}] which yields in this case -1. This is presumably WolframAlpha's way of saying that for this situation there is no such number because one of the building blocks is 1 and that can be used to build any positive integer. Let's look at another Frobenius equation:$$12x_1+16x_2+20x_3+27x_4=123$$Here FrobeniusSolve[{12, 16, 20, 27}, 123] yields:

{{0, 1, 4, 1}, {0, 6, 0, 1}, {1, 4, 1, 1}, {2, 2, 2, 1}, {3, 0, 3, 1}, {4, 3, 0, 1}, {5, 1, 1, 1}, {8, 0, 0, 1}}

The Frobenius number in this case (found using FrobeniusNumber[{12, 16, 20, 27}]) is 89. The actual output is {}. It can be noted that FrobeniusSolve[{12, 16, 20, 27},73]) also produces {} as output but 73<89 and so 73 is not the Frobenius number for the equation. What this means is that the Frobenius equation \(12x_1+16x_2+20x_3+27x_4=m \) always has a solution provided \(m>89\).

Here is a link to the MacTutor archive page about Ferdinand Georg Frobenius. I haven't yet found how to solve Frobenius equations or calculate a Frobenius number in SageMath. The only reference I've found is the following (link) which is clearly not the same:

We can compute the Frobenius coordinates and go back and forth: 

sage: Partition([7,3,1]).frobenius_coordinates()
([6, 1], [2, 0]) 

sage: Partition(frobenius_coordinates=([6,1],[2,0]))
[7, 3, 1] 

sage: all(mu == Partition(frobenius_coordinates=mu.frobenius_coordinates())
....:     for n in range(30) for mu in Partitions(n))
True

2017: A New Year

As the new year begins, it seems appropriate to look at the mathematical character of the number that will identify it: 2017. As a start, this number is prime, in fact the 306th prime. It's nearest neighbours are 2011 and 2027. Thus there will not be another prime year for a decade. Working through the Online Encyclopaedia of Integer Sequences (OEIS), I was made aware initially that the number was linked to one circle and 23 ellipses, each with the major axis equal to the diameter of the circle and major/minor axes coincident with the x and y axes.

The equation all twenty four shapes have in common is \(ax^2+bxy+ cy^2 = 2017\).

In the case of the circle, \(a=1, b=0, c=1\) and so \(x^2 + y^2 = 2017\). The integer solutions are \(x=9\) and \(y=44\). This is the point A on the circle c in the diagram below.

For the ellipses, \(a=1, b=0\) and so \(x^2 + cy^2 = 2017\). The values of \(c\) for which there are integer solutions to \(x\) and \(y\) are 2, 3, 7, 14, 21, 24, 27, 31, 33, 42, 46, 56, 66, 81, 84, 87, 88, 93, 112, 232, 253, 462 and 1848. I've plotted the cases of \(c\)=2, 3, 7 and 14 in the diagram below. The associated points are (37, 18), (17, 24), (15, 16) and (1, 12).

 

For the equation \(ax^2+bxy+ cy^2 = 2017\), when \(a\) is not equal to 1 but \(b\) is still 0, the following \((a, b, c)\) values give integer solutions to \(x\) and \(y\): (4, 0, 9); (2, 0, 41); (5, 0, 17); (2, 0, 65); (2, 0, 95); (8, 0, 65). All these ellipses lie inside the circle \(x^2+y^2=2107\) and have been graphed below (with part of the surrounding circle visible):

If \(xy\) terms are allowed, then there is another whole series of ellipses of the form \(ax^2+bxy+cy^2 = 2017\) where the following values of (a, b, c) yield integer solutions for x and y: (1, 1, 1); (9, 6, 1849); (1, 26, 1); (1, 8, -8); (1, 10, 1); (2, 1, 3); (1, 1, 8); (1, 1, 22); (4, 1, 4); (4, -1, 4); (4, -3, 5); (2, -1, 10); (2, 1, 12); (2, -1, 12); (1, 20, 1); (1, 31, 1); (8, 8, 97); (37, 4, 37); (28, 12, 57); (57, 18, 193). A few of these I've plotted below (with the circle included for comparison):

2017 can also be written as the sum of three (not distinct) cubes, namely \(7^3+7^3+11^3\). Thus it seems that 2107 is unusual in that it can be linked in 2-D space to a large numbers of ellipses. Of course, ellipses are the orbits followed by celestial bodies within the solar system trapped by the gravitational forces of larger bodies such as the Sun and planets.

Consider one of the ellipses above, say \(4x^2+9y^2=2017\) and a general point \((x, y)\) situated on it. The only integer values of \(x\) and \(y\) that satisfy this equation are (±22, ±3) as shown below:

Diophantine Equations

Today I turned 24521 days old and it turns out that this number is a non-negative value of x in the solution (x, y) to the Diophantine equation: 

x^2+(x+16807)^2=y^2

The corresponding y value is 48055 so the solution is (24521, 48055). So I thought that this would be an appropriate time to include some information about Diophantine equations. According to Wikipedia:

The word Diophantine refers to the Hellenistic mathematician of the 3rd century, Diophantus of Alexandria, who made a study of such equations and was one of the first mathematicians to introduce symbolism into algebra. The mathematical study of Diophantine problems that Diophantus initiated is now called Diophantine analysis.

 The definition given in the same article is:

In mathematics, a Diophantine equation is a polynomial equation, usually in two or more unknowns, such that only the integer solutions are sought or studied (an integer solution is a solution such that all the unknowns take integer values). A linear Diophantine equation is an equation between two sums of monomials of degree zero or one. An exponential Diophantine equation is one in which exponents on terms can be unknowns.

Today's Diophantine equation is quadratic and it's equivalent to finding integer solutions to the right-angled triangle with hypotenuse y and arms x and x + 16807. As it turns out, the requisite triangle is 48055, 24521 and 41328. In general terms, finding all right triangles with integer side-lengths is equivalent to solving the Diophantine equation a^2 + b^2 = c^2, except in this case we know that b = a + 16807.

Diophantine analysis is a big topic to which I may return at a later date but that's enough for now.