This story began yesterday, with this blog-post: https://robertlovespi.wordpress.com/2014/12/10/pythagorean-and-fermatian-triples-and-quadruples/ — but it hasn’t ended there. When discussing this with my wife (who, like myself, is also a teacher of mathematics) while writing that post, she speculated that more interesting things might happen — such as a “no solutions” situation, as is the case with Fermat’s Last Theorem — with a search for a Fermatian quadruple, if the exponent used were larger than three, the exponent I checked yesterday.
Tonight, therefore, I modified the program I used for the last post on this subject. Instead of searching for whole-number solutions to an + bn + cn = dn with n = 3, as I did yesterday, I looked for solutions to a4 + b4 + c4 = d4. As I did yesterday, I started with a search of all possibilities with numbers from 1 to 10, and was unsurprised when that quick, preliminary search yielded no solutions. I then ran the program again, but used it to search all possibilities using numbers for a, b, c, and d from 1 to 100. This took a while, for, with loops nested four deep, my computer had to check 1004 = 100 000 000 possibilities. The results are — tentatively — exciting, for this search, indeed, yielded no solutions, which is reminiscent of Fermat’s Last Theorem:
We are now wondering if Fermat’s Last Theorem can be generalized indefinitely. Andrew J. Wiles proved that an + bn = cn has no solutions if n > 2. I’ve now written a program, and checked, and know that an + bn + cn = dn has no solutions, with n = 4, for values of a, b, c, and d up to 100.
Could it be that an + bn + cn = dn has no solutions for any value of n > 3 — like Fermat’s Last Theorem, but with one more term added, and the exponent simply bumped up one place? If that is true, then, might it also be true that an + bn + cn + dn = en has no solutions if n > 4? Might it be possible to extend this idea indefinitely, so that, with an equation containing k terms added together, to equal a single term, there are no solutions if n > k?
I know this much, at this point:
- I can either find a counterexample, to disprove one of these conjectured “sequels” to Fermat’s Last Theorem, if that counterexample is of reasonable size, provided a “smallish” counterexample actually exists, or
- With the assistance of friends of mine whose ability with mathematics, and computer programming, exceeds my own, we can extend this search for counterexamples to much higher limits, or
- This set of conjectures is, in fact, true, in which case we will find no counterexamples — and, if that is the case, I’m going to need to find some major-league help for this problem, for, well, if true, it’s going to be one monster of a job to prove it is true, especially in the general, unlimited form.
- Finally, I know that the prospect of playing any role, whatsoever, in extending Fermat’s Last Theorem to new levels is tremendously exciting.
I’m looking forward to seeing where this goes.
[Update: I’d like to thank my friend Andrew for finding the answer to this puzzle for me. Counterexamples have indeed been found for the four-term and five-term cases, one of which is 26824404 + 153656394 + 187967604 = 206156734. For six terms or more, this remains an unsolved problem. For more information related to this, please visit https://en.wikipedia.org/wiki/Euler’s_sum_of_powers_conjecture.]
Those who have taught geometry, when teaching triangle congruence, go through a familiar pattern. SSS (side-side-side) triangle congruence is usually taught first, as a postulate, or axiom — a statement so obvious that it requires no proof (although demonstrations certainly do help students understand such statements, even if rigorous proof is not possible). Next, SAS (side-angle-side) and ASA (angle-side-angle) congruence are taught, and most textbooks also present them as postulates. AAS (angle-angle-side) congruence is different, however, for it need not be presented without proof, for it follows logically from ASA congruence, paired with the Triangle Sum Theorem. With such a proof, of course, AAS can be called a theorem — and one of the goals of geometricians is to keep the number of postulates as low as possible, for we dislike asking people to simply accept something, without proof.
At about this point in a geometry course, because the subject usually is taught to teenagers, some student, to an audience of giggling and/or snickering, will usually ask something like, “When are we going to learn about angle-side-side?”
The simple answer, of course, is that there’s no such thing, but there’s a much better reason for this than simple avoidance of an acronym which many teenagers, being teenagers, find amusing. When I’ve been asked this question (and, yes, it has come up, every time I have taught geometry), I accept it as a valid question — since, after all, it is — and then proceed to answer it. The first step is to announce that, for the sake of decorum, we’ll call it SSA (side-side-angle), rather than using a synonym for a donkey (in all caps, no less), by spelling the acronym in the other direction. Having set aside the silliness, we can then tackle the actual, valid question: why does SSA not work?
This actually is a question worth spending class time on, for it goes to the heart of what conjectures, theorems, proof, and disproof by counterexample actually mean. When I deal with SSA in class, I refer to it, first, as a conjecture: that two triangles can be shown to be congruent if they each contain two pairs of corresponding, congruent sides, and a pair of corresponding and congruent angles which are not included between the congruent sides, of either triangle. To turn a conjecture into a theorem requires rigorous proof, but, if a conjecture is false, only one counterexample is needed to disprove its validity. Having explained that, I provide this counterexample, to show why SSA does not work:
In this figure, A is at the center of the green circle. Since segments AB and AC are radii of the same circle, those two segments must be congruent to each other. Also, since congruence of segments is reflexive, segment AD must be congruent to itself — and, finally, because angle congruence is also reflexive, angle D must also be congruent to itself.
That’s two pairs of corresponding and congruent segments, plus a non-included pair of congruent and corresponding angles, in triangle ABD, as well as triangle ACD. If SSA congruence worked, therefore, we could use it to prove that triangle ABD and triangle ACD are congruent, when, clearly, they are not. Triangle ACD contains all the points inside triangle ABD, plus others found in isosceles triangle ABC, so triangles ABD and ACD are thereby shown to have different sizes — and, by this point, it has already been explained that two triangles are congruent if, and only if, they have the same size and shape. This single counterexample proves that SSA does not work.
Now, can this figure be modified, to produce an argument for a different type of triangle congruence? Yes, it can. All that is needed is to add the altitude to the base of isosceles triangle ABC, and name the foot of that altitude point E, thereby creating right triangle AED.
It turns out that, for right triangles only, SSA actually does work! The relevant parts of the right triangle, shown in red, are segment DA (congruent to itself, in any figure set up this way), segment AE (also congruent to itself), and the right angle AED (since all right angles are congruent to each other). However, as I’ve explained to students many times, we don’t call this SSA congruence, since SSA only works for right triangles. To call this form of triangle congruence SSA (forwards or backwards), when it only works for some triangles, would be confusing. We use, instead, terms that are specific to right triangles — and that’s how I introduce HL (hypotenuse-leg) congruence, which is what SSA congruence for right triangles is called, in order to avoid confusion. Only right triangles, of course, contain a hypotenuse.
This is simply one example of how to use a potentially-disruptive student question — also known as a teenager being silly — and turn it around, using it as an opportunity to teach something. Many other examples exist, of course, in multiple fields of learning.
In how many ways can different numbers of right angles appear in convex heptagons?
Heptagons do not have to have right angles at all, of course. If a heptagon has exactly one right angle, only one arrangement is possible: a right angle, and six oblique angles (“oblique” means non-right, so it includes both acute and obtuse angles).
With two right angles, there are three possibilities. In the first one shown, the right angles are consecutive. In the second, one oblique angle appears between the right angles. In the third, two oblique angles appear between the right angles. Increasing two oblique angles to three is simply a repeat of the third three-right angle heptagon, so this set stops with exactly three members.
With three right angles, I have found four possibilities: (1) all three right angles appear consecutively; (2) two right angles are consecutive, and one of them has one oblique angle between it and the third right angle; (3) two right angles are consecutive, and each of them has two oblique angles between it and the third right angle; and (4) None of the three right angles is consecutive. In this heptagon, the number of oblique angles which appear between the three different right-angle pairs are one, one, and two.
I have found no others, and, after searching to exhaustion, I do not think any other arrangement of right and oblique angles in a convex heptagon is possible. However, this is a conjecture, not a proof, and may, in fact, be incorrect. If you can provide proof that this listing of possibilities is complete, or a counterexample to show that it is not, please leave a comment with details.
Any triangle may be named triangle ABC. Each of its sides will contain exactly two points — called “tridpoints” — which divide that side into three segments of equal length. In triangle ABC above, the tridpoints are named in such a way that two of them, E and F, are encountered, in that order, if one moves from A to B. On the way from B to C, two more tridpoints are encountered: first F, and then G. Finally, going from C back to A, the last two tridpoints are found: first H, and then J. If a polygon is formed using those six tridpoints in alphabetical order (matching the order of their placement), that polygon is a convex hexagon, DEFGHJ. Another name for it is hexagon DJHGFE, which I mention only because Geometer’s Sketchpad called it that, in the picture above, when I asked it for the area of this hexagon, shown in green. The original triangle, ABC, includes both the yellow and green regions, and I asked Sketchpad for the area of this triangle, also, as well as the hexagon-to-triangle area ratio, which is shown above as the familiar “decimalized” version of the fraction 2/3.
A nice feature of Sketchpad is that you can do things like this — and then move points around, to see what effect that has on measured and calculated values. When I move points A, B, and/or C, the triangle and hexagon areas, of course, change. Their area ratio, however, remains at a decimal which is a rounded-off version of 2/3. It doesn’t change at all, no matter where A, B, and C are placed. Any triangle’s tridpoint-hexagon has an area exactly 2/3 that of the original triangle.
This is not yet a theorem — because what is written above is an explanation, not a proof. I’ve started working on a proof for this conjecture in my head, and will post it on this blog when/if I successfully complete it.
[Later edit — one of my readers provided a proof, so now it’s a theorem. For his proof, see the first comment on this post.]
There is something about the definitions of prime and perfect numbers that always struck me as rather odd. Prime numbers are those which no factors other than themselves, and one. Perfect numbers, on the other hand, equal the sum of all of their factors, excluding themselves, but including one. The first two examples of perfect numbers are 6 (which equals 1 + 2 + 3) and 28 (which equals 1 + 2 + 4 + 7 + 14). Perfect numbers are far more rare than primes.
The thing I find annoying is the exclusion of one, as a factor, from one of these definitions, but not the other. I therefore decided to give a name to a new type of number: one which equals the sum of its factors, excluding itself AND the number one. The first name I thought of, “exceptional numbers,” turns out to have already been taken, so I thought of another, and called these numbers “paraperfect numbers” instead.
Having done that, it was time to start searching for them. I have a reasonably fast mental calculator, but it didn’t take long to figure out that I wasn’t up to this task, so I wrote this program to search for paraperfect numbers:
It’s written in BASIC, an archaic computer language I learned in high school, and, as you can see, I am a horribly sloppy programmer. A better programmer would have written a program for the same purpose, but with only about half this length. Be that as it may, though, the program does work. I’ve had it running for a few minutes now.
It’s gotten past 22,000 — and has found no paraperfect numbers at all. This is not what I expected . . . and now I am wondering if any exist. Right now, of course, “no paraperfect numbers exist” is a mere conjecture. If I can prove it, it will be a theorem. However, I don’t know nearly enough about number theory to write such a proof.
I could use some help. If anyone does find a paraperfect number, please leave a comment on this post identifying your find. If anyone can prove — or simply explain to me — why there are no paraperfect numbers, if that is the case, please let me know that as well.
The program is still running, and has now passed 25,000 without a paraperfect-number find. I guess I’ll leave it running for a while. Any help with this puzzle would be much appreciated.
[Later: see the comments for the rest of the story on these elusive numbers.]
Roger Penrose is famous for many things, including the discovery of aperiodic tilings, the most familiar of which involves two types of rhombus:
I think I have made a minor discovery about this Penrose tiling, and that is that one can add regular pentagons to it, in varying levels of pentagon-density, as shown in the first image, without it losing its aperiodicity. (I created only the first image, not the second.) I have not, however, proven this, and doubt I will.
Is this conjecture provable? I think so, but I lack the ability to write such a proof myself.
As you can see, these circles intersect on the sides of the triangles. I did not expect that, nor have I proven it. I have moved the triangle around to check to see if this remained true, and it did pass this test. If I can figure out a proof for this, I’ll post it; if one exists already, please post a comment letting me know where to find it.
Later edit: I found out that these points of intersection are the altitude feet. Here’s a diagram showing the lines containing the altitudes, meeting at the orthocenter. These blue lines also contain the angle bisectors of the brown triangle defined by the altitude feet.