An Interesting Sequence

Here's an interesting question that was posed by PUZZLE A DAY on 30th October 2024: What number comes next in this sequence and why? 1, 7, 11, 27, ...

Mathematically, there doesn't seem to be any pattern and I needed the clue provided to proceed any further. Here is the clue:

Clue: Say the number aloud.

Immediately we know that the sequence is about how the number sounds and not about anything to do with the numbers per se. When we sound the numbers out, we have:

• one ... single syllable
• seven ... two syllables
• eleven ... three syllables
• twenty seven ... four syllables

Clearly we are looking for the next number that has five syllables and that numbers happens to be 77 or seventy seven. So we now have 1, 7, 11, 27, 77, ...

That's where the puzzle ends but the question then arises as to the next number in the sequence. This will take us into the hundreds and we need to decide whether to sound out the "and" or not. For example, 101 can be read as "one hundred and one" or "one hundred one". Let's go with the latter and search for the next number. 

As far as I can determine this must be 107 or one hundred seven that has five syllables. After that it would be 111 or one hundred eleven with six syllables. So now we have: 1, 7, 11, 27, 77, 107, 111. 

After this the next number with seven syllables is 127 or one hundred twenty seven. For eight syllables, it would 177 or one hundred seventy seven. Let's stop there and then list what we have so far is 1, 7, 11, 27, 77, 107, 111, 127, 177, ...

The puzzle's focus on the sound of the number rather than the number itself reminds of aban, eban, iban, oban and uban numbers. See my blog post Iban Numbers. It's also reminscent of the Look and Say Sequence. Finally, the Mathematics section of PUZZLE A DAY can be found here.

Multipurpose Algorithm using SageMath

Given the recent problems with Numbers Aplenty (see blog post Trouble with Numbers Aplenty), I've continued developing a SageMath program that will generate information about a number in a manner similar to Numbers Aplenty. At the moment, for any number that is input, the following output is generated:

• factorisation of the number or declaration that it is prime
• number of divisors
• list of divisors
• sum of divisors and factorisation of the sum
• sum of proper divisors and factorisation of the sum
• determination of whether a number is deficient, perfect or abundant
• if abundant, determination of whether the number is pseudoperfect or weird
• if abundant, determination of whether the number is a Zumkeller number
• if abundant, determination of whether the number is practical
• determination of whether number is Duffinian
• totient and its factorisation
• cototient and its factorisation
• determination of whether number is cyclic
• sum of squares of digits and factorisation of the sum
• sum of cubes of digits and factorisation of the sum
• determination of whether the number is admirable
• determination of whether the number is an interprime number
• determination of whether the number is a Bogota number
• determination of whether the number is a Curzon number
• determination of whether the number is a plaindrome in any base from 2 to 16
• determination of whether the number is metadrome in any base from 2 to 16
• determination of whether the number is a nialpdrome in any base from 2 to 16
• determination of whether the number is a katadrome in any base from 2 to 16
• determination of whether the number is a xenodrome in any base from 2 to 16
• determination of whether the number is an additive Fibonacci-like number of the second type in any base from 2 to 16
• Gray Code equivalent of the number and its factorisation
• absolute difference between the number and its Gray Code and factorisation of this difference
• determination of whether the number is an Ulam number and, if so, display the two numbers that add together to form it
• binary complement of the number and its factorisation
• absolute difference between the number and its binary complement and factorisation of this difference
• determination of whether the number is energetic
• number expressed as a sum of two squares if possible
• number expressed as a sum of two cubes if possible
• sum of digits (SOD) and product of digits (POD)
• number + SOD and factorisation of this sum
• number - SOD and factorisation of this sum
• number + POD and factorisation of this sum
• number - POD and factorisation of this sum
• sequence generated by repetition of Number + SOD - POD
• sequence generated by repetition of Number + SOD - POD but with zeros ignored
• Collatz trajectory of the number
• aliquot sequence determined up to a maximum of 101 steps
• antidivisors of the number
• arithmetic derivative of the number
• determination as to whether number is a D-number
• determination as to whether number is a de Polignac number
• sequence produced by Maximum-Minimum recursive algorithm
• Minimum Goldbach Decomposition
• sequence generated by Reverse and Add 
• Home Prime calculated to a maximum of 50 steps
• Additive Persistence
• Multiplicative Persistence taking zeros into account
• Multiplicative Persistence NOT taking zeros into account
• sequence generated by repetition of Number + Sum of Odd Digits - Sum of Even Digits
• Circulant Matrix
• Determinant of Circulant Matrix and its factorisation
• absolute difference between the numbers and the determinant and factorisation of this difference

I'll keep adding to this program over time using my Jupyter notebook. The program will run in SageMathCell although the creation of a Permalink in not possible. Here's a link to a Google Doc bookmark however:

https://docs.google.com/document/d/1xwJ2A0dtWN_i3e3ns2YJzmOGyeCaQ-COIwBBS74LFFQ/edit?tab=t.0#bookmark=id.o38tcw5ek6w7

The code located at this bookmark can be copied into SageMathCell and run.

POSTSCRIPT: 31st October 2024

The day after I made this post the Numbers A Plenty website was down again so all the more reason to keep refining my SageMath algorithm.

Number + SOD - POD Sequences

In this post, I will be using SOD as meaning the Sum Of Digits of a number and POD as meaning the Product Of Digits of a number. Consider a number like 27602 that is the number associated with my diurnal age today. What happens if I repeatedly add the sum of its digits and subtract the product of its digits? Because it contains the digit zero, its POD is initially 0 but this soon changes as I repeat the process. Here is the 27602's trajectory that consists of 26 steps (permalink):

276002, 276019, 276044, 276067, 276095, 276124, 275474, 267663, 258621, 257685, 240918, 240942, 240963, 240987, 241017, 241032, 241044, 241059, 241080, 241095, 241116, 241083, 241101, 241110, 241119, 241065, 241083

As can be seen, a loop is reached. What about 27610? Its trajectory consists of 27 steps (permalink):

27610, 27626, 26641, 26372, 25888, 20799, 20826, 20844, 20862, 20880, 20898, 20925, 20943, 20961, 20979, 21006, 21015, 21024, 21033, 21042, 21051, 21060, 21069, 21087, 21105, 21114, 21115

There is no loop here so why does the sequence terminate with 21115. The answer is that this number has a SOD (2 + 1 + 1 + 1 + 5 = 10) equal to its POD (2 x 1 x 1 x 1 x 5 = 10). So for every number, the corresponding sequence will have a finite number of steps ending in either a loop or a number with SOD = POD. 

What's of interest now are the progressive record lengths of the sequences as we consider larger and larger numbers. Table 1 shows the record lengths up to 110,000. 

Around 10,000 it can be seen that there is a big jump from 39 (9541) to 85 (9980) and again aroud 100,000 there is an even bigger jump from 98 (98907) to 631 (99970). If I were to continue there would be another big jump around one million. The sequence of numbers with record lengths is:

1, 10, 400, 417, 432, 482, 730, 2200, 2217, 2232, 2282, 2800, 2903, 3610, 4601, 5177, 6821, 7248, 9380, 9541, 9980, 9990, 10002, 98907, 99970, 99980, 99990, 100008

This sequence is not listed in the OEIS and I certainly won't be proposing it for inclusion. Variations on this theme are of course possible. For example, the product of digits could exclude the digit 0. This digit makes no difference to the sum of a number's digits but it has the effect of always sending the product of its digits to zero.

This makes a big difference to the sequence of record lengths and we no longer get those big jumps around 10,000, 100,000 and 1,000,000. Firstly though, let's go back to the original number 27601 and see how its trajectory is affected. The new trajectory is:

27601, 27533, 26923, 26297, 24811, 24763, 23777, 21745, 21484, 21247, 21151

The initial product of digits is now not zero but 2 x 7 x 6 x 1 = 84. The sum of its digits is still 16 but the SOD - POD now becomes 16 - 84 = -68 and so the next number in the sequence is 27533. The sequence ends with 21151 which has its SOD = POD.

The record lengths are shown in Table 2.

Table 2

The sequence of numbers with record lengths with zero not counting in the POD is now as follows:

1, 11, 100, 810, 1918, 1931, 2614, 2831, 2905, 3149, 3226, 3638, 3943, 4116, 4228, 4290, 4543, 6242, 7504, 7600, 7730, 8152, 8405, 9714, 19911, 23191, 23318, 23470, 25364, 27001, 29270, 44573, 45552, 46680, 47163, 47730, 49434, 54181, 54641, 55418, 56100, 57135, 71620, 73191, 73302, 74620, 75210, 78543, 81200, 82452, 83292, 87294, 88803, 95900, 97630, 98180, 100000, 100002

On the subject of numbers from 10 up to 100,000 in which SOD = POD, it is only these combination of digits that satisfy (permalink):

[2, 2], [1, 2, 3], [1, 1, 2, 4], [1, 1, 1, 2, 5], [1, 1, 1, 3, 3], [1, 1, 2, 2, 2]

There are 68 numbers in the range from 1 to 100,000 that satisfy SOD = POD and they are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 123, 132, 213, 231, 312, 321, 1124, 1142, 1214, 1241, 1412, 1421, 2114, 2141, 2411, 4112, 4121, 4211, 11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512, 11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311, 15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112, 22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211, 52111

However, if we don't include zero in the product of digits then there are 164 numbers that qualify because we can insert zeros into all the above numbers with impunity. Of course this means that some numbers will exceed 100,000 and so will not be included in the list below (permalink).

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 22, 30, 40, 50, 60, 70, 80, 90, 123, 132, 202, 213, 220, 231, 312, 321, 1023, 1032, 1124, 1142, 1203, 1214, 1230, 1241, 1302, 1320, 1412, 1421, 2013, 2020, 2031, 2103, 2114, 2130, 2141, 2301, 2310, 2411, 3012, 3021, 3102, 3120, 3201, 3210, 4112, 4121, 4211, 10124, 10142, 10203, 10214, 10230, 10241, 10302, 10320, 10412, 10421, 11024, 11042, 11125, 11133, 11152, 11204, 11215, 11222, 11240, 11251, 11313, 11331, 11402, 11420, 11512, 11521, 12014, 12030, 12041, 12104, 12115, 12122, 12140, 12151, 12212, 12221, 12401, 12410, 12511, 13020, 13113, 13131, 13311, 14012, 14021, 14102, 14120, 14201, 14210, 15112, 15121, 15211, 20103, 20114, 20130, 20141, 20301, 20310, 20411, 21014, 21030, 21041, 21104, 21115, 21122, 21140, 21151, 21212, 21221, 21401, 21410, 21511, 22112, 22121, 22211, 23010, 24011, 24101, 24110, 25111, 30102, 30120, 30201, 30210, 31020, 31113, 31131, 31311, 32010, 33111, 40112, 40121, 40211, 41012, 41021, 41102, 41120, 41201, 41210, 42011, 42101, 42110, 51112, 51121, 51211, 52111

If we try reversing the order of the subtraction so that we have number + POD - SOD, the POD soon causes the sequences for many numbers to increase without bound. In the case of POD with zeros counted, the number 23 was the first to exceed a sequence length of 700. When zeros were not counted, 516 was the first number to exceed 700. 

Xenodromes

Ben 10 Ultimate Alien: Xenodrome is a fighting
mobile game that was released on mobile devices.

This post has nothing to do with the above mentioned game instead I noticed that the number associated with my diurnal age today, 27594, has no repeating digits and I wondered if there was a word to describe such a number. Well, the OEIS uses the term "xenodrome" and, in base 10, there are 8,877,691 of them with the first being 0 and the last being 9,876,543,210. These numbers form OEIS A010784. Numbers of this sort are not listed in Numbers Aplenty. In the range up to 40,000, there are 14,346 xenodromes.

With so many numbers, it's best to apply some sort of sieve and one that comes to be mind is the metadrome, a number in which the digits are in strictly increasing order. If we look at numbers that are both xenodromes and metadromes in the range up to 40,000, we find that there are only 375 of them:

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89, 123, 124, 125, 126, 127, 128, 129, 134, 135, 136, 137, 138, 139, 145, 146, 147, 148, 149, 156, 157, 158, 159, 167, 168, 169, 178, 179, 189, 234, 235, 236, 237, 238, 239, 245, 246, 247, 248, 249, 256, 257, 258, 259, 267, 268, 269, 278, 279, 289, 345, 346, 347, 348, 349, 356, 357, 358, 359, 367, 368, 369, 378, 379, 389, 456, 457, 458, 459, 467, 468, 469, 478, 479, 489, 567, 568, 569, 578, 579, 589, 678, 679, 689, 789, 1234, 1235, 1236, 1237, 1238, 1239, 1245, 1246, 1247, 1248, 1249, 1256, 1257, 1258, 1259, 1267, 1268, 1269, 1278, 1279, 1289, 1345, 1346, 1347, 1348, 1349, 1356, 1357, 1358, 1359, 1367, 1368, 1369, 1378, 1379, 1389, 1456, 1457, 1458, 1459, 1467, 1468, 1469, 1478, 1479, 1489, 1567, 1568, 1569, 1578, 1579, 1589, 1678, 1679, 1689, 1789, 2345, 2346, 2347, 2348, 2349, 2356, 2357, 2358, 2359, 2367, 2368, 2369, 2378, 2379, 2389, 2456, 2457, 2458, 2459, 2467, 2468, 2469, 2478, 2479, 2489, 2567, 2568, 2569, 2578, 2579, 2589, 2678, 2679, 2689, 2789, 3456, 3457, 3458, 3459, 3467, 3468, 3469, 3478, 3479, 3489, 3567, 3568, 3569, 3578, 3579, 3589, 3678, 3679, 3689, 3789, 4567, 4568, 4569, 4578, 4579, 4589, 4678, 4679, 4689, 4789, 5678, 5679, 5689, 5789, 6789, 12345, 12346, 12347, 12348, 12349, 12356, 12357, 12358, 12359, 12367, 12368, 12369, 12378, 12379, 12389, 12456, 12457, 12458, 12459, 12467, 12468, 12469, 12478, 12479, 12489, 12567, 12568, 12569, 12578, 12579, 12589, 12678, 12679, 12689, 12789, 13456, 13457, 13458, 13459, 13467, 13468, 13469, 13478, 13479, 13489, 13567, 13568, 13569, 13578, 13579, 13589, 13678, 13679, 13689, 13789, 14567, 14568, 14569, 14578, 14579, 14589, 14678, 14679, 14689, 14789, 15678, 15679, 15689, 15789, 16789, 23456, 23457, 23458, 23459, 23467, 23468, 23469, 23478, 23479, 23489, 23567, 23568, 23569, 23578, 23579, 23589, 23678, 23679, 23689, 23789, 24567, 24568, 24569, 24578, 24579, 24589, 24678, 24679, 24689, 24789, 25678, 25679, 25689, 25789, 26789, 34567, 34568, 34569, 34578, 34579, 34589, 34678, 34679, 34689, 34789, 35678, 35679, 35689, 35789, 36789

If we were to consider katadromes instead of metadromes, then in the range up to 40,000 the highest number can only be 9876 so that's a little too restrictive. Katadromes are numbers in which the digits are in strictly decreasing order. See blog posts Metadromes and Katadromes.

Of course numbers can be xenodromes in other bases and 27594 serves as a good example because not only is it a xenodrome in base 10 but also in bases 8, 9, 11 and 12 and others as well no doubt:$$ \begin{align} 27594_{10} &= 41760_{\, 9} \\&= 65712_{\, 8} \\&= 19806_{11} \\&= 13b76_{12} \end{align}$$

An Interesting Prime

Being born on the 3rd April 1949, my date of birth is often represented as 3 - 4 - 49. These numbers when concatenated form the prime number 3449. I was reminded of this number because of the factorisation of the number associated with my diurnal age today, 27592.$$27592 = 2^3 \times 3449 = 8 \times 3449$$So today my life can be divided into exactly eight equal parts, each of them 3449 days long which is about 9.44 years. The previous multiple$$7 \times 3449 = 24143$$occurred on May 10th 2015 when I was still working at the Shanghai Singapore International School. The next multiple$$9 \times 3449 = 31041$$ will fall on March 29th 2034, shortly before my 84th birthday (if I make it that far).

3449 forms the initial prime of a Cunningham chain of the first type with length exactly 3 and so: $$ 3449 \rightarrow 2 \times 3449 + 1 = 6889 \text{ (prime)} \\ 6889 \rightarrow 2 \times 6889 +1 = 13799 \text{ (prime)}$$Primes with this property form OEIS A059762. Another prime-related property of 3449 qualifies it for membership in OEIS A088483:

A088483: primes \( \textit{p} \) such that \(p^2+p-1\) and \(p^2+p+1\) are twin primes.

For 3449, the twin primes are \(11899049\) and \(11899051\). 

3449 is also a home prime with a homeliness of 3 because:$$ \begin{align} 611 &= 13 \times 47 \rightarrow 1347\\ 1347 &= 3 \times 449 \rightarrow 3449 \end{align}$$Not all primes are home primes of course. Take 613 as an example of a prime that is not a home prime because it cannot be formed by the concatenation of the prime factors of any number (the prime factors need to be concatenated in ascending order).

3449 is also a member of OEIS A153116:

A153116: primes \(p\) such that \(p^2 +12\) and \(p^2-12\) are also primes.

Here the two primes are \(11895589\) and \(11895613\). Additionally:$$ \text{period of}\frac{1}{3449}=\frac{3449-1}{8} = 431$$This property qualifies 3449 for membership of OEIS A056213:

A056213: primes \(p\) for which the period of reciprocal = \(\dfrac{p-1}{8}\).

3449 is a Sophie Germain prime because:$$2 \times 3449+1=6899 \text{ is prime}$$3449 also features in so-called "Golden Semiprimes" and this qualifies it for membership in OEIS A108544:

A108544: primes that are factors of distinct golden semiprimes (A108540).

I posted about these types of semiprimes in Semiprime Factor Ratios way back on the 26th August 2016. In that post I said that:

A golden semiprime is a number that factors to:

• \(p \times q\) where \(p\) and \(q\) are prime
• \( | \,p \, \phi - q \,| <1\) where \(\phi=\dfrac{\sqrt{5}+1}{2} \)

In the case of 3449, it is the \(q\) and \(p=2131\) and the golden semiprime is:$$7349819=2131 \times 3449$$The OEIS mentions 314 sequences in which 3449 makes an appearance and I've only dealt with a few of them here. However, I see 3449 as an important number in my life and didn't want its current occurrence to pass unnoticed.

LEGO Mathematics

Figure 1 shows a 1 x 4 LEGO brick.

Figure 1

It turns out that if you have another three of these bricks and you stack them in every possible way on top of this base brick, then there are 27591 ways that you can do this. Today I'm 27591 days old and this number appeared in OEIS A123782:

A123782: number of ways to build a contiguous building with n LEGO blocks of size 1 x 4 on top of a fixed block of the same size. The base block is not counted among the \(n\) and must be the only block in the bottom layer of the building

The sequence begins 23, 679, 27591, 1227556, 59212274, 2990304700, 156407426393, ...

So there are 23 ways to stack a single 1 x 4 brick on top of another 1 x 4 brick. It took me a little sketching and visualisation but there are indeed 23 ways to do it. One way is to place the brick directly on top, then there are six more ways to do it with the long axes parallel and then sixteen ways to do with the long axes at right angles. Rotational symmetries are ignored here and only the strict physical orientation of the bricks are considered. Thus we have 1 + 6 + 16 = 23. So the famous number 23 pops up even in the context of LEGO.

There are other OEIS sequences relating to LEGO blocks and they are: 

• A007575: number of stable towers of 2 X 2 LEGO blocks.
  
  
• A007576: number of maximally stable towers of 2 X 2 LEGO blocks.
  
  
• A112389: Number of ways, counted up to symmetry, to build a contiguous building with n LEGO blocks of size 2 X 4.
  
  
• A112390: number of ways of building a building of height k using n 2 X 4 LEGO blocks, counted up to symmetry.
  
  
• A123762: number of ways, counted up to symmetry, to build a contiguous building with n LEGO blocks of size 1 X 2
  
  ... right through to 
  
  A123827: number of ways to build a contiguous building with n LEGO blocks of size 2 X 3 on top of a fixed block of the same size so that the building is symmetric after a rotation by 180 degrees.
  
  
• A123829: number of ways, counted up to symmetry, to build a contiguous building with n LEGO blocks of size 2x4 which is symmetric after a rotation by 180 degrees
  
  ... right through to
  
  
• A123849: number of ways to build a contiguous building with n LEGO blocks of size 5 X 5 on top of a fixed block of the same size so that the building is symmetric after a rotation by 90 degrees.
  
  
• A272690: number of buildings with n 2 X 4 LEGO bricks of maximal height.

Another Look At Elliptic Curves

Source

It was on the 24th February 2021 that I made a post title Elliptic Curves and I've not made mention of them since despite their importance in high level mathematics. However, I was reminded of them again by a recent article in Quanta Magazine titled Big Advance on Simple-Sounding Math Problem Was a Century in the Making on the 14th October 2024. The article begins:

One morning last November, the mathematician Hector Pasten finally solved the problem that had been dogging him for more than a decade by using a time-tested productivity hack: procrastination.

He was supposed to be writing a final exam for his number theory class at the Pontifical Catholic University of Chile in Santiago. To avoid the task, he started pondering, for the umpteenth time, one of his favorite sequences: 2, 5, 10, 17, 26 and so on, the list of all numbers of the form \(n^2 + 1\) (where \(n\) is a whole number).

Mathematicians have used this sequence for over a century to probe the fraught relationship between addition and multiplication, a tension that lies at the heart of number theory. Fundamental problems about multiplication — about, say, how numbers factor into primes — suddenly become much deeper and more challenging as soon as addition enters the picture. One of math’s biggest open questions, for example, asks whether every even number larger than 2 is the sum of two primes; another asks whether there are infinitely many pairs of primes that differ by only 2, such as 11 and 13.

The \(n^2 + 1\) sequence offers a good starting point for investigating the relationship between addition and multiplication, because it combines one of the simplest types of multiplication (squaring a number) with one of the simplest types of addition (adding 1). That doesn’t mean the sequence itself is simple. Mathematicians still can’t answer elementary questions about it, such as whether it contains infinitely many primes. “It doesn’t take far to get to the boundary of our knowledge,” said Andrew Granville of the University of Montreal. When mathematicians do manage to shift this boundary even a little, the techniques they develop often illuminate much broader questions about addition and multiplication.

Pasten was trying to show that the numbers in the sequence must always have at least one prime factor that is fairly large. On the morning when he should have been writing his final exam, he finally succeeded, by figuring out how to embed information about the prime factors of \(n^2 + 1\) in the structure of an equation called an elliptic curve.

Over lunch that day, he described his proof to his wife, the mathematician Natalia Garcia-Fritz. Given the surprising strength of his result, she “told me that I should probably check this many times,” Pasten said. “That afternoon I did so, and the theorems were still there.”

The elliptic curve that Pasten was focused on is \(y^2 = x^3 + 3x + 2n \) with the variable \(n\) being any integer. Curves of the form:$$ y^2=x^3+Ax+B$$have a discriminant given by:$$ \Delta=-16(4A^3+27B^2)$$Thus when \(A=3\) and \(B=2n\) we have:$$ \Delta=16 \times 108 (1+n^2)$$It was the \(n^2+1\) factor in the discriminant that caught Pasten's attention. The level of mathematics that he then applied to make his discovery is beyond my comprehension unless I take the Krell Brain Boost but here is an excerpt from the article that explains what he found:

Applying his theory of Shimura curves to this particular elliptic curve, he could show that the product of the exponents of \(n^2 + 1\) must be fairly small. This didn’t necessarily mean that all the exponents must be small, but it gave him enough control over them to be able to bring in Stewart and Yu’s older method from transcendence theory. By using the two techniques together, he was able to prove that the largest prime factor of \(n^2 + 1\) must be at least about \( (\log(\log n))^2\) — the square of the estimate Chowla and Mahler discovered in the 1930s. Pasten’s new growth rate is much higher than the previous record, though mathematicians suspect the true growth rate is higher still.

Figure 1 shows what the curve looks like when \(n=7\):

Figure 1

In my earlier post on Elliptic Curves, I mentioned that SageMath provides a way of finding the integer values of \(x\) and \(y\) that satisfy the equation of elliptic curves of the form \(y^2=x^3+ax+b\), namely:

E=EllipticCurve([a,b])

E.integral_points()

In the case of \(n=7\) there is only the point where the curve \(y^2 = x^3 + 3x + 14 \) crosses the \(x\)-axis and thus \(x=-2\) and \(y=0\). However, other values of \(n\) yield more points. For example, if \(n=11\) we have the curve \(y^2 = x^3 + 3x + 22 \) and the points:$$(2 : \pm \, 6) \text{ and } (6 : \pm \, 16)$$Figure 2 shows the graph of this curve:

Figure 2

So that's just a brief refresher on elliptic curves and a reminder of how important they are in higher mathematics.