I keep feeling like there should be a simpler and more obvious proof but this is the best I’ve come up with:

So

But

and QED.

]]>- ???
- ???

That prompts a few questions: Are there solutions for 4 and 5? Are there solutions for *all* integers? Or for an infinite number of integers?

The answers to the first two questions are, no and no. Start by observing that the cube of any multiple of 3, , is divisible by 9 (in fact 27). For numbers equal to 1 mod 3, so is equal to 1 mod 9, and for numbers equal to -1 mod 3, so is equal to -1 mod 9. So all cubes are equal to 0 or ±1 mod 9. From this you can figure out all sums of three cubes are equal to 0, 1, 2, or 3 mod 9. But 4 and 5 equal -4 and +4 mod 9. So they cannot be written as sums of three cubes; nor can 13, 14, 22, 23, 31, 32, ….

As for the third question, it’s conjectured all other integers can be expressed as sums of three cubes. But it’s not proved.

The integers up through 11 are easy. In fact you might realize 10 can be done another way: . For that matter, and so on for an infinite number of solutions.

But 12 might take you a little longer: . Worse, . Things are pretty easy up through 29, and then 30 is hard. *Very* hard.

Good luck coming up with the simplest known answer:

.

Yikes!

So it goes. Lots of easy ones

interspersed with slightly tougher ones

But as of a few days ago, there were solutions known for all but two numbers less than 100 (other than the ±4 mod 9 impossibilities).

That was then. Andrew R. Booker has just published this result for the smallest hitherto-unsolved case:

I hardly need add that this was not found with a Python script during Booker’s lunch hour. The paper goes into the number theoretic gymnastics involved in his computer search, which, he reports, “used approximately 15 core-years over three weeks of real time.”

So that’s 33 solved. We’re down to one remaining unsolved number under 100. And I’m ready to give it to you. Now.

Though I don’t think that you’re going to like it.

All right.

You’re really not going to like it.

All right. The smallest for which no expression as the sum of three cubes is known is…

is…

Forty-two.

]]>Check out this grafting number, the second biggest I’ve got at the moment:

There’s some structure to understand here. This is generated using , , and . You can read those values off the numbers above, though. There are 22 digits in the integer on the left, and that’s . On the right, those 22 digits recur starting 7 digits left of the decimal point; that’s . Before them are the digits , which is .

Likewise, examine this one:

You can see and , so , and .

Though here’s an evil one:

Here and , so , and . Or… wait, is it? If you write the integer on the left as then and so and . Or should you write it as and conclude and so and ? Turns out all three interpretations are valid; they lead to three different equations yielding the same grafting number.

What about this miserable attempt at a grafting number?

Well, and , so , um, ? And ? Those give (with the sign in the quadratic formula, unlike all normal grafting numbers, because the sign gives ) and the equation or . Yes, that sort of fits the profile, but in an ugly way.

Likewise, and have similarly pathological analyses. Those I think really do need to be counted as grafting numbers, but not normal ones.

Two things I have not figured out yet is why it’s always the ceiling, not the floor, that gives the grafting number, and why and gives grafting numbers for, apparently, all values of while other values of and don’t. I suspect these are related. Observe

(with braces denoting fractional part) which is, to first order

and is whatever it is, but it’s in and for an approximate result we can take so

.

By taking the square root of the ceiling of a number instead of itself, we add approximately to the square root. By the same token, using floor subtracts about .

That’s sort of an average, though, and in the case of grafting numbers, it’s often an overestimate. For instance, for , , we get . Then with we end up with grafting number . Clearly the ceiling in this case changes the number by several orders of magnitude less than . But why don’t numbers looking hypothetically like work out?

]]>We could use other values for the coefficient of , though. Any power of 10 in fact. (For grafting numbers in base 10. If you want grafting numbers in another base, use powers of that base.) We have . Solutions are . Real solutions are obtained up to but we want . You can figure out this means .

For instance, with , we can use . With , . Now = 2 (), 103, 10206, 1020515, 102051444… not one of which, sorry to report, is a grafting number. With , though, and , we get and that is a grafting number, though not very impressive. and gives us . Yes! That’s what I’m talking about! And with gives , , and .

And on it goes. Those two I started off the first post with, 60,755,907 and 63,826,090,875, arise from , , and and , , and , respectively. Here’s another: , ~~and that’s the only one I’ve found so far using floor instead of ceil.~~ *Edit:* This does not use floor; I must have been fooled by a rounding error. This comes from , , and . Then , and when rounded *up*. In fact I’ve found no cases where floor gives the grafting number, and exactly one case where both ceil and 1+ceil work: , , and giving grafting numbers 764 and 765. Otherwise it’s always ceil. Which suggests other questions to ask. Sometime.

A grafting number is a counting number whose digits appear (consecutively, and in order) in its square root, starting at or to the left of the decimal point.

Matt Parker excludes cases where the digits appear further to the right of the decimal point, which may seem an arbitrary restriction, but it makes sense if you think about it. We don’t know that square roots of nonsquare counting numbers are normal numbers, but it does seem plausible. A normal number is an irrational number in which every finite string of digits occurs with equal density to every other digit string of the same length. If square roots are normal then including numbers whose digits appear *anywhere* in their square root as grafting numbers would mean *every* nonsquare number is a grafting number! For instance:

That would be a bore, so we just allow numbers whose digits appear beginning at or to the left of the decimal point in their square roots.

The above definition allows and , for instance, and it probably shouldn’t, but let’s leave it as is for now.

We came up with a grafting number by first finding , then multiplying both sides by 100, and then rounding up the number under the radical sign to an integer.

100 was an arbitrary choice, though. What if we used ? That is, candidate grafting numbers would be .

Let’s check:

Seems to work every time. Notice the floor function *doesn’t* work, though: doesn’t match that last digit, for instance. It’s not obvious why the ceiling function works and the floor function doesn’t. One might think both could work, or floor but not ceiling, or neither. For that matter there’s the… superceil? The second integer higher? That works in one case: . There’s stuff to dig into here. But let’s set it aside.

We have here a series of grafting numbers, none of which is 60755907 or 63826090875 so clearly there’s more exploration to be done.

We got the above by considering , where . What about other values of , though? Expanding gives and so . Then when one solution is . (We want solutions in , so the other solution is discarded.)

So which looks promising. Let’s check:

- — nope.
- — nope.
- — nope.
- — nope.
- — finally!

And no, the floor function doesn’t work where the ceiling fails here (or where it succeeds). Looks like we have a way of generating *candidate* grafting numbers, but not necessarily good ones.

That was and . For , and there are no real solutions. So that’s that.

Except we can vary the right hand side of . The coefficient of can be any power of 10: . We’ll do that next time.

The grafting numbers less than a million (excluding the dubious 1, 100, and 10000) are:

n | sqrt(n) | |
---|---|---|

8 | 2.8284… | |

77 | 8.77496… | |

98 | 9.8994… | |

99 | 9.9498… | |

764 | 27.64054… | |

765 | 27.65863… | |

5711 | 75.5711585… | |

5736 | 75.736384… | |

9797 | 98.9797959… | |

9998 | 99.98999… | |

9999 | 99.99499… | |

76394 | 276.394645… | |

997997 | 998.997997995… | |

999998 | 999.998999… | |

999999 | 999.999499… |

Check out what’s to the right of the decimal point. Crazy, right? Here’s another similar number:

Here the digits of the integer on the left appear in the real number on the right starting in the 100s place. Matt Parker calls these things “grafting numbers“. What’s going on with them? This isn’t just weird coincidence, is it?

It isn’t.

Consider good old . It’s a solution of the equation . So and . There a number and its square (or, looked at the other way, a number and its square root) have digits in common; an infinite number of them, in fact. There’s a hint here.

Now let’s look at solutions to . Suppose . Now if is an integer, then in decimal form is just the concatenation of and and is just the digits of shifted one to the left.

For instance, ; then . One solution is , and . Or looked at another way, .

This starts to look like the grafting numbers idea, but grafting numbers are integers. But hang on. Multiply both sides by, say, 100: . Round the number on the left *up* and you find

There you go, a grafting number. How about more? We’ll see…

]]>This design is made of three 2×1 rectangles. What fraction of it is shaded?

Label vertices and draw in a couple line segments:

Considering triangles CFJ and CKJ, angles CFJ and CKJ are right, line segment CJ is common, and CF = CK = 2, so the two triangles are congruent and FJ = JK = 1. But then angles CGE and JGK are equal, angles CEG and GKJ are right, and CE = JK, so those triangles are congruent and CG = GJ.

So JK = 1, and if KG = x, GJ = CG = 2-x, so 1²+x² = (2-x)². The solution is x = 3/4. The area of each shaded triangle is 3/8. The area of the whole pattern is three rectangles minus the shaded area: 6-3/4 = 21/5. The shaded area is 1/7 of the area of the whole pattern.

Notice the shaded triangles are 3:4:5.

Can the shaded area be dissected into pieces, seven of each of which will fill the pattern? Yes.

Here are the relative sizes of the pieces, in case you’re interested:

- Blue: 9:12:15
- Green: 12:16:20
- Red: 9:25:30:38
- Magenta: 10:16:38:40
- Shaded triangles (blue+red and green+magenta): 30:40:50

Here’s a square inscribed in a unit circle:

Or to keep down the clutter, here’s just one quadrant of that:

One side of the square along with the radii at each endpoint forms a triangle. The length of one side of the square we’ll call and it is of course . That means the perimeter of the square is .

Now consider an inscribed octagon. Again, here’s just one quadrant.

One side of the square is labeled ; what’s its length? Well, drop a perpendicular from the vertex in the middle:

And now notice the right triangle with hypotenuse 1 and sides and is just half of a quadrant of an inscribed square, which means . Then . From that you can get and from that, . The perimeter of the octagon is then .

Well, that was so much fun, let’s do it again. Here’s a quadrant of a 16-gon:

The right triangle with hypotenuse 1 and sides and is just half of an eighth of an inscribed octagon, which means . Then , , and . The perimeter of the 16-gon is then .

You know the rest, right? From here you can do the 32-gon, 64-gon, 128-gon, and so on, getting , , , and perimeter . For , the perimeter is . As increases, the perimeter gets closer and closer to the circumference of the circle, so this gives a way to calculate .

What about circumscribed polygons? If you look at the figures above, you can see the ratio of the size of a circumscribed square to an inscribed square is . Likewise the ratio of the size of a circumscribed octagon to an inscribed octagon is , and so on. So the perimeter of a circumscribed -gon is . As increases, approaches from above. And the average of and is a better approximation to than either, though not by a lot.

n | h_n | x_n | y_n | s_n | p_n | P_n | average |

4 | 1.41421356 | 5.65685425 | 8.00000000 | 6.82842712 | |||

8 | 0.70710678 | 0.70710678 | 0.29289322 | 0.76536686 | 6.12293492 | 6.62741700 | 6.37517596 |

16 | 0.38268343 | 0.92387953 | 0.07612047 | 0.39018064 | 6.24289030 | 6.36519576 | 6.30404303 |

32 | 0.19509032 | 0.98078528 | 0.01921472 | 0.19603428 | 6.27309698 | 6.30344981 | 6.28827340 |

64 | 0.09801714 | 0.99518473 | 0.00481527 | 0.09813535 | 6.28066231 | 6.28823677 | 6.28444954 |

128 | 0.04906767 | 0.99879546 | 0.00120454 | 0.04908246 | 6.28255450 | 6.28444726 | 6.28350088 |

256 | 0.02454123 | 0.99969882 | 0.00030118 | 0.02454308 | 6.28302760 | 6.28350074 | 6.28326417 |

512 | 0.01227154 | 0.99992470 | 0.00007530 | 0.01227177 | 6.28314588 | 6.28326416 | 6.28320502 |

1024 | 0.00613588 | 0.99998118 | 0.00001882 | 0.00613591 | 6.28317545 | 6.28320502 | 6.28319024 |

This is something like the way Archimedes calculated around 250 BC (I *told* you this was ancient), although he started with a hexagon rather than a square and only went up to a 96-gon. He didn’t have Google Sheets, though.

Now consider , that is, the number formed from the first two pairs of digits in . Its square root is a little more than (or equal to) where is the largest digit which, appended to , gives you a number whose square is no greater than . That is, . But and so . The number on the left is just the remainder from the previous step with the next pair of digits from appended, and the first term on the right is twice the result from the previous step with a new digit appended, multiplied by that new digit. Their difference gives our new remainder, .

Iterate that process and you get each digit of the square root.

]]>Recently I saw a YouTube video explaining the method, so I got it back. And I’m writing it up here because I have an easier time keeping track of this blog than all the YouTube videos I’ve seen. In fact I’ve lost track of the video already.*

So let’s find the square root of 7252249.

This is going to look kind of like long division. Kind of. Start by putting the number under a square root sign, and separate it into pairs of digits, right to left. (If the number is not an integer, start from the decimal point.)

Now look at the leftmost pair of digits. (In this case the number of digits is odd, so the leftmost “pair” is just the 7.) Above it, write the largest single digit whose square is no greater than that pair. Here it’s 2, because 2² < 7 but 3² > 7. Write the square of that digit (4) under the pair, and subtract (3=7-4).

Next bring down the next pair, 25, to make the number 325. Take the result so far (2), multiply by 2 (4=2×2), and write that to the left, leaving room for another digit.

Now we want the largest single digit such that if we append it to the number we just wrote down on the left, and multiply the result by that digit, it’s no greater than the number on the right. Here 7 is just barely too large, because 7×47 = 329 which is >325. But 6×46 = 276, which is <325, so put 6 on top and after the 4, put 276 under the 325, and subtract (49=325-276).

Lather, rinse, repeat. Bring down the next pair (22) and to the left write the result so far times 2, with room for another digit (52=2×26).

The largest digit that will work next is 9, because 9×529=4761 which is <4922. Write that down and subtract (161=4922-4761).

Again, bring down the next pair (49), double the result so far and write that on the left (538=2×269).

Looks like 3 should work for the next digit and in fact, 3×5383 = 16149 which is exactly what we have, so 7252249 is actually a perfect square, 2693².

If it were not a perfect square we could continue by bringing down pairs of zeros after the decimal point and carrying on as before to get as good an approximation as we need.

It’s not as easy as punching a calculator, but that’s not the point. Being able to do this on paper means you’re that much less dependent on machines. It’s amazing the feeling of power it gives you.

* I lied. It’s here: https://www.youtube.com/watch?v=nAZvUnWbS8c

]]>