a very aesthetic equation

[@hyperboloidofonesheet1036]

12=1²+2² (base 3)

[@johnchessant3012]

Fun fact: 1/17 = 0.05882353 (when rounded) and 588^2 + 2353^2 = 5882353

[@pierreabbat6157]

I was once in a meeting about a battery charger, which measured voltage from 0 to 15 in 255 equal steps. It had a threshold of something like 9.4 volts, and I rattled off 9.4117647, and the others wondered where I got all those digits from. 9412 works instead of 588 in the equation.

[@stickfiftyfive]

@MyOneFiftiethOfADollar That's pretty presumptive of you.. maybe they just shared it because it relates to the video in a sense, they found it interesting, and figured that we, people interested in math, might also find it interesting. Meanwhile you have shared nothing worthwhile in this thread, except for alerting us to your tendency to be rude to people for no reason. 😂

[@stickfiftyfive]

@@MyOneFiftiethOfADollar My parents are dead.

[@stickfiftyfive]

@ you. Thank you for demonstrating my point. My parents are neither here nor there, and you have the social and communication skills of an asphalt slab.

[@shruggzdastr8-facedclown]

I wonder if it would be possible to construct a 17-gon* with this math factoid? See Numberphile's* videos on the constructa(-i-)bility of 17-gons, ftg Prof. Eisenbud (*on both the main Numberphile channel and Numberphile2)

[@minamagdy4126]

An even more general equation here is, for 0

[@pepebriguglio6125]

0.05882353 ≈ 1/17 and 5882353 = 588²+2353² Where: 2353² is 1/17 of: 94122353 = 9412²+×2353²

[@mrgold4678]

I suggest looking into the similar problem: x1…xny1…yn = (x1…xn + y1…yn)^2 Didn’t tried it myself, but coming 2025 is a solution! 2025 = (20 + 25)^2

[@hxc7273]

1233 and 8833 new favorite numbers

[@miraj2264]

At 5:20, he explains this in greater detail here: kzbin.info/www/bejne/lYfGdpehd791g8k You can factor 10,001 into 137 and 73. These are both primes of the form 4k+1, which means you can uniquely represent each of these primes as the sum of two squares. They're small enough you can just guess it. Importantly, if two numbers can be expressed as the sum of squares, than their product can be as well (note that you don't have uniqueness anymore). This can be seen by FOILing: (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 So applying this formula to 10,001: 10,001 = 137*73 = (11^2 + 4^2)(8^2 + 3^2) = (11*8 - 4*3)^2 + (11*3 + 4*8)^2 = (88-12)^2 + (33+32)^2 = 76^2 + 65^2 or 10,001 = 137*73 = (4^2 + 11^2)(8^2 + 3^2) = (4*8 - 11*3)^2 + (4*3 + 11*8)^2 = (32-33)^2 + (12+88)^2 = 1^2 + 100^2

[@Happy_Abe]

I think we get only possibilities for the sum of squares decomposition because these numbers 10001 and 145 are a product of two primes

[@konraddapper7764]

You are one Word of one the factors should be a gausian Prime

[@Bodyknock]

So using the generalization, there are no solutions with two digits numbers (base 10). 101 = 10² + 1² is the only factorization, and there are no integer solutions for 2A - 10 = +/-10 or +/- 1 with 0

[@stickfiftyfive]

I looked at the problem generalized in a different way (but only in base ten) out of curiosity. 12^2 + 33^2 = 1233 88^2 + 33^2 = 8833 990^2 + 100^2 = 990100 9412^2 + 2353^2 = 94122353 17650^2 + 38125^2 = 1765038125 25840^2 + 43776^2 = 2584043776 I'll fix this algorithm too, if I can. It's too slow.

[@stephenyip5827]

Can anyone help to share back the video to express any integer to sum of two square mentioned in video?

[@ozguryldrm64]

kzbin.info/www/bejne/lYfGdpehd791g8ksi=w5mtpQGMW4uo5n14

[@MyOneFiftiethOfADollar]

@@stephenyip5827 that is in every number theory textbook ever written. Consider learning to read and comprehend. Good Luck

[@coc235]

​@@MyOneFiftiethOfADollarWhy are you spending time on youtube and not reading then?

[@dan-florinchereches4892]

I greatly appreciate your videos professor Penn. Can we please have the link to splitting a number into perfect squares in the video description as you said it exists in one of your older posts. Thank you

[@shruggzdastr8-facedclown]

Um, isn't the "12/twelve" in base-12 actually expressed graphically as "10" (say "one-zero" in your head rather than "ten" so as to avoid confusion) with ten and eleven being expressed using singular letters or other graphic symbols for each ("A" for ten, and "B" for eleven as one common method -- and "X" for ten, and "Ɛ" for eleven with respect to another common method)?

[@falkranduhm10]

Is it just a coincidence or do the two solutions for A add up to the base every time?

[@petersievert6830]

not a coincidence at all: 2A - b = +-x results in A1 = b/2 + x and A2 = b/2 - x, thus A1 + A2 = b

[@aleksmich8928]

Wow!!! That is amazing, since I was thinking of this problem (the 4-digit one) as a candidate for math contest organized by KTU (Kaunas University of Technology, Lithuania). I settled on a similar one i.e. find all 4-digit xyzw such that xyzw = (xy + zw)^2. Maybe you could make a video of the solution to that one in a month. Our contest will take place on Jan. 25-th.

[@notnr]

How do we know that A & B are greater than 0 and less than 100?

[@gileadedetogni9054]

Because they are two digit numbers

[@notnr]

@@gileadedetogni9054 thank you

[@gileadedetogni9054]

@notnr you're welcome bro :)

[@CautionRamen]

For the base b case, why is it Ab + B, and not A(b^n) + B?

[@rainerzufall42]

More general! A and B are not interested in the number base! The number of digits is only distracting, wlog we can consider 1-digit numbers!

[@CautionRamen]

@rainerzufall42Are you saying we can think of A and B as one symbol rather than an n-digit number. Like if A was 25, would it mean to think of ‘25’ as a single character, kinda like F in hexadecimal?

[@rainerzufall42]

@@CautionRamen A is xy and B is zw, just for naming conventions. It's either a block of digits or a single digit, but that doesn't matter for this problem: We are putting equal length strings of digits together and compare the value to the sum of both strings squared (the base doesn't change a thing either, as long there's no overflow and carry). So to make the problem as simple as possible, we can skip any considerations about multi-digit strings, we can use single-digit numbers to solve the problem. It will immedietely work for concatenated digits. In your example, it doesn't matter, if A is "15" (base 10) or "F" (hexadecimal), the problem is the same!

[@CautionRamen]

@rainerzufall42 So is considering x₁…x_n as one digit what you mean? Like if A = 25 (base 10) and B = 50 (base 10), does considering ‘25’ as a single digit in base 100, and ‘50’ as a single digit in base 100 equivalent?

[@CautionRamen]

Sorry I’m asking so many questions, I just find it confusing

[@talberger4305]

10:00 b need to be b^n

[@DavidSavinainen]

No, the base in question is indeed b. In his first example, b = 100, and in the slight generalisation, b = 10ⁿ.

[@jorgechavesfilho]

No. Note that he wrote at the beginning “In base, b 0 < A,B < b”. In other words, it's a different approach to the previous base 10 resolution. In this approach, you have to think that in the previous problem, base 100 would be used.

[@Exarchate]

This could definitely have been made clearer. For instance in base 100, the number 8833 is expressed as a 2-digit number: (88)(33). And it is of course still equal to 88^2+33^2 (as this fact doesn't depend on the base). So he is finding one-digit solutions in base b=c^n where c is the true base of the given initial numbers. Now it's still fuzzy to me why a solution in base b always remains a solution when expressed in base c.

[@maxhagenauer24]

No the thumbnail is not true. 25 ≠ 2^2 + 5^2.

[@TheEternalVortex42]

True in base 12

[@maxhagenauer24]

@TheEternalVortex42 But we always assume base 10 when not being said it's base 12 like in the thumbnail, so it's still wrong.

[@stickfiftyfive]

The video explains how to generalize to other bases, so if you watched the video, you can see it's a clever way to check if we're paying attention and to catalyze views and engagement. Based on the content of the video, you would be wrong to assume the equation is in base 10.

[@maxhagenauer24]

@stickfiftyfive I did watch the video and actually didn't see much explination into generalizing to other bases even in the end. But it's still true that the thumbnail is wrong because we always assume base 10 unless told otherwise.

[@stickfiftyfive]

Ok

[@PetraKann]

So 2^2 = 0. ?????

[@samueldeandrade8535]

Base 12. 25 in base 12 is 29, which is equal to 2²+5². Great thumbnail, huh?

[@hello_hi1]

This is a comment

[@undercoveragent9889]

This is _not_ a comment and thus your comment is cancelled out.