is this my new favorite number??

[@Mr_Happy_Face]

Happy 311th birthday this year

[@insouciantFox]

Are you saying he's 311 or are you wishing him happy birthday for the 311th time

[@cameronbigley7483]

​​​@@insouciantFoxPenn's made jokes that he was born in the 1700s, and a few other years, so likely the former. I can't recall a specific example. Edit: He made the joke at 13:34. He's joked about 1807 being his birth year as well in Sylvester's Sequence.

[@iWilburnYou]

13:34 Michael nonchalantly admitting to be the oldest recorded person by almost 200 years is crazy

[@coreyyanofsky]

he has also admitted to being a vampire and, separately, has claimed that his birth year is 1790 (in "an aesthetically anti-symmetric formula for Euler's constant") since vampires are notorious liars i don't think we can really trust him on this

[@braydentaylor4639]

@@coreyyanofsky Wait, so he admitted to being a vampire, which he isn't, and you said that all vampires are liars, but Michael isn't a vampire and yet you're saying we shouldn't trust him. My brain hurts.

[@zh84]

@@braydentaylor4639 Logically, if all vampires are liars, than anyone who says they are a vampire isn't, otherwise it would be true.

[@braydentaylor4639]

@@zh84 DAMN THIS CIRCULAR LOGIC!

[@Alan-zf2tt]

Well - that is surprising! I thought he was older 🙂

[@12100044]

10:52 Why is 1996 congruent to 0 mod 6?

[@ProactiveYellow]

It isn't?? 1996=332(6)+4

[@miraj2264]

It's a typo. He meant 1998 = 0 mod(6). From this, he concludes that 1997^1998 will be 1 mod(7). Specifically: 7 and 1997 are coprime so Fermat's Little Theorem gives you 1997^6 = 1 mod(7). Therefore, 1997^6n = 1^n mod(7) = 1 mod(7) where n is a natural number. Since 1998 = 0 mod(6), there exists n s.t. 1998 = 6n so we conclude that 1997^1998 = 1 mod(7). Personally, I think it's a little opaque to phrase it that way, but I guess for people that have done number theory this fact is maybe obvious. I had to write it out to understand his rationale.

[@Alan-zf2tt]

@@miraj2264 Bravo @ "I had to write it out to understand his rationale." I think that may have been a good teacher's intention

[@goodplacetostop2973]

18:57 Good Place To Sto-

[@JOSHUVASRINATH]

You made this account for this 😂

[@aerglo9721]

Hey Michael I just wanted to appreciate your videos and the efforts you are making. The problems you introduce are often very interesting. Thanks

[@MothRay]

“Alexis Claude Clairaut (French pronunciation: [alɛksi klod klɛʁo]; 13 May 1713 - 17 May 1765) was a French mathematician, astronomer, and geophysicist.”

[@pietergeerkens6324]

At 11:07, note that 1996 is congruent 1 (mod 3) - just add the digits to see this, getting 1+9+9+6 = 25 and 2+5 = 7.

[@petermayes8764]

I can follow the solution process. But I have absolutely no idea how somebody could set such a problem in the first place, without knowing how it is going to work out. Would love Michael to give some insights into how people set questions for Olympiads, competitions etc.

[@romajimamulo]

Someone started with "huh, 5 mod 7 is not a perfect square", and came up with some wild thing that's 0 mod 7 to add to it, then progressively worked backwards to turn it into the original sum

[@supratimsantra5413]

Sir wonderful discussion, it is like a thunderstorm in mind to get spark of motivation towards magic of mathematics, the top of all sciences

[@tahirimathscienceonlinetea4273]

Hi micheal you deserved million thumbs up Excellent job

[@mathhack8647]

simply Amazing. best thing to start the day with this Demo. .

[@TomFarrell-p9z]

Had no idea why you started doing modular arithmetic until the last few seconds!

[@Fred-yq3fs]

It's a technique he uses routinely. Squares only have specific residues mod p. But how to choose p, that's the trick. He did not show, he invites us to experiment and "you'll see that 7 looks good".

[@pacolibre5411]

0:43 What is this “closed form” because I’ve never heard of such a thing for finite power sums.

[@АндрейДенькевич]

consider relativity. if we look at '12' from below then we see 12 different parts. if we look at '12' from above then we see '1'.

[@АндрейДенькевич]

a^b is a carry for a-ary number system. Infinity, being center of interior of shape, is a carry. Process of MOVING carry form right to left digits is a "uncurving of center of interior of digit". We must turn things inside out like a glove. So Exterior-Interior inversion happens, except center of interior(infinity) , wich pulled from interior and then becomes uncurved 1 times, so we see "1". So carry(infinity) can't live without motion. "Infinity is a motion" Aristotel.

[@roberttelarket4934]

1713 is your birth year in what base?

[@xinpingdonohoe3978]

Base 10.236?

[@jamesfortune243]

Best puzzle so far! I've started working on it. (1713) ^ 1.02 didn't make sense.

[@jamesfortune243]

It's possible that a two digit year format is used. I tried 1713 mod m where m > 74 and got some possibilities. The context was removing the count of years congruent to 0 mod 7.

[@АндрейДенькевич]

​@@jamesfortune243maybe 311 is between 44*7 and 45*7.

[@davidgillies620]

I have a hankering to write a bit of C or C++ code to evaluate this exactly. I'm not entirely sure how many digits it has but a quick bit of experimentation suggests 600 thousand or so, although possibly a LOT more because the partial products get very big around n = 1998/2 (similar to how the middle of a row of Pascal's triangle gets very big). UPDATE: actually it's not all that big. The parentheses round the inner sum tame it quite well. It's 13186 digits.

[@emanuellandeholm5657]

Naive [ 1997**(n-1) * [ m**1998 for m in range(1, n+1) ] for n in range(1, 1998+1) ] crashes my Jupyter Notebook with a memory error. :D

[@davidgillies620]

@@emanuellandeholm5657 You have to careful about intermediates. If you remove the parentheses around the inner sum you get a very different result.

[@charleyhoward4594]

does the 666 at 16:55 mean anything to Micheal ??? Hmmm ??

[@mskellyrlv]

You don't look a day over 300. 🙂 Every time my wife walks in while I'm watching one of your videos, she remarks "Oh, watching math porn again?" Yep!

[@kkanden]

guys this is not the first time that michael hints at being 300 years old, he's trying to send a message

[@jamesfortune243]

Carefully and skillfully done!

[@drssimonhottentot]

1996 is not divisable by 6, so not congruent 0 mod 6.

[@АндрейВоинков-е9п]

Good visualization of multi-sum

[@StanleyDevastating]

what happens if 7 divides m?

[@doraemon402]

Is that a reference to the 200-300 years of history missing?

[@qedmath1729]

Shifting the index was unnecessary and in fact slower than just using the formula for a geometric series, but nonetheless good demonstration on how to shift indexes in summations. Really nice video and problem!

[@rockysmith6105]

So is it common for exceptional mathematicians to pick a secondary birthday that might be in accordance with a mathematician in history that has a lot of importance to the formerly mentioned mathematicians life? This isn't the first time that I've noted this behavior~ I think it's awesome and I always get a kick out of joking about vampirism, but in all sincerity it's actually a beautiful thing, to commemorate mathematics in such a way the more I think about it the more it works. I might even have to do it myself(:

[@rockysmith6105]

Although... I don't think I'm the caliber necessary to really do it

[@59de44955ebd]

If I'm not mistaken, m^1998 is 1 (mod 7) for any m, and therefor, when calculating mod 7, the inner sum can simply be replaced with n. But then I get as result that the total sum is 3 (mod 7) instead of 5 (mod 7). This would still prove that it can't be a perfect square, but where is my error?

[@DylanNelsonSA]

If m is divisible by 7 then m^1998 is 0 mod 7, not 1 mod 7.

[@59de44955ebd]

@@DylanNelsonSA Ah, ok, that's it, thanks!

[@rickostidich]

Since the beginning of the video I was wondering if either 1585 or 1999 was going to be the year of your birthday, Michael, but you fooled us all. By the way, 7 (C.E.) is my own birth year! 🍻

[@chayapholtopar5992]

Using modular arithmetic to check whether a number is prime is a good idea (perhaps it's commonly known to others). However, can we use another prime number to perform the check, such as 3, 5, or 11?

[@japanada11]

It turns out that the number is congruent to 0 mod 3, 0 mod 5, and 9 mod 11; all of these can come from squares, so you can't show it's not a square this way. But there are other primes that you can reduce by to get a non-square; the next one that works is 23.

[@MacHooolahan]

Just *knew* he was a vampire :)

[@tomholroyd7519]

Let's draw a graph! I love it when you draw graphs, do it more.

[@Sunlessilver]

Hello love your vids. Is there anything interesting about an equation with infinite positive and negative powers set equal to 0. It would look like ...x^2+x^1+x^0+x^-1+x^-2+...=0

[@david-melekh-ysroel]

Are you sure that 1996 divides 6?

[@JanJannink]

I think it’s a typo and he meant 1998 (based on the flow of the video, and 1998 is 0 mod 6)

[@quazzydiscman]

What's your favorite number? Me: 17 Michael Penn:

[@charleyhoward4594]

the way he messes with the indexes of summation always amazes and confuses me

[@VideoFusco]

The sum symbol becomes smaller and smaller as Micheale writes further down the board :D

[@yoav613]

1713😂😂😂😂

[@anestismoutafidis4575]

=> ΣΣ= 1×1=1 1 is the perfekt number for a function. Thank you, Michael.

[@greennumber7aimcom]

Ahh, vampire!