Become a MILLIONAIRE by winning The Parker Prize (extra footage)
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- Опубліковано 7 жов 2021
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The Parker prize is where you almost, but not quite, receive the prize money.
I laughed way too hard :D
It's the intrinsic value we met along the way.
@@mrwalter1049 Me too 😄 Nice one, Per
And there is no first place. Only second.
I think I just woke up my family laughing at this
The prize will be £1 million ... mod 29.
Best one yet
So it would turn out to be £100000 %29 so £22...
@@livedandletdie so 29$?
@@LtLollo Nope. £22 according to @The Major
Yes, £1 million mod 29 is £22. And according to Google, current exchange rates say that is worth $29.95 in the United States.
Most people: A MILLION DOLLARS!
Matt Parker: *ten to the sixth currency units*
Can you think of anything less than a satoshi (making the price about a nickel). I guess he didn't specify actively valuable, German Marx got pretty bad but I don't think they were divisible enough, plus some historical value.
Oh I know, knots in rope. You'll receive a thread with precisely 1 million marked with knots and possibly colors, ancient Egyptian style. (If you want to cash them in though, you'll have to learn necromancy and head to Egypt)
Yeah, that would be easier than cowlery shells, but the shipping might be more than 5 satoshis
"currency units to be determined by the awarder"
The currency units being, ofc, Parker Dollars, which you can spend on the Standup Maths merch store! 😅
I was thinking Prussian francs. But it does beg the question:
What's the equivalent of Parker Dollars to Stanley Nickels?
more likely Parker Pence, which aren't quite equal to actual pence in value
How about Zimbabwe dollars? I have a five billion Zimbabwe dollar bill in my wallet right now. I don't know precisely how much that's worth in US currency, but rounded to the nearest US dollar, I'm pretty sure it's worth $0.
@@PhilBagels A first I was impressed because I thought it would be rare, but then I realized the entire issue of Zimbabwe's inflation is that it wasn't rare. You can buy a five billion Zimbabwe dollar bill for like, $3 online but the actual value is probably much less. XD
Parker Fun Bux
Brady's done so many Numberphile videos that I think he's got a pretty sophisticated grasp of abstract maths now. His questions here for Matt really show it! What a great and fun way to get an education in mathematics.
Didn’t he also study maths?
I assume that when the Parker Prize is awarded only 87½% of it will go to the right person.
I'd guess all of it will go to the correct person, but it will be split into 8 payments and only 7 of them with the correct amount of money being transferred.
Well then I hope I’m the wrong person.
Why would it be mod eighty seven and a half
I will be surprised if governments take less than a half:p
😂
I think Monopoly money would be best because the game is from Parker Brothers. Thus, Monopoly money _is_ legitimately called Parker Dollars.
That'd be really funny as there's still only about '$'20,000 in each game set, so the winner would get about 50 sets of Monopoly. It could even be a nice collectable prize if the sets were all different variations.
Lol!
Zimbabwe Dollars will be the prize sounds like
I have 10 billion Zim dollars 😎🤑💰💸
@@robertaries2974 Beware, they stroke off a few dozen zeroes in the past years
@@HagenvonEitzen They struck off ALL the digits a few years ago. Z$ is no longer valid.
It's no longer currency, though, so he'll have to find something else. The Venezuelan Sovereign Bolivar (VES) is at about 4,000,815 VES to 1 USD. That's what I'd go for.
The Zimbabwe madness is long over.
If the "Parker Prize" is actually going to pay out a bounty, it would have to be called the "Non-Parker Prize" wouldn't it ? (Because it works)
The Fields medal has a notoriously stingy payout. It's 15,000 Canadian dollars, which at time of writing is about £8800. Having said that, if you win a Fields you'll make fifty times as much in the first year just on the lecture circuit.
It's a lot by many standards (not many academic institutions would offer prizes that large), but it's definitely nowhere near the 10 million Swedish krona (840,000 British pounds) for the Nobel Prize.
I don't think this "lecture circuit" is as lucrative as you suggest.
How about making the Parker prize a crowd funded trust and as the value goes up at each million pennies another problem is created to be solved
Give it a crypto currency spin and half the worlds computing power could be used to crunch it.
I loved Matt's reaction to the idea of Parker Prize, he almost spit out his drink!
Corollary: if a non-Parker magic square of squares exists, then there are finitely many Parker finite fields.
And if the numbers are small enough, you could just check the remaining ones with a computer.
I think you can only say that there then are finitely many Parker finite *prime* fields. I don't see immediately how a non-Parker magic square of squares in Z prevents GF(p^n) from being Parker if p is small and n arbitrarily large
@@HagenvonEitzen That would be an interesting related problem: show the finite field F_p^n is non-Parker for sufficiently large n. I would guess that it is false.
That's already the conjecture in the paper though? That there are finitely many Parker finite fields.
That's true, because once you no longer need to mod, every finite field bigger than the largest number in the non-parker magic square will have to be non Parker, because the non-parker integer magic square will fit inside them.
As to Brady's point, if there is a proper integer non-Parker Square of Squares, not only will it satisfy *one* of the prime fields, it will satisfy the conditions in an infinite number of fields (in fact all of them higher than the lowest solution). The Cain paper itself makes much the same point in its concluding paragraph: "Note a proof of the existence of infinitely many Parker rings of the form Z/nZ would prove that no magic square of distinct squares exists over the integers. If, instead, such a square existed and N were its largest entry, then Z/nZ would be non-Parker for n > N. And therefore only finitely many such rings could be Parker.
"
It takes a bit to get your head around the problem but other than that it is quite the mystery.
i think it's also natural to say that the first N which satisfies it can also be modded down to lower fields and still work as a solution, no? This is because the modular value can be interpreted as simply the remainder on division, so if the math works for the integers, then it must work for every prime number, with the exception that duplicate entries would be allowed *after* modding, because multiple numbers may have the same remainder value.
I believe the ending should be "...then Z/nZ would be non-Parker for n>3N...", not that it changes stuff
@@Czeckie probably 3 N^2 actually, but i'm not sure that's a requirement, it may be that N is sufficient
@@Czeckie No, just N. In fact, it could be strengthened from n > N to n ≥ N.
If none of the entries are reduced, their magic sums will _still_ all be the same. And then those magic sums will be reduced mod N+1, and they will again still be the same, because if a = b, then of course a = b (mod N+1). And it even works mod N, because any multiple of N added in will reduce to 0 in the sum anyway. The reason we can't use a modulus < N is because then we might have redundant entries. And for a square of squares, we also need N prime, because otherwise two different numbers < N could have the same square.
The Parker Prize should be a prize for the best "negative results"/"best effort" in science.
If you find out that something dos not work, this can help many people so that they know that the thing you tried did not work, and they can try something else.
And if you find out way it does not work, this may help a lot to find something that does work.
If I were an arbitrarily wealthy person I would definitely bankroll an awards ceremony for notable lack of results in scientific research. Encourage those highly necessary but non-glamorous and therefore non-funded research topics that verify results or look at things everyone assumed was the case.
Sort of like the ignobel prizes!
time to quit everything and start pursuing magic squares for a bottle of champaz and a tenner
Brady's insightful questions have never ceased to amaze!
"Do we live in a Parker universe?"
(enter the Spider-Verse)
Venezuelan Bolivars are 4 million to 1 USD, so that would equal about a US quarter...
i might be wrong, but i think you are off by a decimal
The Venezuelan bolivar is no longer legal tender
The Bolivar, having just been redenominated, is 4.14 to the dollar as of time of writing (08 Oct 2021). But taking into account this redenomination and the previous one it is actually 414 billion to the dollar.
1,000,000 parker dollars. They're almost worth something.
Contrasted, of course, with the non-Parker dollars, which consists of most currencies commonly used.
If you can prove that the set of integers is Parker (meaning no such non-parker square exists), then Matt will pay you in Parker dollars, which can be worth however much money he decides. However, if you can prove that the set of integers is non-parker, then you get 1,000,000 non-Parker dollars, which means there is a lower limit in how much money you will win. :D
Everyone's thinking too obvious for the currency units. What if the prize is 1,000,000 micropounds?
You mean Parker Pounds™️
1,000,023 just to be Parker about it
What if the currency unit is $0?
With the implications of the determiner "Parker", I don't know if I'd like to receive a Parker Price 😂
Also that's why our universe is totally Parker
How do you think I feel?
While the value of a non-parker prize may be worth more, to win a parker prize is STILL, winning! Or the flip side to this NP vs P coin, mean that you either win 1 non-parker prize and lost an infinite number of parker prizes, or win 1 parker prize and lost an infinite number of non-parker prizes. I think that the equivalency of valuation of non-parker/parker prize winning is equal and just as undecidable as NP/P (in computer science).
There's some parker spelling in your comment!
Given our difficulties with constructing a coherent unified system of mathematics, and the trouble Physicists have had conciliating various accepted fundamental theories, maybe we DO live in a Parker universe.
Now that there is a Parker Prize, they should just rename the Fields Medal to be the Non-Parker Prize, or the Non-Parker Fields Medal.
The point at the end is quite interesting. If you would find a non parker square of integers, it would mean all finite fields of mod higher than the sum of that square have that square as their solution. Which means (if I understand this correctly) that if one can proove somehow that no two non-parker fields share a solution, they would also prove that there is no real square of squares in the ring of integers
He's even managed to Parker his own Parker award with a Parker sum of money.
All I'd want is to be in a few Numberphile videos and a medal hand-crafted by the man himself. Too bad all the maths involved are undoubtedly way too complicated for me.
If Parker money is anything like the Parker Square...
I'm both excited and afraid
Why not just give one million parker pesos? If they want to convert it to some other non-parker currency, they can figure it out
Edit: ooo, parker promissory notes sounds much better.
The Parker prize should be to prove that such square can not exist and so proving that integers are Parker.
"currency units" i love it lol
1,000,000 Iranian Rials is currently worth $23.81, £17.88, or €21.
1,000,000 Vietnamese Dongs, yes really, is currently worth $44.09, £32.50, or €38.79
(Conversions done on December 3rd, 2021)
Shhh. Don't give him any more ideas!
Iranian Rials are the lowest valued currency right now, unless he uses obsolete or crypto currency.
You don't want to make the mistake of committing to the currency now though.
ParkerCoin. Hmm… based on non-fungible finds of almost-but-not-quite-magic squares… To be honest, I’ve heard worse pitches.
And a million of them are currently worth 23.70 USD.
@@unvergebeneid 🤣 it's been a few minutes hope it hasn't gotten worse. All in this together
@@unvergebeneid Or £17.41 for those of us using the same currency as Matt Parker's current residence...
Everybody gangsta until the currency units turn Zimbabwean
Explanation: zimbabwe is sadly undergoing hyperinflation, so even one trillion zimbabwean dollars is worthless.
These have been the best videos in ages. Hillarious!
A solution for the ring of integers will also be a solution for the field of rationals, and by clearing denominators, the converse also holds.
1000000 Sen of Indonesian Rupiah (not to be confused with the former unit Sen of Japanese Yen) is currently 0.61€. I think you could afford to pay out such a price. :D
Huh, started tracking this problem before even knowing there was a bounty.
the sentence "currency units to be determined by the awarder" got me. and i looked into the list of exchange rates and in my list was the "Albanian Lek" (ALL) the first one. so it thought, hang on don't take one currency, take ALL of them! take 1mio. pennys/cents/kopeki divide this by the number of currencies (the list i found got 153 (excl. Silver/Gold)) and you have a value per currency (65.36 CU) convert this back to €/GBP/USD and sum it up!
And, what is the value when converted back?
At the time of writing, 1 million Iranian Reals is worth 23.70 USD.
one million _iranian rials_ is what a person I know did recently for their little scale thing.
Sounds like trying to have a Parker universe communicate with a non-Parker universe, will... Parker.
Something to look forward to. :)
I wonder what it would tell you if you plotted the lowest maximum value of an NP magic square for all the finite fields of prime order. Depending on what line you get as a result, you may be able to estimate where it goes and calculate where that line may drop far enough to create a magic square of squares in the integers.
They would probably be mostly zeroes and ones, like the example with mod 29, so not very useful, sadly.
The problem with magic squares of squares is that squared numbers are very spaced out, and using bigger numbers just doesn't help because those are even more spaced out between them, so their sums are very hard to match. With finite fields, on the other hand, you can pick very big numbers, because the powers will just loop around, and so you won't get as many gaps between them
The Parker prize is when you win the lottery but forget where you put the ticket.
Does this mean THERE IS a magic square of squares that does not need to wrap around using MOD, we just haven't found it yet?
OR
We DON'T KNOW weather there could be such a magic square of squares?
And by the way, are ZERO squared and ONE squared excluded?
Don't know whether there is or not.
And no 0^2 and 1^2 are allowed to be used in a magic square of squares, but there's a no repeat clause so you can't fill a grid with the same number squared. (ie: no "all 1^2" grid)
Just to add on to what Elen said, the value of the squares themselves have to be different, so 1^2 and (-1)^2 is not allowed.
The Parker medal with the Parker prize of a million Parker money.
"10 to the 6 curency units" is the most math(s) -- I'm an American-- way to say millions, evah.
If there is a magic square of squares it will still have constant sums of you reduce it by any prime. That means, that if you reduce it to a Parker field, some of the numbers must be equal in that field.
So you get the following lemma:
If there is a magic square of squares, then for every Parker prime p, two entries of the square are congruent mod p.
I just like your energy
If you win this price you become a Imaginary Parker Crypto Currency Coin (short IPCC Coin) millionaire.
The IPCC Coin is a block chain and only so far unknown Parker Squares are valid keys. The imaginary part means that we just do this for fun and the money isn't real but integer.
in all seriousness, like $1000 is a fair prize for a mathematical bounty like this. I don't know which currency unit to choose for 1M of it to become $1000.
There exists already a 1000$ prize for a working square of squares or a mathematical sound proof it isn't possible
The Parker Prize would have a value of $999,999
can't wait to get my prize of a million bitcoins
I found this cool way of making magic squares which requires a sequence of 9 numbers such that they have common difference and I don't see how that is possible with square numbers of whole numbers. So, I concluded it was impossible to make magic square of squares. If there is any other way to make the magic squares let me know.
You don't need numbers with a common difference to make a magic square.
You're making a mistake in your logic here; your conclusion would only hold if the ONLY way to make a magic square would be with common differences, which is false
@@omerd602 can you please tell me about other methods that you know about
@@pimcoenders-with-a-c1725 can you please tell me other methods that you know about
@@aayamshrestha5982 Consider the following square:
24 49 2
3 25 47
48 1 26
The numbers aren't in an arithmetic sequence, yet they still work.
In general, the following works:
a+b a-b-c a+c
a-b+c a a+b-c
a-c a+b+c a-b
Hey Matt, you could have made the prize so much larger! A few years back, Zimbabwe underwent hyperinflation, and during that time, produced bills denominated at $100,000,000,000,000 ZWL (100 trillion Zimbabwe dollars, third re-denomination). Those bills are available on eBay as collectors' items at fairly reasonable prices. Regrettably, they are no longer legal tender in Zimbabwe, which switched to using US Dollars a few years ago, among other regional currencies. But still, you could make the prize be ten of those bills and totally make someone a Parker Quadrillionaire with the Parker Prize! Do it!
I have a new purpose for the next week, giving it a go and being happy with being a Parker
So if the integers turn out to be Non-Parker, so that a perfect square of squares exists, then all finite fields of order higher than the common row/column/diagonal sum of the smallest such perfect square of squares in the integers will also contain that perfect square of squares (no modding required) and will thus be Non-Parker, so this will prove the conjecture that there are only finitely many Parker fields.
The real Parker Prize is the Parker Square you find along the way
Now he must win a millenium price to pay his own price. Maybe that way Riemann is solved?
Maybe that's the purpose of these prizes. So you actually have to work to pay off student debt!
Venezuelan Bolivar suddenly has an use again... Matt handing away $2.5 as it's slightly more than 1 million Venezuelan Bolivar.
"Currency units" 😂
Parker square: Into the Parker-verse
I guess we could make this a Crowdfunded Price for Proving the Conjecture. We could srap some Money as encouragment together.
If you can correctly identify the currency unit, you will receive the prize . . .
Sadly, the Fields Medal does not include cash prize of order 10^6 currency units, unless your currency units are on the order of 100 times less valuable than the U.S. dollar.
"Orders of ten to the sixth currency units"
You can use, as a currency unit, the Brazilian Cruzeiro of 1990. Look it up, see how much that is today.
Shhh. Don't give him any more ideas!
Parker millicents (which can be spent in the Merch store)
It should be The Non-Parker prize actually.
"A million units"
What if he does 1 million yen, that would only be about 10k dollars xD
Approximate squaring problem quite interesting... , for example 200/199 becomes integer after 1444 iterations , and it is really really HUGE integer, about the size of the 1400th Fermat number (!!). Does there exist a positive integer n such that (n+1)/n never reaches an integer from the iteration scheme A* ceiling (A) , starting from (n+1)/n ?
I don’t want to live in a non-Parker universe.
WOW! ... That was unexpected.
I’m looking forward to my Korean Won XD
I rewatched Sound of the Drums yesterday and then 1:36 happened... Matt Parker == Harold Saxon? o.O
So if I solve P vs NP I can be a millionaire? Sweet!!
the currency units are Russian Kopeks (1,000,000 Kopeks = £100)
Oh this is interestening. The solution to this if you dont completely bruteforce it should be closely related to the Gaussian Circle Problem i believe
Sorry Matt, but Ponzi already discovered the Parker Prize lol
A million "currency units?" Pretty vague. I'll be looking for a million farthings when I win.
1 Million "Currency Units". Chooses Vietnam Dong, gives you $50. :rofl:
Wouldn't it be the case if you had a magic square of squares in the integers, then reducing it's entries modulo any prime would give a magic square of squares mod p for all primes? So the existence of parker fields implies that no magic squares of squares of the integers exist
It might just be that there are only a finite amount of Parker fields
@@cartatowegs5080 But you only need one for this
I thought so at first, but then I realised: the no-duplicates clause. It's still possible that there's a proper square-of-squares in the integers, but that whenever you reduce it mod any of the Parker primes, you end up with two (or more) of the values in the grid being equal.
@@mrphlip On yeah, I forgot about that one. That could save it after all
@@mrphlip disregarding the non-duplicate clause, the squaring of squares using squares, in a magic square would be trivial. 1² exists..
What do you suppose are the odds that Ramanujan found one or more such squares, and either failed to write about it, or did but the ms. is lost?
Fred
The currency units to be determined.. He is gona probably pick the VES (Venezuelan sovereign bolivar) which is like
1 USD = 4 million ves, or
1 EUR = 4.7 million ves
So you’ll a quarter 🤓😂
Does the paper say anything about the sum always being zero in those squares?
The trickster is back!!
Does that mean if one is found where they didn’t need modulo arithmetic then there is infinitely many, as you could just take the modulo n to be any arbitrary n larger than the minimum value? As you are not needing to do the modulo arithmetic, the value or n does not matter.
Yes, if you had a square in which all the sides sum to n, and n is smaller than the size of the field, the same square would work in any larger field.
In fact, the is also means if there is a square using the natural numbers, there are infinitely many non-Parker fields as any field larger than the sum of the square will be non-Parker.
Interestingly, the paper actually proves that there are only parker fields up to Z/67. Which perhaps hints that maybe Z/71 has a solution? (I doubt it, but hey)
@@niwasox3 well actually you just need a field which is larger size than smallest number before squaring, because yes they will still loop around but the way mod works they still guaranteed to work, your condition is not when they all must work but when they also all stop looping around...
@@djsmeguk, the paper poses a conjecture that this is the case. It does not prove that.
One million ParkerCoins.
Solution: Make it all 0s.
I will take my prize money now.
One million bitcoin? I'm in.
not requiering mod? but what about the standard mod10? it's still a mod ;)
One million South Korean won, a nice $838 USD or 725 euros.
It's the new currency unit!
does this mean there's a Group of Parker Fields? if there is homology between the set of the fields?
This should be the Non-Parker Price
Well if it’s in USD, then we can get at least 1,000,000 pennies, which is $10,000.
what if all possible magic squares of squares need partially the modulus operation, and then it will not work outside of the non parker field
1 million ParkerCoins !!
After sorting this mess out and publishing, I suppose I will need to keep my 1*10^6 ParkerCoins in my ParkerCoin digital wallet.
Give me a million satoshi