these are the only perfect squares

[@ShaunakDesaiPiano]

The title should have been “these are the only absolutely perfect squares”.

[@bearantarctic5843]

But then it wouldn't be clickbait and then no one would watch the video

[@oflameo8927]

@@bearantarctic5843 I don't know anything about math. I would have clicked on that title anyways because I don't know the difference between an absolute square and a perfect square, and it would have been technically correct when someone who knows what they are looking at, looks at it.

[@pepebriguglio6125]

Why is a list of the first 24 squares (up to 23²) a sufficient list? Already at 24² we get a new set of last two digits, 76. So, when do we know that the list is complete?

[@gerryiles3925]

Yes, you actually need to keep going until you see them start to repeat again backwards after 25^2 (625): 26^2 (676 repeat of 24^2), 27^2 (729 repeat of 23^2) etc.

[@pepebriguglio6125]

@gerryiles3925 Ah, thanks, I didn't know. Nice fact!

[@MrRyanroberson1]

he's getting really lazy lately with these videos, which indicates this channel might not continue for very long

[@pepebriguglio6125]

@@MrRyanroberson1 I think it's his typical mischief. He likes it. And I usually catch it when he does it. But not this time 😅😅

[@Alan-zf2tt]

At some of the earlier replies: hmmm I do not know and please let me explain why. The comments in this post suggest math is being done as well as being a "spectator sport". And that seems a very good thing to happen no?

[@zygoloid]

If m²=...ab is an absolute square then n²=...ba is also square, and their difference is 10(a-b)+(b-a) = 9(a-b) ≤ 81. The difference of two distinct squares m²-n² = (m-n)(m+n) ≥ m+n, so m < m+n ≤ 81, so there are no absolute squares > 6400. Checking those remaining 80 cases is left as an exercise to the reader :-)

[@zygoloid]

For completeness, two other cases must also be considered: - if m² doesn't have at least two digits, m² is an absolute square. - if a=b for all reorderings of m², then m² is a repdigit. But the only repeated digit that a square can end with is 00, which means m=0.

[@evreatic3438]

Agreed: m²-n² = 9(a-b) ≤ 81. Observe: 41²-40² = 81 , so |y²-x²| > 81 for all distinct natural x, y greater than 41. This immediately excludes all 3 or more digit numbers and leaves only 42 (counting from 0) cases to check.

[@howardthompson3543]

26^2 -> 76 Note that this does not effect the (gist of the) proof

[@bertsierra]

Indeed. I spotted the 24^2 -> 76 case as well.

[@disgruntledtoons]

I guess today's solution was not a good place to stop.

[@Qermaq]

Never stop!

[@randomjin9392]

Consider all digits. 2,3,7,8 are out (no square ends with those). Then, we can't have more than one 1,6 or 9 since 11, 66 and 99 fail the mod 4 square test. 5 is out: such square is 0 mod 25, thus must end with 25 or 75 but 2,7 are out. 0 is out: a square must end with an even # of 0s. 4 is fine (so far). Suppose we got a single 1: 4,9 are out: 14 and 91 fail mod 4 test. 6 can only be single, but 61 isn't a square. So 1 is out. For a single 6: only 4,9 are left. 4 fails (46 mod 4), 9 can only be single, but 69 and 96 aren't squares. So 6 is out. For a sole 9, only 4 is left, but 94 fails mod 4 test. What remains is repeated 4. That fails too: 4...4 = 4x(1...1) but 1...1 ends in 11 and fails mod 4. Thus only sole 0,1,4,9 are left. All hail mod 4.

[@fahrenheit2101]

Whoa. A bit messy but very nice. All hail mod 4 indeed.

[@DragonOfThePineForest]

interestingly, while there are no non trivial examples of these perfectly permutable perfect sqares in base 10, there are in other bases. such as base 12. (4*12+1)=49=7^2 (1*12+4)=16=4^2 I'm currently looking for more in this base and others.

[@DragonOfThePineForest]

33 in base 11 is a trivial case but fun nonetheless (3*11+3=36) base nine is the first case where it could be possible to find a number that has more than one 1 and is written entirely with 1's and 0's

[@Alan-zf2tt]

Thank you for sharing your enthusiasm with us - well the world of KZbin really - . It is great to see math being done. My hope is that I absorb these skills subliminally

[@goodplacetostop2973]

Not today

[@MrRyanroberson1]

you had a really good run, it's a shame this won't be a regular thing any more.

[@goodplacetostop2973]

@@MrRyanroberson1Oh don’t worry, it will come back 😂

[@txikitofandango]

24^2 = 26^2 = 76 (mod 100) but the point still stands because 67 is not on the list

[@maxhagenauer24]

Yeah 24^2 totally equals 26^2...

[@Hiltok]

@@maxhagenauer24 mod 100 it totally does. In fact, (25-n)² is congruent to (25+n)² mod 100. I don't know why Michael was sloppy with this (other than to generate comments that feed the algorithm). (25-n)² = 625 - 50n + n² (25+n)² = 625 + 50n + n² (25+n)² - (25-n)² = 100n The same symmetry is in play for squares of 50+/-n and 75+/-n, so the squares of 1 to 24 do indeed cover all possible cases for the last two digits.

[@maxhagenauer24]

@Hiltok I didn't know he was talking about mod 100 and congruence, he still said 24^2 = 26^2. But (25 - n)^2 is not congruent to (25 + n)^2 if n is not a natural number. I wish Michael is the video was less sloppy as well and more regular and "true" because it wasn't at some parts.

[@Hiltok]

@@maxhagenauer24 Convention is 'n' is used for variable name when it is a natural number, just as we use 'x' for variable name when it is a real number. Most people reading maths will know than 'n' means a natural number (or integer) without it being explicitly stated.

[@maxhagenauer24]

@Hiltok That's not always true though, I seen people use n as any real number a couple times but it's generically used as a natural number while x is anything like you said.

[@acelm8437]

Happy new square year! (2025 = 45^2)

[@weeblol4050]

this is the last one you are gonna witness unless you are a kid

[@fahrenheit2101]

@weeblol4050 Oh shit. And they're only gonna get less frequent... At some point, there'll be people living their whole lives between squares :(

[@landsgevaer]

​​@@fahrenheit2101 Sorry to break it to you, but there have already been millions of such people. Sadly, they are all dead already, or dead very very soon... 😢

[@fahrenheit2101]

@@landsgevaer Well yeah, I figured that much, it's just gonna happen all the more often, and eventually with near certainty, whereas before you'd be fairly likely to live through a square year... except... It might be kinda interesting to compare life expectancy with the square gaps over the years. I may be far more wrong than I think.

[@weeblol4050]

@@fahrenheit2101 difference grows linearly every as 2n+1 for every year n ^2 till next its 81 years and then 83 and so on. If Jesus wasnt crucified he had a potential to witness 10 perfect square years

[@minamagdy4126]

I'm now wondering what if there was a restriction on allowable permutations that creating leading zeros aren't allowed. That would trivially add {1,4,9} times powers of 100 as possible candidates, and (per an extension of the proof in the video) nothing else? Also, "a" in the case for zero didn't need to be the largest, but only the first. That would lead you to a contradiction if n>0 there.

[@aadfg0]

It doesn't. For example, 400 can be rearranged to 040.

[@minamagdy4126]

@aadfg0 that's exactly the type of permutation I'm trying to avoid. By saying "avoid permutations that create leading zeros", I mean you can't put a zero left of all non-zero digits (if any, there's the obvious special case of 0) and call it a valid permutation by my extra rules.

[@aadfg0]

@@minamagdy4126 In that case, you're right. If there's a 0, there can't be more than 1 non-zero digit because of a(...)b0, so we get 1, 4, 9 * 10^2n. If there's no 0, proceed as usual.

[@MacHooolahan]

Thanks for another year of maths madness! Stellar content all year thru, as ever x. Merry Xmas from Yorkshire.

[@petersievert6830]

01:44 Pretty funny, that there is exactly one square (24^2=..76) missing before the list would have been exhaustive 😀 (others have already pointed out, why there are no new numbers beyond 25)

[@michaelruiz9451]

Excellent proof!

[@petermcgowan1705]

There's something I love about this type of problem: proving 0, 1, 4 and 9 are absolute squares is easy; proving they're _the_ _only_ absolute squares is not.

[@phatguardian]

I don’t understand the point of these like digit permutation properties… do they provide any deeper meaning or practical applications

[@felipevasconcelos6736]

It's base-dependent, so of course not. Mathematics isn't really about practical applications, though, sometimes it's about making something up and seeing where that takes you.

[@fahrenheit2101]

Almost never. But that's just a slight on recreational mathematics, and an invalid one. There wasn't meant to be a point, so there's no failure to meet some standard. Sometimes it gives you useful tools for other problems, but mostly people do it because it's interesting, fun, and all knowledge is good knowledge.

[@SirNobleIZH]

It helps sharpen your mathematical mind for other, more practical problems

[@natepolidoro4565]

I thought that this would be about squares that are also perfect numbers.

[@matematicacommarcospaulo]

Where do I get this Math Magazine?

[@Bodyknock]

You can maybe streamline the solution a bit by noting that all the digits must be included in the list of digits that are possible for squares mod 100, i.e. all the digits are either 0,1,4,6, or 9.

[@fahrenheit2101]

You still need to go through all those remainders and find the ones which are possible first, which I'd argue is the most laborious bit. And the cases to eliminate would still be the same as the ones he went through. The only bits streamlined are the arguments within those cases, but I don't see how much you'd gain from this, since you'd probably just make roughly the same arguments anyways? (e.g. "00" isn't possible because any other digit could be put in the tens place for a contradiction. those digits being restricted doesn't change anything for this case) Or were you considering a different approach from that point onwards?

[@samdaman2510]

1:40 you wrote 121 instead of 21 on 11^2

[@jackkalver4644]

In base 12, “14” is a perfect square.

[@derwolf7810]

I'm not sure, but ad hoc It seems to be inconsistent, to accept the one digit squares as absolute squares on basis of the trivial permutation (identity), while not accepting to use the identity for all other permutations.

[@JavierSalcedoC]

just reading the absolute square description my brain goes "hell no that doesn't exist" (except for the single digit ones)

[@mrvmurray]

The choice of base is significant. Squareness is independent of base, but permuting digits isn’t. Are there bases which are “better” for absolute squares?

[@mrvmurray]

E.g. 16 with any base >16 is an absolute perfect square.

[@skylardeslypere9909]

Can anyone explain why case 4 is only n=44...4? Why not n=55...544? We only exclude 0,1,6,9 from the digits, why also 2,3,5,7,8?

[@Bodyknock]

3:29 Also 10 should have been boxed since 00 swapped is still 00. (It’s technically the same as 0 being boxed.)

[@Qermaq]

Since all perfect squares end in 0, 1, 4, 5, 6 or 9, you could truncate your starting list considerably as these are the only digits allowable.

[@angeldude101]

Either every perfect square is an absolute square in a sufficiently large base, or 10, 40, and 90 aren't squares, so only 0 is an absolute square. (I'm generally not a huge fan of digit-dependant stuff.) Probably the most interesting bastardisation of this problem is the situation where 4, 16, 64, and 256 are all absolute squares... in base 4. When the digits are all 0s except for a single 1, and the base itself is a square, then every permutation of those 0s and 1 also makes a square, meaning that, if you include leading 0s, then square bases all have infinitely many absolute squares, while non-square bases only have one, which is 0.

[@manudude02]

shouldn't the question have stated if ALL permutations are square. As stated all squares are perfect squares since the square itself is a permutation.

[@Epyxoid]

1:48 "I only went up to 23, because *I think* we get all of the perfect squares mod 100 just by looking at that list" and then later you based a ton of things off this list, which lacks any deeper evidence of being meaningful.

[@MGoebel-c8e]

He’s just getting worse. There was always a tendency on this channel to misfocus, especially in number theory, but this really tops it. Instead of nitty-grittily proving that a string of 4 is not a square number (genius!), he should have focused on the one-and-only clear core of the entire argument, which is the periodicity of the final digits mod 100. This is literally the only thing that is interesting about this. I find it absolutely flabbergasting how Penn manages to copy other people’s arguments without at least making sure to do that properly. Why does he insist on making videos about a subject matter that he clearly has no grasp of?

[@fahrenheit2101]

A tad harsh. The core of this argument is already inevitably nitty gritty by needing to exhaust all the quadratic residues mod 100, and then further exhaust a bunch of cases. That's exactly what he did, albeit somewhat laboriously with not much extra value beyond the first example (since all others are effectively an identical argument). And how would he avoid the string of 4s in this argument? That also seems an inevitable extra hurdle in this approach. So are you suggesting an alternate approach, or that the string of 4s should have been skipped for some reason, or am I missing something?

[@Noconstitutionfordemocrats1]

Poppycock!

[@petersiracusa5281]

In case 1, i miss why checking 0a0 is exhaustive.

[@KeimoKissa]

Is n is congruent to 0 mod 100, it will end in 00. For the number to be absolutely square, every permutation, including 0a0, will have to be a square. If it's not, not every permutation is a square, so the number can't be absolutely square.

[@tangoyip2016]

43^2=1849

[@wesleydeng71]

Audio is quite low.

[@fahrenheit2101]

I think the presentation could've been better, but I like the argument. Firstly, yet another conveniently obvious mistake - you missed 24^2 = 76 (mod 100). It also would've been very nice to see WHY we're allowed to stop there, or else we're essentially just relying on a very exhaustive check. As it happens, you can show with not too much effort that the remainders mod 100 are symmetrical around 25, and also around 50 (you always get symmetry around the halfway point, so 25 is the main thing to show). It would've been more convenient to write 00, 01, 04, 09 for the remainders mod 100, since that makes it obvious why you can't swap digits for 1, 4 and 9. It was also sort of tedious to go over the exact details for each case after 00, since they all boiled down to the same idea - you could easily leave that as an exercise to the viewer.

[@marcellosalis5063]

All numbers in base 2 are absolute squares, I guess.

[@fahrenheit2101]

No? Not at all? I mean, you require the number to be a square first of all. Being a square is not base-dependent, the very same numbers are squares irrespective of what base you put them in. So in binary, 1 is absolutely square, but 10 already isn't, since it isn't even square. The next square is 100, but this isn't absolutely square, since 010 isn't square. Considering things mod 4 (still "mod 100", but just in base 2 now), the allowed remainders are 00 and 01, and this is easy to check. However, 01 already fails at being swappable, so you can only have 00. However, unless the number is 0 itself, it can now be rearranged to end in 10, which isn't allowed. So, in summary, 0 and 1 are the ONLY absolute squares in base-2. Any number with 2 or more digits either has a 0 and a 1, or consists of only 1s. The latter will end in 11 which isn't allowed, and the former can be rearranged to end in 10, which isn't allowed.

[@marcellosalis5063]

@fahrenheit2101 Sorry, I meant all "square" numbers in base 2 are absolute squares.

[@fahrenheit2101]

@@marcellosalis5063 But I just proved that this is also false. In particular, it fails for 100, or 4, whatever you prefer to call it.

[@angeldude101]

In base 4, every power of 4 is an absolute square.

[@fahrenheit2101]

@@angeldude101 Lol, this one actually is true, finally. And of course, this applies in square bases in general.

[@bwoy12345]

I dont get it.

[@topquark22]

Any permutation of digits is a red flag, becauwe it depends on the base. So, this falls into the realm of recreational mathematics, not pure mathematics. That said, I haven't yet watched the entire video.

[@supratimsantra5413]

Again a great motivation you have made through your unique style lecture series vidioes.... ..it is almost heroine like adventure rush when we get your notification at our end

[@fernandobeiroa]

16*16=256 25*25=625

[@millwrightrick1]

How about 526? Or 652? They don't work.

[@Hiltok]

That is only two of the six permutations of 2,5,6. What about 526, 562, 265, 652?

[@fahrenheit2101]

All permutations, not any.

[@jacemandt]

The definition is ambiguous as stated. The word "any" here could mean "every" or it could mean "there exists a". I interpreted it the second way at first, in which case this proof doesn't make sense.