Can you guess the trick for this problem from the thumbnail?

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[@goodplacetostop2973]

0:06 Phoebe Bridgers 13:30 T-shirt change 19:26 Good Place To Stop

[@advaykumar9726]

U won by 6 seconds

[@JalebJay]

13:30 has a magic trick in it.

[@kostasbr51]

You are the fastest gun west of Pecos.

[@violetasuklevska9074]

My initial guess was to multiply by 4 to get some perfect squares and then a difference of squares, but I now realize that that's unnecessary because the 4s anyway cancel out and you can factor n(n+1)-q(q+1)=(n-q)(n+q+1). It amazes me that I haven't seen that before.

[@violetasuklevska9074]

Seeing that he's using triangular numbers I'll show my proof since I've already gone in a different direction. Sol. We reorder the terms as follows: p(p+1)=n(n+1)-q(q+1)=(n-q)(n+q+1) Since the LHS is positive the RHS must also be positive, therefore n-q>0 Set k=n-q>0 and substitute for k instead of n in the above equation to get: p(p+1)=k(2q+k+1) If p l k then p==2q+k+1>= 2q+p+1>p+1 contradiction. => p l 2q+k+1 and p= kt=p+1, now we have the following: (pt)t=(2q+k+1)t=2qt+p+1+t => 2qt=(pt^2-p)-(t+1)=p(t-1)(t+1)-(t+1)=(t+1)(p(t-1)-1) => q=(t+1)(p(t-1)-1)/2t - t=1 gives q=-1 which isn't possible - t=2 gives q=3(p-1)/4; since q is prime => p-1=4 and we get a solution (p,q,n)=(5,3,6) - t=3 gives q=2(2p-1)/3; since q is prime => 2p-1=3 and we get a solution (p,q,n)=(2,2,3) - t=4 gives q=5(3p-1)/8; since q is prime => 3p-1=8 and we get a solution (p,q,n)=(3,5,6) - We'll prove that there are no solution for t>4 Since GCD(t,t+1)=1 and t>1 no factor of t divides t+1. In order for q to be an integer we must have: t l p(t-1)-1 and t= q=(t+1)(p(t-1)-1)/2t=(t+1)tr/2t=(t+1)r/2 We get that (t+1)r/2 must be prime. Since a product of two integers greater than 1 gives a composite number and t+1>2 then r=1 or r=2. - r=1 gives t=p(t-1)-1>=2(t-1)-1=t+(t-3)>t contradiction. - r=2 gives 2t=p(t-1)-1 If p=2 we have: 2t=2(t-1)-1=2t-3 contradiction. If p>2 we have: 2t>=3(t-1)-1=2t+(t-4)>2t contradiction. We deduce that the only solutions are: (p,q,n)=(2,2,3); (3,5,6); (5,3,6).

[@pratikmaity4315]

2:14 another way to show that nn(n+1). Or (p+q)(p+q+1)>n(n+1) which obviously shows p+q>n so n

[@riadsouissi]

Write n=p+q-x. x > 0 because otherwise, by replacing n in the original equation, we get 2pq=q+1 then q+p-x>=q+1 which gives x

[@yoav613]

Short solution: wlog p>= q and n>p so n= p+m (m is natural number). Then p(p+1) +q(q+1) =(p+m)(p+m+1) so we get q^2+q=m^2+m+2pm. We see that m=q) if p>q then p=q+m+1 and 2m=q-m so q=3m since q is prime m=1 q=3 and p=5 (or the symmetric solutionp=3 q=5) if p=q so m=p-1=1 so p=q=2 is the solution

[@goblinkoma]

I really thought the trick was to reduce the equation mod(bigger prime), but i wasnt sure where to go from there.. Interestion solution you have there!

[@pierrechardaire8525]

At 10:50 for the case n=1 he gets n=p+q+1, but he has already shown before that n < p+q+1. So, no need to do anything more to show that m = 1 does not work .

[@DanielWalvin]

Even more obviously: n + q + 1 = mp, m = 1 => n + q + 1 = p. But p < n, so we have a contradiction.

[@BiscuitZombies]

Yeah I was thinking that

[@tahirimathscienceonlinetea4273]

Thanks Michael you did a great job with this monster problem keep going man

[@s4623]

8:56 m = 4 gives 4p

[@bobh6728]

The inequality on the board is strictly less than, not or equals. He misspoke, I think.

[@TrimutiusToo]

Yeah it has to br strictly less, less than or equal was second step but first step was strictly less

[@gavinxm]

yep, this is exactily where I went

[@matthieumoussiegt]

at 5:26 for case 1 : you just need to say that if p|n-q then pp+q and we have a contradiction

[@samwalko]

From the thumbnail I can see that a solution set can be found by finding 1 + 2 + ... + p = q + 1 = n. Basically, find a p'th triangular number that is one more than another prime q, add it to the q'th triangular number, and you get the nth = (q+1)th triangular number. But finding p,q that fulfill this and proving the exclusivity of this class of solutions is a whole other matter.

[@somasahu1234]

And here's another amazing NT prob.

[@manucitomx]

Why the wardrobe change? Thank you, professor!

[@thatmathnerdteachesphysics5180]

He did it in two go's

[@matthieumoussiegt]

at 9:22 for subcase 1 you have if m=1 n+q+1=p so p>n contradiction

[@OscarCunningham]

8:55 "We can't have 4p be less than or equal to 3p+2 when p is a prime." This is false, and then later we find (p,q,n) = (2,2,3) in the m=3 case?

[@bobh6728]

I think he misspoke because the expression on the board is strictly less than.

[@OscarCunningham]

@@bobh6728 Thanks, that makes sense.

[@jacobgoldman5780]

For subcase 1, once you get that n=p+q+1 can’t we discard this case as we previously proved that n

[@hugovandersanden3058]

I kinda enjoyed this, but I feel it missed an opportunity to convey a much more visceral appreciation of the role that the relations between the factors play. It could perhaps have started with a worked example (p=11, say) - writing out the factors of p and p+1, showing that p cannot divide n-q *because* n+q+1 is more than (n-q)+1, and therefore for some m we must have n-q=(p+1)/m, n+q+1=mp. Then ask what m can be, and proceed much as before (though some of the later parts now become simpler).

[@weonlygoupfromhere7369]

That intro is awesome

[@hassanalihusseini1717]

I hate number theory because always I feel so stupid. But I am addicted to Prof. Penn's videos, so I cannot skip them. :-)

[@robertfoskin9668]

Michael needs to hire a scene photographer to maintain continuity

[@CM63_France]

Hi, For fun: 4:39 : "ok, great", 11:44 : "ok, great", 13:30 : good place to change t-shirt.

[@kevinmartin7760]

It might have saved a bit of time to do the first two steps of each subcase all at once by keeping m as an unknown a bit longer, even after determining which values were valid. You get two equations: p = m(n-q)-1 and n+q+1 = np Multiply the first by m, making its LHS equal to the RHS of the second, and then equate the two other half-equations: n+q+1 = (m^2)(n-q)-m Combine terms and isolate the q term: (m^2+1)q = (m^2-1)n-m-1 = (m-1)(m+1)n-(m+1) = (m+1)(n(m-1)-1) If m is even, GCD(m^2+1, m-1) = 1 so (m+1)|q but that requires q=m+1. If m is odd, GCD(m^2+1, m-1) = 2, so you can divide both sides by 2 and you then get ((m+1)/2)|q but that requires q=(m+1)/2. I'm not sure if this can be pushed further to derive the limits on the value of m independently of the first 3 or 4 minutes of the video. It certainly eliminates m=1 because that yields q=1 which is not a prime. Actually, yes, (perhaps without needing any assumptions about the relative sizes of p, q, and n). In the even-m case you plug q=m+1 back into the combined equation, divide both sides by (m+1), and isolate n: n=(m^2+2)/(m-1) Do the long division on the RHS, yielding n=m+1+3/(m-1) For the second term to be integral m must be 2 or 4, yielding p=3 or 5 and (in both cases) n=6. These lead to the two non-symmetric solutions. In the odd-m case you plug q=(m+1)/2 back into the combined equation, divide both sides by (m+1) and isolate n: 2n=(m^2+3)/(m-1) 2n=m+1+4/(m-1) For n to be integral, the 4/(m-1) term must be even, and so m must be 2 or 3 but the 2 is not odd so 3 is the only possibility. This yields q=2, n=3 which leads to the symmetric solution.

[@emanuellandeholm5657]

3:02 I can kind of see that 1 + 2 + ... + n (>= p + q + 1) is >= 1 + 2 + ... + max(p, q), but I need a proof for the inequality used here.

[@natepolidoro4565]

8:50 But m=4 is possible if p=2, which means q=2 and therefore n=3.

[@neilgerace355]

Is the trick something to do with the triangle inequality?

[@xCorvus7x]

3:31 I get that here in the sum of 1 up to n the next q numbers after p (p+1, p+2,..., p+q) are replaced with 1+2+...+q, after which the sum is cut off. Since n is assumed bigger than p+q, we have more than p+q terms in this sum and we are only making existing terms smaller, yielding the stated inequality. But doesn't this also rule out the case of n = p+q? Why would we need an additional p+q+1st term, which is discarded in that step, for this inequality to hold? _Edit: This is further supported by the fact that there's no solution for n=p+q as is later verified (the m=1 case)._

[@thomashoffmann8857]

Maybe if p = q, then the equality would be possible. +1 will make sure that the strict inequality holds

[@xCorvus7x]

@@thomashoffmann8857 But even then, for n=2p, 1+2+...+p+(p+1)+...+2p is already strictly greater than 1+2+...+p + 1+2+...+p, the exact difference being p^2.

[@thomashoffmann8857]

@@xCorvus7x you are right. q must at least be 2,thus strict inequality should hold

[@xCorvus7x]

@@thomashoffmann8857 Well, you'd have to at least include 0 to the numbers p and q could be so that 1+...+2p is not strictly greater than 1+...+p + 1+...+p. p and q could both be 1 and the inequality holds still: 1+2 > 1+1. For the fun of it: 1+...+2p sums to a triangular number equal to 2p(2p+1)/2 = p(2p+1) = 2p^2 + p. For all p>0 this is bigger than p(p+1) = p^2 + p (combining twice p(p+1)/2) by a difference of p^2.

[@leif1075]

I don't see why anyone would EVER deduce that n is less than p plus q plus 1..there's no way to see that ..

[@Happy_Abe]

Instant shirt change

[@scebsy6524]

love the phoebe bridgers shirt

[@joyofmath654]

13:29 why the wardrobe change?

[@andychow5509]

I don't think I could have ever solved this problem by myself... sad.

[@shahinjahanlu2199]

Great

[@adityaekbote8498]

Nice

[@eeee123478]

5:46 not really 5 × 6 = 10 × 3 10 is devided by 5, but 3 is not devided by 6

[@tomatrix7525]

You’re example isn’t pertinent however. If p divides n-q, and we know n+q+1 > p (see first inequality he proves by contradiction and it is clear) that means that since n+q+1 must divide LHS, and cannot divide p as it is bigger, it must divide p+1

[@Yxiomel]

Actually, he has the division the other way around, so it would be 6 is divided by 3, which is true.

[@dbmalesani]

Careful: using your example, Penn is not saying that 6 divides 3, but that 3 divides 6, which is correct. We start from p(p+1) = (n-q)(n+q+1); what Penn says is, if p divides (n-q), then (n+q+1) divides (p+1) (not "p+1 divides n+q+1"). His statement is correct: if p divides n-1, then (n-q)/p = m is an integer, hence m(n+q+1) = p+1, that is (n+q+1) divides (p+1).

[@eeee123478]

oh sorry my English is so poor :(

[@dbmalesani]

@@eeee123478 I had also got confused the first time I heard it - don't worry :)

[@udic01]

9:02 we can't have 4p equal 3p+2 when p is a prime?! Well i just think that p=2 is a prime...

[@muhammedmrtkn]

But mp

[@udic01]

@@muhammedmrtkn i know. But that's not what he said. He should have been more precise.

[@notnilc2107]

I took a break from uni this year because of how they handled online learning last year. I am supposed to be reviewing maths in preparation for next year. I have forced myself to isolate myself from all social interactions outside of family because I keep putting it off. I have bought a bunch of digital textbooks. Yet I still do not study. I am considering buying physical hard copies because that's how I studied in high school but I feel like it'll be a waste of money. I think this is a cry for help. I love maths but a part of me wants to abandon it all and just become a farmer like my dad always wanted but I genuinely feel like I will suicide 10 years down the line if I do so.

[@goodplacetostop2973]

Efforts make the path, not talent. Time will recognize you. If you don't feel comfortable with digital books and digital copies, you can try to buy physical books and study in places like univerisity library or anything that cean remind you the period you were able to focus. Solutions exist, even when you feel they don't. Seek help from your friends, family, any person you trust, or a mental health professional if needed.

[@stefanalecu9532]

You don't have to think so pessimistically, all hope is not lost. You have a year to prepare, not like a few weeks. I know how it feels to struggle with studying because I've been there. In your mind, you have this goal of reviewing everything in one year, which is A LOT of work, an insurmountable amount. You should instead focus on what you can do daily, have daily goals like idk, today I will review complex numbers and I will do some exercises. If you track your progress in small steps, it's easier to graph out and conceptualize in your mind your progress, whereas if you have big goals like reviewing Calculus 1 and 2 this month, then there will be big spikes as far as your progress is concerned and you can't look back and appreciate the small consistent efforts that will get you to your ultimate goal. Mathematics isn't a chore and it shouldn't feel like it and I feel bad for you getting to the point of considering suicide in a few years. Look, even Michael keeps making videos daily even though his hand isn't in a really good shape. If he can make such wonderful videos for us consistently, you can do some exercises daily as well. Keep your head up, do easy exercises, take it easy. I believe in you and your progress and I truly hope my words have been helpful.

[@barakfriedman1262]

I've been where you seem to be. I struggled with motivation, and barely went through my B.Sc, in Engineering, with more inertia than anything else. Had I taken a year a break, I probably wouldn't come back. What I realized later, as life took it's usual turns and gave it's usual surprises, is that when I found my true calling, which was in a different area altogether, I found my motivation at last. Maybe the lack of motivation to continue studying is a temporary thing, and maybe all you need is someone who can council you on how best to study on your own. This is very different to studying in class or with friends. But also consider the deeper reasons for how you feel, and do not ignore it. You could benefit from a talk with someone who can advise you on shifting your focus to some tangent field that you have more motivation to pursue. All in all, remember that we build ourselves from hardships and struggle, not from success.

[@somasahu1234]

Chill man! After a few years u will realise that u were just overthinking n all these r bullshit , Everything will be fine!

[@ollisd7230]

I didn't unterstand the conclusion at 3:00. Maybe it's easy, but... Can anybody help me 🤗?

[@phiefer3]

the assumption here is that n >= p+q+1, which means that the sum of 1+2+2...+n numbers has at least p+q terms (since n is larger than p+q), so the first p terms is just (1+2+...+p), then the next q terms would be (p+1) + (p+2) + .....+ (p+q), which are all strictly larger than 1+2+....+q, so by replacing them like that, the result must be strictly smaller than the original sum of n numbers. As an example, if p was 8, and q was 6 (I know they're not prime, but that doesn't actually matter for this part), this means that we're asserting that n >= 15, so n(n+1) = 2(1+2+3+4+5+6+7+8+9+10+11+12+13+14+15) now if we break this off after 8 (which is our p) and start recounting we get 2(1+2+3+4+5+6+7+8+1+2+3+4+5+6) see how this must be strictly smaller than the first since all of the terms either stayed the same or got smaller (we actually lost one as well...), this is where the inequality came from, then just as we turned n(n+1) into the other sum, we can split this into 2 sums (1to8, and 1to 6) and turn them back into p(p+1) and q(q+1), and see that the inequality violates the base equation that we started from, which means that n cannot be >= to p+q+1 so it must be less than them.

[@ollisd7230]

@@phiefer3 got it 😀! Thanks a lot for your fast answer 👍

[@nevokrien95]

U missed a case by asuming m isny 4 if p and q are 2 then n is 3 and the eqution holds