The smallest such prime... 

If anyone is wondering: a=6083, b=78, p=23

4:58 "See what I did there?" mp = Michael Penn?

For subcase 1 you didn't need to involve x at all because (p-1)^2

Your presentation style is unbeatable. So enthralling, It’s usually you find a good solution but bad presentation, or the inverse, but rarely both.

That was a fun one, but the hint totally gave the game away. :-) So the problem statement is equivalent to p³ = b⁴ - a² = (b² - a)(b² + a). Since p is prime, p³ can only be split up two ways, p * p² and 1 * p³. And since a and b are positive integers, b² - a < b² + a, so the factors can be mapped to the terms in only two ways. Case 1: b² - a = p, b² + a = p² Add up both equations yields 2b² = p(1+p). The case p=2 can be discarded quickly, so p must be odd. Therefore 1+p is even, so we can divide the equation by 2 and get b² = p (1+p)/2 (where the last term is an integer). Therefore b² is divisible by p. Since p is an integer, it follows that b is divisible by p. So there is some integer n with b = np, so n² p² = p (1+p)/2 n² p = (1+p)/2. So (1+p)/2 is divisible by p. But (1+p)/2 < p for all p >= 3, so that can't be right. So there can be no solutions in this case. Case 2: b² - a = 1, b² + a = p³ Add up both equations yields 2b² = 1 + p³ --> b² = (1 + p³)/2. This again discounts p=2 as possibility, since in that case the RHS is not natural. I had no idea how to proceed from here, so I just wrote down a table of prime numbers, their cubes, the RHS of the equation, and their square roots. By checking every known prime number, I was able to find that p=23 leads to b=78, and a=6083. A quick check later confirmed this to be a solution. And since the other case found none, and this case found none before this one, the smallest solution is: 6083² + 23³ = 78⁴

Once one has p³ = 2b² − 1, a quick analysis modulo 8 yields p ≡ ±1 mod 8. Then one just needs to test p = 7, 17, then 23 works. That's much faster, but Michael's solution could have been faster if the minimum p were much larger.

National mathematics day. 22 dec.🇮🇳🇮🇳🇮🇳

p = 2 doesn't work for case 2 because if b^2 + a = 8, and b^2 - a = 1, then 2a = 7. But a is an integer: contradiction.

National mathematics Day :)

The relatively prime case is easier. You get (p-1)^2

Thanks to the guy who invented internet. I get to watch and learn such cool stuff.

Happy national mathematics day! :) Nice problem!

14:30 *Michael Penn:* Tony, there was no other way. We're in the endgame now...

So I made a spreadsheet of the possibilities as we found in the only subcase that works. x, p=6x^2-1, y=sqrt((p^2-p+1)/3) As it turns out, x=2 is the only x

Case 1 can be made easier by: since p | b, p² | b², but p² ∤ p(p+1), so we have a contradiction. Also, I thought of finding all values of p that satisfy this equation and there may actually not be so many solutions. I tried to narrow down the possibilities of p and got this: p + 1 = 6x² p = 6x² - 1 3y² = p² - p + 1 = (36x⁴ - 12x² + 1) - (6x² - 1) + 1 = 36x⁴ - 18x² + 3 y^2 = 12x⁴ - 6x² + 1 Since y² is odd, y ≡ 1 (mod 8), thus x is even. Let 2w = x, then p = 24w² - 1 (this means that p ≡ 23 (mod 24)) y^2 = 192w⁴ - 24w² + 1 (this polynomial cannot be factorised without going into complex numbers) Additionally, since for any perfect square k^2 ≡ -1, 0, 1 (mod 5), w ≡ -1, 0, 1 (mod 5). (And in particular, y ≡ 1 (mod 100) if w ≡ 0 (mod 5) and y ≡ 69 (mod 100) if w ≡ ±1 (mod 5), though I'm not sure how helpful this is.) In fact, I plugged in values for w = 4, 5, 6, 10, 11 (9 gives a composite value for 24w² - 1) and none of them gave a perfect square for 192w⁴ - 24w² + 1.

In subcase 1, we need not consider x at all to see that y^2=p^2-p+1 implies (p-1)^2 < y^2 < p^2

p=23 is the only prime that works for p < 1.000.000.

37,002,889+12,167=37,015,056 or 6,083^2+23^3=78^4

Penn has become a Teller - magical problem!

..amazing ..so many layers of concepts revealed as Michael Penn went on..

Find all m,n natural number such that (3^m) +23=2^n

I really enjoyed this problem, Michael, thank you! I would like to suggest another solution, that doesn't need the use of the gcd's tricks. Same procedure up to minute 06:55, so we know 2b²=p³ + 1. Since p is odd, let's write p=2n-1. Then we have: 2b² = p³ + 1 = (8n³ - 12n² + 6n - 1) + 1 = 8n³ - 12n² + 6n b² = 4n³ - 6n² + 3n = n × (4n² - 6n + 3) Let's factorize n in prime numbers: If c is a prime dividing n, then c also divides 4n² - 6n, so c divides (4n² - 6n + 3) ONLY IF c=3. Any prime d≠3 dividing n IS NOT a factor of (4n² - 6n + 3), so d must have an even exponent in the factorization of n, else n×(4n² - 6n + 3) CANNOT be a perfect square. In other words, the only prime number than can appear with an odd exponent in the factorization of n is the number 3. So we must search for n such that 2n-1 is prime and: 1) n is a perfect square, OR 2) n is 3 times a perfect square Then n is in the set {3, 4, 9, 12, 16, ...} The first value for n in this set that works is n=12: n = 3×2² (4n² - 6n + 3) = 4×144 - 6×12 + 3 = 507 = 3×169 = 3×13² b² = (3×2²) × (3×13²) = (2 × 3 × 13)² That yields p = 2×12 - 1 = 23

The way I discarded the first case where 2b² = p(p+1) you have b² = p(p+1)/2. Geometrically that can't work since a stick of length p that can't be broken down (because p can't be factored) to fit into a square of size b², because square root of b is necessarily smaller than p (geometric mean lies between p and (p+1)/2 ). There you go. Not as elegant as how you gave it, but it's what I came up with.

You eliminated p=2 only in respect to the first factorization, cannot bring that result into another case I think

I was delighted to notice that for Case 1, we find that b^2 equals p(p+1)/2, which is 1+2+...+p (a.k.a. the triangle number of p). This harkens back to the balancing number video. It's easy to show that balancing numbers are the square roots of triangle numbers (though this wasn't mentioned in the video). Thus for Case 1, b has to be a balancing number! Now, it appears that the sequence of numbers whose triangle number is a perfect square (8, 49, 288, 1681, 9800, 57121, ...) contains only even numbers alternating with odd perfect squares -- i.e. no primes! Thus, Case 1 yields no solutions. For my next trick, I need to prove that last assertion...

My first thought is p^3 = b^4 - a^2 = (b^2)^2 - a^2 = (b^2 - a)(b^2 + a). Examine small primes in order to find a solution. (b^2 -a) is obviously less than (b^2 + a) so it has to be either 1 or p. If we assume p is 2, we get b^2 - a = 1 or 2 and b^2 + a = 2 or 4. This gives 2 b^2 = 3 or 6, which has no solutions in Z. So p must be an odd prime. Just check all of the small ones until you get a hit. p|b^2 => p|b, this is only true for prime p. Obviously 50|100 but 50 does not | 10.

We've got the smallest solution, but is it the only one? I found no other below 10^15 (Python is fast!). I suspect it's the only solution.

He always knows a good place to stop

I feel no shame in admitting I wasn't able to solve this on my own. Ok maybe a little.

13:29 I find it sort of confusing bounding y^2 in terms of x. Could have just written as (p-1)^2

One of the best videos I have found!

We proved that IF the solution exists, thus it has to be 23. But we could have that the solution doesn't exist, right?

Looks like my approach was basically the same as Michael's: a²+p³=b⁴ p³=(b²+a)(b²-a) Case 1: b²+a=p² b²-a=p 2b²=p(p+1) p would make RHS ≥ (p+1)p, so n+1=p, but then RHS = p(p-2)). Case B: gcd = 3: p+1=6m² p=6m²-1 Try values of m: p=5: no p=23: yes, 23²-23+1=507=3.13² p=23, b=78, a=6083

I need help Suppose p>2 and a and b are coprime with p then Rearrenging we get that p^3=(b^2)^2-a^2 By LTE we get that 3=v_p((b^2)^2-a^2)=v_p(b^2-a)+v_p(2). Since v_p(2) for p>2 is 0 then v_p(b^2-a)=3. In other words p^3|b^2-a and that b^2-a >= p^3 Since p^3=(b^2-a)(b^2+a)>=p^3(b^2+a) which means that b^2+a=1 but this is ridiculous. What did I do wrong?

Lazy way: taking b^2-a=1, b^2+a=p^3, solve for b^2 = (p^3+1)/2. p=23 is the first prime that works, b=78, a=6083

Great problem!

This one required some patience, but nonetheless nice problem

nicely donee !! :) great job !

16:42 My comment was shadowbanned by RU-vid so let’s try this again... Geometry homework today! Good luck and have a good day... Squares are constructed externally on the sides of an arbitrary quadrilateral. Show that the line segments joining the centers of opposite squares lie on perpendicular lines and are of equal length.

ok so (I'm omitting all the exponentiation symbols : p3 means p^3 etc.. but 2p still means 2*p.) p3 = b4 - a2 = ( b2-a)(b2+a) So either: 1) b2-a=1 and b2+a=p3 or 2) b2-a = p and b2+a = p2 Since b2+a == b2-a + 2a --> For 1) we get that 1+2a = p3 so a = ( p3-1)/2 and b2 = (p3+1)/2 For 2) we get p + 2a = p2 --> a = (p2 - p) /2 = p (p-1) / 2 and b2 = p (p+1) / 2 Now, by successively trying for p=2,3,... ; I do get that the smallest solution for p is p=23, a=6083, b=78 I can get to that solution with just pen and paper but it is tedious. I used a spreadsheet and got the solution in minutes. Using a simple calculator would also get me the solution in just a few more minutes. I don't really know how to reduce the amount of manual calcul to do to get to the solution.

So 6083^2 + 23^3 = 78^4. Yay science!

Sounds like you chopped off some explanation at the beginning there!

Prof Michael Can you check this problem out Define a positive integer $n$ to be a factorial tail if there is some positive integer $m$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails? It's such a beautiful question :) also involves floor functions

Is the symbol Michael uses for "contradiction" a standard one? I've never seen it before. 🤔

After I got 2*b^2=p^3+1, I just checked b and p mod 3,5,8. And had the following conditions: p must be 1,5,7 mod(8) p cannot be 1 mod(3) p cannot be 2 mod(5) the first prime that verifies these 3 conditions is p=23 which also verifies the equation.

When I came to case 2 I just started to calc sqrt((p^3+1)/2) for small primes which gave me 23 as the smallest solution. I had more test cases but got the solution in less time. btw: I as far as I can see 23 is also the ONLY prime number which solves this problem.

4:58 mp michael penn

I ran a Python script to check all possible combinations of a, b, and p up to a certain b-value, and I only found one combination: p=23, a=6083, and b=78 when searching up to a b-value of 3000. The numbers are getting quite large and the script runs so slow at that point, even with all 16 threads running on my Ryzen 5800X.

Very nice problem involving lots of elementary number theory!

Honestly one can just start putting in primes after we get the relatively simple criterion that (p^3+1)/2 has to be a square, since we're just looking for the smallest solution... not as elegant, but way faster

Wow this is something beautiful and incredible thank you for this video

Lol why I think you should put your silver play button above the blackboard

any idea about any other possible solutions? Exhaustive search for x up to 10^8 yielded none.

best video ever! i could actually see myself solving this myself

that was amazing

I like to write my thoughts before seeing too many hints: p^3=(b^2 - a)(b^2 +a) so b^2-a must be 1 and b^2+a must be minimized. I imagine just picking the smallest square other than 1 to be equal to b^2 would work. b=2, a=3 so p=7 Edit: I see I just forgot the cubed ... so b^2-a does not have to be 1. Sorry to everyone who has to read my bad ideas

p=23, OK, but what about a and b?

Would it not be faster to write a script that tests different primes till it finds a solution?

what is the inspiration to look at the GCD of the 2 numbers?

Aren't there 2 more cases; b^2-a being p^2, b^2+a being p and b^2-a being p^3, b^2+a being 1?

i'm convinced the 23 enigma is real

Wait a minute at 2:25 why can't he switch the cases he forgot..you can have p divide the bigger term and p squared divide the b squared minus a..I dont see why not. Same with p cubed and 1 factors..

9:48 isnt gcd of p+1 and 3 = 1?

Jesus, I did this thinking before the video, and thought I had to prove each prime. Glad to know I’m not the only one totally lost on that front. Also, this is my solution using the last digits of numbers as my contradictions. The first step is the factoring then proving that bb-a = 1 not p Since that would say p(p+1)=2bb which means p2M=2bb or pM=bb so p=b since m doesnt have p as factor since ita smaller than p, but p=b means p+1 = 2p Ppp= bb + a And bb - a = 1 So ( ppp+1 )/2 = bb, a square A square only ends in 014569 But primes only end in 1379 Then, this function shows that the only primes that can work are those ending in 13 or 9, and bb ends in 014 Ppp -1 )/2 = a But bb - a = 1 Therefore a can only end in 903 The only primes that do this end in 0 or 3 This was tricky: if a ends in 0 Then p ends in 1 But then appp = ???10 = 2aa +a = ??00 + a Therefore a ends in 10 But if a ends in 10, bb ends in 11 Since bb-a = exactly 1 But no number squared ends in 11 Proving that is confusing but generally the first two digits of square are created by {aa+20ab} where a and b are the digit of the ones and the tens places. There is no combination of the digits a and b that make that end in 11, since a must be 1 or 9, but that means 1+20b ends in 11, so 2b ends in 1, or 81+180b ends in 11 which brute force shows simply doesnt. Therefore only primes ending in 3 will ever work. There is no way to show what the next prime beyond 23 would be, but I tried: p is z10 +3 And putting that into ( ppp+1) /2 equals a square, perhaps you could show which values of z are possible and more importantly why.

How do you prove the GCD trick?

We could also factorize p as p^(1 / 4 ) x p(3 /4) , can't we ?

Great!

8:30 I think this is the proof of the property used but is it correct? Let gcd(a,b)=k and let gcd(a,b+an)=m where n is an integer By definition of the gcd, we know k is the greatest natural number such that k|a and k|b simultaneously. Similarly, m is the greatest natural number to divide both a and b+an at the same time. But if m|a, then m|b+an implies m|b as a*n is a multiple of m -and therefore m is the greatest† number to divide both a and b and so is their gcd and therefore m=k- QED But I think just because m|a and m|b doesn't imply it is the _greatest_ number to do so and it might not be the gcd of a and b necessarily. Can someone please help me? *EDIT* : The crossed-out part is the non-sequitur part of the original attempt at a proof †We know both m and k divide a and b and k is the greatest possible number that can do so. So m

MP = Michael Penn

please someone explain this 14:17

Good

Like it. Especially because the answer isn't 0,1 or 2.

Hi everyone, I'm not sure if this is the place, but I've wanted to ask for advice. I like maths, and I really do find them enjoyable, through videos or pop-sci books, and I think I'd be pretty good at it; however, due to personal issues I can't study that (or related topics) in college right now. What are some books, videos, lectures to help me get started learning mathematics? I'm not sure if I'm going to major in it, but I really do want to learn more of it! P.S. Does anyone know about career options?

I hate this problem. Relentess hopeless handwaving that makes no sense. Just like the worst problems from school.

Why would anyone think to break up p into p squared times p ..i dont see anyone thinking of that..

Good!

Which semester is this?

Micheal you are my drug 🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩

Can anyone help mewith the GCD trick....can anyone help..plsssss

Beautiful ❤️

that was not a good places to stop. You have tested the values in the equation.

Are there infinitely many such primes?

So don t be late 🙏

33333rd view, lol

Bruh why am I wasting my early teenager days with calculus.

First comment! Nice vid!

Wrong, if p is natural, then equality holds for p = 8, then a = 28, b = 6, if p belongs to the set of integers, then for p = -9, a = 45, b = 6

Bruh why am I wasting my early teenager days with calculus.