It should be m+1 at 4:58

@krisbrandenberger544

Жыл бұрын

That's exactly right ✅! And it would have been okay to prove by induction that m+1=2.

"n equals 1 is zero" Michael Penn, 2022

@emanuellandeholm5657

Жыл бұрын

Deepity or Derpity, you be the judge

@sladebaker9882

Жыл бұрын

Nx4-4 is easier for finding squares

Just for fun: the limit as n approaches -∞ is 1, which is a square.

@HershO.

Жыл бұрын

I suppose he missed a solution then lmao

@rocky171986

Жыл бұрын

2^n yes, but not (n-1)*2^n, the limit is still -inf

@jursamaj

Жыл бұрын

@@rocky171986 2^n grows much faster than n, so the division goes to 0, leaving the +1. That's a standard limit technique.

@DavidSavinainen

Жыл бұрын

@@jursamaj Alternatively, let m = -n, so the limit is -(m+1)2^(-m) as m approaches infinity. = -(m+1)/(2^m) as m approaches infinity L’Hopital now gives -1/(ln(2)2^m), which clearly approaches 0 as m approaches infinity.

@jursamaj

Жыл бұрын

@@DavidSavinainen Yup, different method, same result.

9:00 The factors *m* and *m+1* are relatively prime, so we can combine both cases: - *2^{n-2}* divides either *m* or *m+1* . In both cases, we have *2^{n-2} ≤ m+1* - The other factor divides *n-1* . In both cases, we have *m ≤ n - 1* For both statements to be true, we need to have *2^{n-2} ≤ n* . Via induction this is only true for *n ≤ 4*

Thanks 😊 for making the video really needed something for my mind to do to help he rationalize.

You're the best, Michael!!!

5:17 Classic Michael Penn induction 14:03 Good Place To Stop

@carpyook

Жыл бұрын

2:15 Spoiler ;)

@tomholroyd7519

Жыл бұрын

Inside out t-shirt

Thanks! Great problem and explanation

Thanks a lot! I liked your solution of the problem!

From 1-n

A bit easier way is to look at (n - 1)*2^n as a product of two consecutive even numbers: (n - 1)*2^n = a*(a+2), n >=2. Then, there are two cases depending on which one of a, a + 2 is divisible by 4. Case 1: a = b * 2^r, b is odd, r > 1. Here r > 1 because a is divisible by at least 4. Then, a + 2 = 2 * (b * 2^(r-1) + 1) Then, we see that (n - 1) * 2^n = 2^(r + 1) * b * (b * 2^(r - 1) + 1) From this it is clear that n = n - 1 We also use that b >= 1, so we have the inequality: (n - 1) * 2^n >= 2^n * (2^(n-2) + 1) This can be simplified to n - 2 >= 2^(n-2) This inequality is false at any n >=2. Case 2: a + 2 = b * 2^r, b is odd, b >= 1, r >=2. After using similar logic we get: n >= 2^(n - 2) This is only possible for n = 2, 3 or 4. Then, by substitution we find that only n = 4 yields the solution among n >=2.

@backyard282

Жыл бұрын

yes but how do you know that (n-1)*2^n is a product of two consecutive even numbers?

@Idran

Жыл бұрын

@@backyard282 assume (n-1)*2^n + 1= m^2 then (n-1)*2^n = m^2 - 1 = (m-1)(m+1) since we know that m^2 has to be odd (as the video showed), then m must also be odd, and so m-1 and m+1 are the two consecutive even numbers on either side of m

@fordiscordfordiscord69

6 ай бұрын

Cool

4:48 1-n=1+m not m-1. Although the proof is the same

@TedHopp

Жыл бұрын

Actually, it should have started with 0 < -1-n < 2^n. So the error was earlier and m-1 is correct.

I've basically figured it out... I have (n - 1)2^(n-2) = k(k+1) for some integer k. Now either k or k+1 must be divisible by 2^(n-2), because 2 does not divide the other factor. This works out for n=4, but if n starts getting too big... For big enough n, n-1 2^(n-2), and the RHS > 2^(2n-4) so we've reached a contradiction. The RHS quickly outpaces the LHS. This is very crude but it works even if n is still quite small. Like, if n=8, then the LHS is 7*2^6, which means k >= 2^6, which means k(k+1) >= 2^12, and this discrepancy grows as n grows.

@leif1075

Жыл бұрын

What about breaking it into difference of squares k minus 1 times k pkus q and set each factor equal to 2^n or (n-1)..you can solve that way

Thanks for the videos. I have watched keenly and enjoyed. Getting good enough to actually finish some of them and I felt super accomplished having gotten to the end of this one! I prefer my solution though for case n>=5 although do appreciate that yours is more algebraically robust. If we accept that 2^(n-2)>n-1 (and the difference differs by more than 1) for n>=5. We can exhaustively extract all 2 contained in n-1 leaving an odd number. The LHS is now the product of two numbers: one is an odd number less than n-1 and the other a power of two that is greater than or equal to n-2. i.e LHS is now the product of two coprime integers. This implies that the smaller of these numbers is equal to smaller of that on the RHS and vice versa. However, the difference between these two numbers is even greater than what we started with (at least 2) and yet the RHS only differs by 1 which is a contradiction!

About the negative case, I wonder what happens if instead of integer perfect squares, we're talking about rational perfect squares ? By that, I mean rational numbers whose square roots are also rational. 4/9 is a rational perfect square but 1/2 isn't.

@chrisglosser7318

Жыл бұрын

Yes - his proof for n

Nice problem. For the size contradiction, instead of splitting by cases you could start with stating that 2^(n-2) divides either m or m+1, so m+1 >= 2^(n-2) in any case, and get the contradiction. It saves a bit of case analysis.

@mcwulf25

Жыл бұрын

Good point

I have a little question: (n-1)2^n + 1 has to be an integer? Because it could be, for example, a number like 16/49 that is a perfect square.

Nice problem. Here's my version of the no solutions where n > 4: (n-1) 2^n = (N-1) (N+1) For larger and larger n, the multiple of 4 on the right dominates. For largest max n, set: N+1 to be a power of 2 (so it can't eat-up any of the odd factors of n-1) and n to be even (so N+1 doesn't eat-up any of the even factors of n-1) This ensures N-1 is as large as it can be. Then compare: N-1 = 2(n-1) with N+1 = 2^(n-1) They differ by 2 when n = 4 but differ by more than 2 (increasing) when n > 4

Using odd numbers to find squares. Or box in a box in a box effect. Use N x 4 - 4

I have decided to use induction to show that n=5. Base Case: 5

@8:18 - the moment "= 1" is written backward in chalk on the back of Michael's shirt

n=-2 would work as plugging in that value you get .25 which is .5²

How does one work and see such hidden inequalities, and I loved how you could conclude om fly that m = 2^(n-2) * x I would have messed up there and notice it much later. I am a Computer Science and Engg student and I really wanna improve with handling such stuff in my mind, I will be glad to have suggestions. I struggle with exercises in technical/mathematical texts very bad

Tshirt is inside out, that proves the solution is correct.

At 3:46, that is not equivalent. I think you mean

Hi Dr.!

Very cool

For a moment I was going to say it was satisfied for n=-2, as it *is* technically (1/2)^2 (-3*1/4+1=1/4), but then I realized they meant perfect squares in the *integers* and not the *rationals*.

wowww cool solution

If you wrote a book with even just the major theorems used for proofs with proofs of those with examples along with a few general techniques such as strong induction, I'd buy it. I'd much prefer that to channel donations.

just a direct way

What if we allowed perfect squares of rational numbers? Are there any solutions for negative n?

I’m not a fan of Number Theory, but I did enjoy this problem very much. Thank you, professor.

CMU shout out! 😄

What's the preliminary to this video by Penn to explain basics about number theory

@luisaleman9512

Жыл бұрын

Check his other channel "MathMajor". He has a full course on number theory there.

Please make a probability cours in mathmajor

10:58 Why doesn’t that inequation hold either for n=1 or n=4 when those values are solutions ???

Its a perfect Square for all n except for the ones it is not.

3:44 I'm struggling to understand why that's equivelant

Does the phrase perfect square imply integers only? Is 9/49 not considered a perfect square?

@notananimenerd1333

Жыл бұрын

Exactly what I thought , isn’t n=-2 a solution too as for n=-2 we get the simplified value as 1/4

@andrewdsotomayor

Жыл бұрын

@@notananimenerd1333 pretty sure the original question specified “the square of an integer” because apparently “prefect square” could mean the square of an integer or the square of a rational number, depending on the context.

@andrewdsotomayor

Жыл бұрын

Moreover, I think it would be difficult to solve cases when n

m+1?

Can't we have like fractional perfect squares?

@Alex_Deam

Жыл бұрын

No, the definition of perfect square is that it has an integer square root

Isn’t the output being a fraction still possible to be a perfect square ex 1/4 = (1/2)^2 Edit: or do perfect squares have to be integers

@vanjansampath5753

Жыл бұрын

yes perfect squares have to be integers

@AmosNewcombe

Жыл бұрын

@@vanjansampath5753 says who?

@vanjansampath5753

Жыл бұрын

@@AmosNewcombe joe

3:43 I did not follow this part. Could someone clarify?

@carstenmeyer7786

Жыл бұрын

At that point, you want to show the following implication *n ≤ -2: f(n) := (n-1) * 2^n ∈ (-1; 0) => f(n) ∉ ℤ* Notice *f(n)* is clearly negative, so you only need to show *f(n) > -1* - that's what happens at 3:45ff. The reason why you do it this way is the graph of *f(x)* with *x ∈ ℝ* .

So - rational numbers can’t be perfect squares(?!)

Ηii,has anyone solve it with square entrapment?? If yes I would like to see the solution, thank you

A bit off the topic, but is 9^m-1 a perfect square?

@kennethforeman6164

Жыл бұрын

Only when m is 0. If m negative, the difference is non-integer. If m>0, a square is either 0 or 1 mod 3 but 9^m-1 is -1 mod 3.

@guidomartinez5099

Жыл бұрын

Well, 9^m is always a perfect square :). So if you subtract one it usually won't be, the only exception being m=0 as Kenneth explains.

3:12 Does anyone see alternating 0 and 1 on the top of the board?

@amireallythatgrumpy6508

Жыл бұрын

Not quite alternating. There are more 0s than 1s. I think that's from the Prime Constant video.

N=-1 solution

Couldn't you have stopped the n≥5 case at 8:53, when the product of two integers of opposite parity (clearly odd) was "equal" to an even integer?

@ultimatedude5686

Жыл бұрын

Nope. 2*3 is the product of two integers of opposite parity, but it is even. In fact, the product of two numbers of opposite parity will _always_ be even, because one of the numbers must have a factor of 2.

For all real x, we have 2 to the x power is greater than x.

@8:20 Remember, don't drink (water) and use math. 🤡

Michael is king of inequalities. Other content providers would probably use other methods to solve this.

Please! :) can we not do induction? sick of it, already!!!! :):):) love you pro!