the wildest exponential equation I have ever seen!

[@shohamsen8986]

From 14:09 to 15:10, the algebra is incorrect. I present the following corrections. v(c^2b)=v(b)+v(c^a+2). This can be simplified using exponent rules to get 2bv(c)=v(b)+v(c^a+2). Since p>=3, we should be able to show that v(c^a+2)=0. This is because. if p|c^a, then p cannot divide c^a+2 (cause p) must divide 2 which is a contradiction. Moreover, v(c)>=1 cause we know p divides it. Thus we have 2bv(c)=v(b). which means that 2bv(b) which we established earlier.

[@Noam_.Menashe]

Yeah this is what I thought of. Because otherwise we didn't really use p>2 did we?

[@boiii2148]

shouldn't there be 2 v(b) terms? but anyway 2b

[@shohamsen8986]

@@boiii2148So we are applying v(.) to ac^{2b}=bc^a+2b=b(c^a+2). Then i am using the factor rule. The original explanation which gives you 2v(b) uses the rule v(a+b)=v(a)+v(b) which is not correct.

[@Happy_Abe]

Well explained thank you!

[@pepefrogic3034]

Yes he is very sloppy all the time. Says greater than 1 means greater or equal, makes mistakes all the time but this is just a mess. Does not affect the result

[@ilonachan]

15:30 small error: we've shown that c has only 2 as prime factor, but that doesn't mean that c=2. So if we keep a and b the same as before, we CAN'T say m=2^b, n=2^a, we have to consider that if c=2^d then m=2^db and n=2^da. We can of course fold this d into a and b each, but then they're explicitly not coprime anymore with gcd(a,b)=d... luckily we don't need the coprime-ness anymore, so that's fine. Just erase that "fact" from the final whiteboard, and everything is okay.

[@goodplacetostop2973]

21:25 Ah the classic « Obvious solutions but prove they are the only ones »

[@PetraKann]

Solutions to what exactly? What practical application does this equation have?

[@juxx9628]

@@PetraKann Not everything is practical, kiddo. Sometimes, when you're doing practical things, you may encounter these types of equations. That doesn't mean every equation like this comes with practical applications. Just do math for fun!

[@TheEternalVortex42]

@@PetraKann This might not be the best channel for you

[@l.w.paradis2108]

Of course.

[@PetraKann]

@@juxx9628 Liston Mr Juice8745 I am not your "kiddo". You should apologise

[@samueldevulder]

13:56 I don't get why we have the "+ nu_p(b)" term on the RHS since "2b" is not a multiplicative term in the source equation.

[@jcsahnwaldt]

Yeah, that's an error. (Doesn't matter much though since we drop the term anyway.)

[@aweebthatlovesmath4220]

13:55 wait but p-adic valuation does not satisfy v(a+b)=v(a)+v(b)?

[@shohamsen8986]

I think i corrected it.

[@eugene_v-ko]

@@shohamsen8986 At what time? I didn't see the correction, you never return to the incorrect deduction.

[@JordHaj]

@@eugene_v-ko the guy you replied to is not Michael Penn, he wrote a comment where he corrects Michael.

[@eugene_v-ko]

The correct logic should be: since p|c, and p>=3, then p doen't divide (c^a+2), so v_p(c^2b)=v_p(a*c^2b)=v_p(b*(c^a+2))=v_p(b), then 2b*v_p(c)=v_p(b)

[@eugene_v-ko]

@@JordHaj Right, I didn't notice that. I've come to the same correction above. :)

[@Minskeeeee]

I'm confused at the inequality at 14:40. we drop terms from the right side, but it gets bigger?

[@quite_unknown_1]

Big mistake yes

[@shohamsen8986]

i think i corrected it.

[@kevinmartin7760]

@@shohamsen8986 I'm not sure I would call it a "correction" but I agree with your demonstration. Michael's mistake with his inequality direction seems to make his path irredeemably broken.

[@shohamsen8986]

@@kevinmartin7760 so what's wrong with calling it a correction?

[@kevinmartin7760]

@@shohamsen8986 There's nothing really "wrong" with it, but for me the idiom of "correcting" something would involves a small change, often where there is only one correct form, rather than replacing a relatively large chunk with an alternative (of which there might be several quite different correct forms) and thus avoiding the mistake. Referring to a large change as a "correction" sort of makes me feel the same as if someone had re-done their bathroom and referred to it as "repairing" a leaky faucet.

[@l.w.paradis2108]

Anyone can spot (m,n) = (1, 1); (2,2). The point is to prove they are exhaustive. Loved this!

[@wes9627]

Makes me hungry for some m&n's.

[@biodreg1332]

Here is a shortened version: Let us take it up at the point where n > 1. Let p be a prime that divides n. Write n = p^a * q , p does not divide q Since p divides m, write m = p^b * r , p does not divide r then p^(a * m^2) * q^(m^2) = p^(b * (n+2)) * r^(n+2) since p divides neither q nor r, by the uniqueness of factorization a * m^2 = b * (n+2) so a * p^(2b) * r^2 = b * (p^a * q + 2) If p > 2 then p does not divide p^a * q + 2 so p^(2b) must divide b which is impossible since p^(2b) > b. This means that 2 is the only prime divisor of m, n, so n = 2^a, m = 2^b and, after a short calculation, m = n = 2.

[@Happy_Abe]

@2:02 why does bigger than 1 mean they’re composite? That can be prime too

[@thomashoffmann8857]

Maybe the wording is not perfect. It just means you can write it as a product of powers of primes. Like 2=2^1 (product with one prime factor)

[@Happy_Abe]

@@thomashoffmann8857 yep!

[@mohdarmanansari290]

Your board looking good. ❤

[@cmilkau]

12:00 12:30 About p|b: If m=n=2, then a=b=1, c=2, and p=2 most definitely does not divide b. Now in that case also p < 3, however it illustrates that it's not obvious at all that p|b, and that proving this requires use of p ≥ 3. Here's a proof: if p|c, then p divides ac^(2b) = b(c^a + 2). If p is prime this means p|b (then we're done) or p divides c^a + 2. In the latter case, since c^a also is a multiple of p, the difference 2 must be a multiple of p. Hence, p being a prime, p=2 (the only other option is p=1, which is not prime).

[@XenophonSoulis]

Thanks! I knew something was missing there, but I couldn't figure out exactly what

[@cmilkau]

15:40 we haven't ruled out that ν_2(c) > 1 yet! Here's a fix: We first have to consider the case a = 1, then m = n^b, so n^(2b) = b(n + 2) from the exponents of the original equation. Now n being a power of 2, n^(2b), b and n + 2 must also be powers of 2. This is only possible when n = 2 (4 and 2 are the only powers of two with a difference of 2). Now consider the same contradiction argument as in the video, with p = 4 instead of a prime p≥3. Note that it only fails when both b and c are even. So let's divide the equation ac^(2b) = b(c^a + 2) by 4, distributing a factor of two to each b and c. Recall that a,b≥1. We get ac^(2b)/4 = (b/2)((c^a)/2 + 1) where all the divisions are integer divisions without remainder. Recall that a>1, so (c^a)/2 + 1 is odd. Since 2^(2b-2) divides c^(2b)/4, this means 2^(2b-2) divides b. Hence 2^(2b-2) ≤ b, i.e. 4^b ≤ 4b. This means b ≤ 1, so b = 1. So n = m^a, and again from exponents of our original equation, am² = m^a + 2. Thus 2 = am² - m^a is a multiple of m. Consequently, m = 2 (or m=1). In the following, assume a > 1. Note that when m=2, n=2 by the original equation, and vice versa. So we don't have to consider a,b>1 anymore, we're completely done. EDIT: should've watched the rest of the video first, could've saved some time writing it in less detail.

[@cmilkau]

14:00 What, no! ν_p(bc^a + 2b) = ν_p(b(c^a + 2)) = ν_p(b) + ν_p(c^a + 2). There is no way to take apart ν_p(c^a + 2), it's a sum! BUT since p|c and p≠2, we can actually conclude ν_p(c^a + 2) = 0. Continuing the proof, this gives us 2bν_p(c) = ν_p(b). But since p|c, the lhs is at least 2b, which we just established is strictly larger than ν_p(b), so the lhs and the rhs can't be equal.

[@HagenvonEitzen]

14:00 That's not how a valuation u_p works! In general, u_p(x+y) does not equal u(x)+ u(y). Rather, we have u_p(x+y) \ge \min\{ u_p(x), u_p(y)\} with equality if (but no t only if) u_p(x) = u_p(y)

[@ZekeRaiden]

Interesting note: If we expand the solution space to the integers, not just the naturals, we only get two other solutions, which are just the negatives of the solutions we already got.

[@teeweezeven]

(-2,-2) is not a solution. If n=-2, then the right side becomes m^0 = 1 unless m=0. And the left side is 1 only when m=0. So we get three solutions: (-1,-1), (1,1) and (2,2). (And if you want to say 0^0 has a value of either 0 or 1, you get a fourth: (m,n)=(0,0) and (m,n)=(0,-2) respectively)

[@tahirimathscienceonlinetea4273]

Hi micheal here is my solution:(n=1,m=1) ,(n=2,m=2) ,for all m>3 ---> m>n because m^2>n+2 so m>n means m=n+k replacing that into eq (m-k)^m^2=m^m^2--->m-k m^2-m+k-2>0 ,let's take f(x)=x^2-x+k-2 -->f'(x)=2x-1=0 --->x=1/2 so base on graph f(1/2)=k-9/40 but we find eq accepts roots because discriment is positive so it is a contradiction means m>3 no solution good luck

[@uncountableuk]

Whenever you get stuck in maths, just call everything you have so far a new letter and press on

[@coolbepis9301]

19:55 the way I did this was to observe that since 2^(a-1)+1 divides a*2^(2b-1), it must divide a, since 2^(a-1)+1 is odd and 2^(2b-1) is a power of 2. Since 2^(a-1)+1divides a, 2^(a-1)+1 is less than or equal to a. But this is not true for any natural number a. That felt slightly more natural to me.

[@loganholdaway4831]

thank you so much for this. this is a relation my brain has been looking for for 13 years.

[@confusedsoul4775]

it seems that we can alternatively use another result of number theory: if x,a,b, n are natural numbers and x^(a/b)=n , then either b divides a, or x=y^b for some natural number y. if this is right,then this problem can be solved very easily.

[@Reza_Audio]

another way. take a log of both sides with base "m" then you get another equation in which in order to get natural number in both sides , log(n) with base m must be natural number. so that n must be equal to m or or must be in form of m^t . in first case n=m will lead you to a quadratic equation n^2-n-2=0 which has only one natural solution n=2=m in second case you get t*m^2=n+2 . in this case because of side is linear and the other side is quadratic (faster growth) the only solution would be acceptable is for t=1 m=n=2 which is already the same as the other case. so beside m=n=1 the only possible solution would be n=m=2

[@carlosayam]

case t > 1 probably better as: t m^2 = n + 2 = m^t + 2 => t m^2 - m^t = 2 => case t=2, 2 m^2. - m^2 = m^2 = 2, bad; case t >= 3, t m^2 - m^t = m^2 (t - m^(t-2)) = 2, again m^2 | 2, bad again.

[@de_oScar]

I love this

[@assassin01620]

How is (2+a)x < x when x,a >= 1? 14:40

[@coolbepis9301]

15:25 I'm confused. Couldn't c be a power of 2? We haven't proved that nu_2(c)=1, only that it has no prime factors greater than 2.

[@Blabla0124]

Hold on, something goes wrong at 13:55 ...

[@1991tnh]

Nice sol Prof

[@williamwarren5234]

Can someone tell me where to look into online about the part where introduced v (nu)? Is that some number theory thing I should know

[@TedHopp]

Look up p-adic valuation.

[@papaha01]

what about n=2, m=-2 ?

[@theuserings]

4:28 I dont get why a_k and b_k has to be natural numbers. Take for example m = 15 and n = 80. Both are divisible by 5 (prime). So according to the video m and n have the same prime factors but with a different exponent with the exponents being natural numbers. But this reached a contradiction... because if we follow the rules... 15 = 5 × 3 80 = 2⁴ × 5 We dont get m and n has the same prime factors. But if we just say a_k and b_k are natural numbers including zero, our assumption holds true because 15 = 2⁰ × 3 × 5 80 = 2⁴ × 3⁰ × 5 Was it a mistake by michael?

[@samueldevulder]

14:53 how to get 2b nu_p(c)

[@pepefrogic3034]

Messy error when computing index of p for a sum, treated it as product. Does not effect the result though. Very sloppy!

[@s80236g]

n > 1, where n is number of ictus.

[@markwalker8846]

Am I missing something? Near the beginning, you say that taking p as a divisor of m proves that p also divides n^m^2, but this is clearly untrue if you take m=2 and n=3. 3^2^2=81, which is not divisible by 2.

[@jcsahnwaldt]

But m=2, n=3 doesn't satisfy the initial equation. We know that m and n have to satisfy the initial equation. That's why we know that if p divides m then p also has to divide n (and vice versa).

[@stefanschroder4694]

Why is b> p^v(b)?

[@Alan-zf2tt]

Hmmm interesting! I think due diligence before proof would be helpful too. Trivially looking at behavior of exponents m to the two equals n plus two too.. This eventually leads quickly to case n equals m equals two too true too, that is n = 2 and m = 2 too Difficulty arises in proving there are only two solution pairs in this entanglement of exponents so just follow Michael's example in this. However exploration of verasity of two's too true too?

[@mathunt1130]

Is this the usual way to tackle these problems? Write m and n in their prime decomposition and just play with the exponents?

[@martincohen8991]

I get m=n=2 as the only solution.

[@GearsScrewlose]

Typical - An over the top solution. Notice that LHS has an m^2 so that motivates rewriting the RHS with a m^2 in it. Now, you can make a cleaver substitution suggests 2^(m^2) is factor of RHS. At this point you can easily show the only solution is exactly what you had.

[@papanujian7758]

Very nice video

[@brian554xx]

lost me when nu came in. will rewatch someday when thinking more clearly.

[@gp-ht7ug]

🤕 what an headache

[@s.l.2227]

Zero is a natural number.

[@debtanaysarkar9744]

Zero is a whole number, not a natural number.

[@mairc9228]

average set theorist

[@DeanCalhoun]

show me zero things

[@egillandersson1780]

In US, most authors seem to consider that 0 ∉ ℕ while, in several European countries, 0 ∈ ℕ. There is no truth, it's just a matter of convention !

[@Bruh-bk6yo]

​@@DeanCalhounhere.