This problem is really cool since it uses many tricks (mod - inequalities - ...) For the inequality part at the end I like doing it this way: q|p-2 , p-1|q-1 thus q(p-1)|(q-1)(p-2), but clearly q(p-1) is greater than (q-1)(p-2) Hence p=2 And so we are done!
If you Dont know thise tricks , it is possible to solve by plugging and tryingdiff valuesn orb replcing p abd q abd r by 2m plus 1 since all primes are odd except 2....so why not do it this way?
Hey, Michael. A request: can we do some topology (maybe some key point-set concepts or an intro to algebraic topology)? You do have a gift for explaining this stuff.
At 7:30 the argument for Ord_(p-1) q = 1 doesn't hold. For example, if q=5 and p=11, as p-1=10, then p-1 and q are not relatively prime and you cannot apply Euler theorem. In fact this same invalidation holds for your argument in 5:09: p and q are relatively prime but not necessarily this holds for q and p-1. All of this only holds if p < q, which you can maintain by simmetry of the terms in the equation proposed as problem.
Sahil says, p-1 and q is a relatively prime means that we can apply the order integer. Otherwise, if it's not a relatively prime then this most definitely doesn't hold.
Good thing I have some backup homeworks 😂 HOMEWORK : Suppose we write x² as the sum of x x's, and then take the derivative: Let f(x) = x + x + ... + x (x times) Then f'(x) = d/dx[x + x + ... + x] (x times) = d/dx[x] + d/dx[x] + ... + d/dx[x] (x times) = 1 + 1 + ... + 1 (x times) = x. We proved that the derivative of x², with respect to x, is actually x. Where is the misteak? SOURCE : Nick's Mathematical Puzzles and Doug Shaw’s blog.
You forgot to differentiate the x in (x times). Should use the chain rule to get f'(x) = d/dx[x + x + ... + x] (x times) + [x + x + … + x] (d/dx x times) = d/dx[x] + d/dx[x] + ... + d/dx[x] (x times) + [x + … + x] (1 time) = 1 + 1 + ... + 1 (x times) + x = x + x = 2x
Missing bracket at 5:43. This is the only time I'll ever be smart enough to spot a mistake in one of Michael's videos lol. Anyway, he fixed it at 7:55, so my excitement was short lived.
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To show no solutions to the final equation if q>5 without calculus. Notice that (q+1)^2+(q+1)+2 = (q^2+q+2) + 2q+2 < 2*(q^2+q+2), if q>1. So when q increases by 1, LHS doubles but RHS is less than doubled.
Another way to show that p=2 mod(q) implies p=2 is to assume p=2+kp with k>=1 then show that p^q-q^p is always negative which is impossible since (r-1)(p+q) is positive. So we want p^q-q^p > 0 with p=qk+2>=q+2, which is equivalent to showing qlog(p) > plog(q). If we consider f(x) =log(x) / x. We can show for x>=3, if y>x and y>=4 then f(y) < f(x). But for q>=3, p=q+2>=5, this means p^q-q^p cannot be positive. For q=2, the equation we get is 2^p=2-p which gives no solution.
The ending explanation is sketched as followed. Diffrentiating 2^x gives 2^x . ln2. Thus diffrentiating the whole function three times removes all the polynomial terms leaving dy/dx(2^x)’’’ which gives 2^x . (ln2)^3 As Michael suggetsed, we use the fact that ln2 > 1/2, since sq rt. of e is is clearly less than 2 b/c 2^2 = 4 > e. We can see that 2^x . (ln2)^3 > 2^x . (1/2)^3 = 2^{x-3) > 0 for all x >=6.
Def an awesome problem. Interesting use of calculus at the end, I just used IVT and single derivative of each: the polynomial and the exponential. Once one is greater and its derivative is larger as well, there will be no more crossings
Make a new function which is the exponential minus the polynomial. IVT, combined with using the derivative, will help you determine that there are no more zeroes
An inductive proof for the last part: For q=7, LHS is larger than RHS Suppose this is true for q=k (k>=7) For q=k+1, we have LHS = 2x2^k > 2 (k^2 + k + 2) = (k^2 + 2k + 1) + k^2 + 3 > (k+1)^2 + k + 3 = RHS
Problem like this : (USA TST 2003) Find all prime p,q,r such that: +) q^r +1 is divisibled by p +) r^p + 1 is divisibled by q +) p^q +1 is divisibled by r By the way , the problem in video use many tricks, kind of number theory problem i like. Good job michael
Viewer suggestion problem. [All-Russian Mathematical Olympiad 2021] Natural numbers n > 20 and k > 1 are such that n is divisible by k^2. Prove, that there are natural numbers a, b, c such that: n = ab + bc + ac
You could. You end up with qr ≡ 0 (mod p). But then you'd still want to break the problem up into the two cases p = q and p = r, so you don't really avoid checking if p = q.
Wouldn't one get away at a contest by stating that "if 2^7>7²+7+2 then obviously 2^q is bigger for all higher values of q" ? I guess, induction over n=q would be rather easy as well, if one doesn't like the calculus.
Assume 2^n>n^2 + n + 2 Now then 2^(n+1)=2*2^n>2(n^2+n + 2)=n^2+ (n^2+2n +1) + 3=(n+1)^2 + (n^2+1) +2>=(n+1)^2 +n+1 + 2 So if the inequality is ever satisfied it is, by this induction step, always satisfied afterwords
Note to f(x) . 2^x -1.75=(x+0,5)^2. Yes there is only 1 solution. But Michael, have you the Proof, that only this 3 numbers are solution of this problem?
I tried to solved, could you anyone give me a feedback (cuz it is so shorter than the video) : p^q - pr + p = q^p + qr - q => p(p^(q-1) - r + 1) = q^p + qr - q => p | (q^p + qr - q) from Fermat a^p is equavelant to a mod p so q^p - q = 0 (mod p), hence q^p - q = 0 (mod p) p | qr. It is clear that p cannot be equal to q, then p = r. Also on the other hand q | (p^q - pr + p) = (p^q - p^2 + p) = (from fermat) (2p - p^2) then here comes 2p is congruent to p^2 => 2 is congruent to p (mod q). So p = 2, or p = q + 2, if p = 2, it is so short to find p = r = 2, q = 5. Otherwise if p > 3 it is clear that p^(p-2)
@@bobh6728 Not always. According to your argument, why study something which is less likely to have application than something such as, say, engineering?
This problem is really cool since it uses many tricks (mod - inequalities - ...)
For the inequality part at the end I like doing it this way:
q|p-2 , p-1|q-1 thus q(p-1)|(q-1)(p-2), but clearly q(p-1) is greater than (q-1)(p-2) Hence p=2
And so we are done!
If you Dont know thise tricks , it is possible to solve by plugging and tryingdiff valuesn orb replcing p abd q abd r by 2m plus 1 since all primes are odd except 2....so why not do it this way?
@Rick Does Math why not it's more intuitive and as logical
Hey, Michael. A request: can we do some topology (maybe some key point-set concepts or an intro to algebraic topology)? You do have a gift for explaining this stuff.
Wow that is definitely an interesting problem
At 7:30 the argument for Ord_(p-1) q = 1 doesn't hold. For example, if q=5 and p=11, as p-1=10, then p-1 and q are not relatively prime and you cannot apply Euler theorem. In fact this same invalidation holds for your argument in 5:09: p and q are relatively prime but not necessarily this holds for q and p-1.
All of this only holds if p < q, which you can maintain by simmetry of the terms in the equation proposed as problem.
Yeah he was a bit sloppy there at 7:30 but I think there was no mistake at that 5:09 part, he just reduced mod p-1, and p^q is 1 mod (p-1) clearly
Sahil says, p-1 and q is a relatively prime means that we can apply the order integer. Otherwise, if it's not a relatively prime then this most definitely doesn't hold.
The order of an integer modulo n says that the a^m is congruent to 1 (mod n) where the gcd(a,n)=1. Thank you R. Maelstorm!
Good thing I have some backup homeworks 😂
HOMEWORK : Suppose we write x² as the sum of x x's, and then take the derivative:
Let f(x) = x + x + ... + x (x times)
Then f'(x) = d/dx[x + x + ... + x] (x times) = d/dx[x] + d/dx[x] + ... + d/dx[x] (x times) = 1 + 1 + ... + 1 (x times) = x. We proved that the derivative of x², with respect to x, is actually x. Where is the misteak?
SOURCE : Nick's Mathematical Puzzles and Doug Shaw’s blog.
That *misteak*
@@pardeepgarg2640 Yummy indeed
It's due to not also taking the derivative at the "x times"; this is pretty much a mistake of not applying the chain rule correctly.
the definition of f doesn't make sense when x is not a natural number
You forgot to differentiate the x in (x times). Should use the chain rule to get
f'(x) = d/dx[x + x + ... + x] (x times) + [x + x + … + x] (d/dx x times) = d/dx[x] + d/dx[x] + ... + d/dx[x] (x times) + [x + … + x] (1 time) = 1 + 1 + ... + 1 (x times) + x = x + x = 2x
Missing bracket at 5:43. This is the only time I'll ever be smart enough to spot a mistake in one of Michael's videos lol. Anyway, he fixed it at 7:55, so my excitement was short lived.
for the last part induction is also very easy.
Hello everyone. My channel offers a lot of videos such as: limits, integrals, more difficult exercises. Please join all the students to learn. Thank you!
I agree. Avoid using calculus in math contests. Markers hate it. You will get spanked
When he argued that p = r the first thing I did was used exactly the same arguments with mod q. The result was a far simpler proof that r=2
To show no solutions to the final equation if q>5 without calculus. Notice that (q+1)^2+(q+1)+2 = (q^2+q+2) + 2q+2 < 2*(q^2+q+2), if q>1. So when q increases by 1, LHS doubles but RHS is less than doubled.
Ok, that's a typical Fermat's Little Therom and congruences problem, cool
We can also argue that p^q - q^p > 0 so p^q > q^p which happens only for p < q (modulo edge cases) but since p = 2 (mod q) we must have p = 2.
Another way to show that p=2 mod(q) implies p=2 is to assume p=2+kp with k>=1 then show that p^q-q^p is always negative which is impossible since (r-1)(p+q) is positive.
So we want p^q-q^p > 0 with p=qk+2>=q+2, which is equivalent to showing qlog(p) > plog(q). If we consider f(x) =log(x) / x. We can show for x>=3, if y>x and y>=4 then f(y) < f(x).
But for q>=3, p=q+2>=5, this means p^q-q^p cannot be positive.
For q=2, the equation we get is 2^p=2-p which gives no solution.
This is such a beautiful problem! Thanks Michael
7:21 Michael became a car for a second there
What do you mean?
@@tonyennis1787 His vibratto sounded like a car horn there
That is a pretty little problem! Thanks Michael.
The ending explanation is sketched as followed. Diffrentiating 2^x gives 2^x . ln2.
Thus diffrentiating the whole function three times removes all the polynomial terms leaving dy/dx(2^x)’’’ which gives 2^x . (ln2)^3 As Michael suggetsed, we use the fact that ln2 > 1/2, since sq rt. of e is is clearly less than 2 b/c 2^2 = 4 > e. We can see that 2^x . (ln2)^3 > 2^x . (1/2)^3 = 2^{x-3) > 0 for all x >=6.
Dont we only need ln2 > 0 since 2^x > 0 for all x?
You only need to check f(x)>0 for x >= 7 since x=q is prime
Def an awesome problem. Interesting use of calculus at the end, I just used IVT and single derivative of each: the polynomial and the exponential. Once one is greater and its derivative is larger as well, there will be no more crossings
Where does IVT apply
Make a new function which is the exponential minus the polynomial. IVT, combined with using the derivative, will help you determine that there are no more zeroes
What is/are the relationships between this and Weyl inequalities?
An inductive proof for the last part:
For q=7, LHS is larger than RHS
Suppose this is true for q=k (k>=7)
For q=k+1, we have LHS = 2x2^k > 2 (k^2 + k + 2) = (k^2 + 2k + 1) + k^2 + 3 > (k+1)^2 + k + 3 = RHS
Problem like this : (USA TST 2003)
Find all prime p,q,r such that:
+) q^r +1 is divisibled by p
+) r^p + 1 is divisibled by q
+) p^q +1 is divisibled by r
By the way , the problem in video use many tricks, kind of number theory problem i like. Good job michael
Viewer suggestion problem.
[All-Russian Mathematical Olympiad 2021]
Natural numbers n > 20 and k > 1 are such that n is divisible by k^2. Prove, that there are natural numbers a, b, c such that:
n = ab + bc + ac
Good Place To Start At 0:01
so why order n/m is less or equal than n?
why do you even multiply the congruency q=(1-r)q by the inverse of q instead of just subtracting q which doesn't require q!=p
You could. You end up with qr ≡ 0 (mod p). But then you'd still want to break the problem up into the two cases p = q and p = r, so you don't really avoid checking if p = q.
Wouldn't one get away at a contest by stating that "if 2^7>7²+7+2 then obviously 2^q is bigger for all higher values of q" ?
I guess, induction over n=q would be rather easy as well, if one doesn't like the calculus.
Assume
2^n>n^2 + n + 2
Now then
2^(n+1)=2*2^n>2(n^2+n + 2)=n^2+ (n^2+2n +1) + 3=(n+1)^2 + (n^2+1) +2>=(n+1)^2 +n+1 + 2
So if the inequality is ever satisfied it is, by this induction step, always satisfied afterwords
This problem was fun to work on. I personally tackled this during the contest and got 1 pt lol
Why isn't first derivative enough?
because you have not yet eliminated all the terms of the polynomial, which are negative
Good solution
Note to f(x) . 2^x -1.75=(x+0,5)^2. Yes there is only 1 solution. But Michael, have you the Proof, that only this 3 numbers are solution of this problem?
It's very smart proof man 👍
11:43 polyal?
poly[nomi]al -> poly'al
_poly'al,_ abbreviation of _polynomial,_ see _o'er_ for _over_
@@swaree Thanks I got it now
@@swaree except o'er is an archaic, literary form, not an abbreviation (it uses the same # of chars as the original).
I tried to solved, could you anyone give me a feedback (cuz it is so shorter than the video) : p^q - pr + p = q^p + qr - q => p(p^(q-1) - r + 1) = q^p + qr - q => p | (q^p + qr - q) from Fermat a^p is equavelant to a mod p so q^p - q = 0 (mod p), hence q^p - q = 0 (mod p) p | qr. It is clear that p cannot be equal to q, then p = r. Also on the other hand q | (p^q - pr + p) = (p^q - p^2 + p) = (from fermat) (2p - p^2) then here comes 2p is congruent to p^2 => 2 is congruent to p (mod q). So p = 2, or p = q + 2, if p = 2, it is so short to find p = r = 2, q = 5. Otherwise if p > 3 it is clear that p^(p-2)
Wait, why can we use the little theorem mod a different prime than that power?
What do you mean? He uses that q^p ≡ q (mod p) and p^q ≡ p (mod q).
@@DylanNelsonSA that must be a different theorem.
As FLT is a^p = a (mod p)
@@romajimamulo What do you get if you take a = q?
@@DylanNelsonSA but he's not working mod... Oh he is working mod p
EXCELLENT thumbnail
Hello l am from Azerbaijan🇦🇿🇹🇷👍 intresting study
JBalkanMO?
Hi,
For fun:
1 "and so on and so forth".
А не сошел ли с ума автор задачи!?
I love math...
I also have no idea what is going on...yet it's intriguing 🤔
Zero is positive. Not related to the question
Zero is not positive lmao
**HI**
Why study mathematics?
Because it's the ultimate truth and ultimate beauty.
@@maxwellsequation4887 mathematics is queen 👑 science
because it is good for your brain.
Things discovered in “pure” mathematics end up being used in the real world in areas such as computer science, quantum mechanics, etc.
@@bobh6728 Not always. According to your argument, why study something which is less likely to have application than something such as, say, engineering?