These problem building videos are really interesting because they force us to see math from a different perspective. As usual this video was pretty great.
You can also eliminate 2*28 by seeing that since one is 3^a-5^b and the other is 3^a+5^b, their difference must be twice a power of 5, which only applies to 4*14
Thanks Michael Penn, your videos are always very educational. I didn't know the fact or having thought about adding or subtracting even numbers and they then become odd numbers. Diophantine solving for the right polynomials occurs quite often in solving adaptive controllers for control purpose.
You can also get that m,n are even by checking mod 4, in general for any exponential diophantine equation checking mod 3, mod 4 usually gives some interesting results.
Because if m = 4a+1, then 3^m = 3^(4a).3 = 3.81^a ≡ 3.1^a ≡ 3 (mod 8) for all natural numbers a. Whereas 5^n = 5^(4b+3) = 5^(4b).125 = 125.625^b ≡ 5.1^b ≡ 5 (mod 8) for all natural numbers b. So you can never have 3^(4a+1) equal to 5^(4b+3) for any natural numbers a, b. It can only work if a and b are even, which is a necessary (but not sufficient) condition.
This channel made me realize just how much math I don't understand. I don't think I've seen a single video in which I could have solved the problem on my own.
Me encantó que propusieras este video enseñando a armar problemas. Muchas Gracias! I loved that you proposed this video teaching how to set up problems. Thank you very much!
Nice video as always In some equations , we need to check the remainder on division by sth . How does the person who makes the problem know which polynomials or numbers give us unique answer or contradiction ?
Can anyone solve this for me:- Show that there are infinitely many pairs (a,b) of coprime integers (which may be negative, but not zero) such that x^2 + ax + b = 0 and x^2 + 2ax + b have integral roots.
Basically you try to find a relation between a and b(in particular try to express b as a function of a) so that the statement is true. I found that for odd a b(a) = - (a²-9)(a²-1)/16
For integer p the functions: x²+(2p+1)x-(p-1)p(p+1)(p+2)=0 has roots: x=-p²-2p and x=p²-1 whilst x²+2(2p+1)x-(p-1)p(p+1)(p+2)=0 has roots: x=-p²-3p-2 and x=p(p-1). Still need to show there are infinitely many ( 2p+1 , -(p-1)p(p+1)(p+2) ) which are co-prime.
I don't know if it's because (to me) you look like Neil Patrick or because you are a great explainer but i love your videos! Thanks for making the boring interesting!
In an other way: n satisfies the equation if n >or = m/2, because 3^x-5^(x/2)-56 is an increasing function that is zero only if x = 4, and also n < or = m -2, because 3^x-5^(x-1)-56 is negative so the only solution seems to be m = 4... the better way seems the other...
![The Most Useful Curve in Mathematics [Logarithms]](http://i.ytimg.com/vi/OjIwCOevUew/mqdefault.jpg)
I love these "make your own problem" style videos! They give twice as much insight on problems
These problem building videos are really interesting because they force us to see math from a different perspective.
As usual this video was pretty great.
You can also eliminate 2*28 by seeing that since one is 3^a-5^b and the other is 3^a+5^b, their difference must be twice a power of 5, which only applies to 4*14
10:02 15 is divisible by 3, it's not a power of three though
That's what they want you to think
Only if you want to be rational about it.
But to be fair, you, I and most of the viewers spotted that and knew what was meant if we were paying attention.
..michael penn ...always a joy to watch your math content.. thnq
10:47
Doing gods work
God is away on holidays
I believe you mean power not mulitple at the end for 3^a = 15
Thank you Michael Penn, you are a legend
Thanks Michael Penn, your videos are always very educational. I didn't know the fact or having thought about adding or subtracting even numbers and they then become odd numbers. Diophantine solving for the right polynomials occurs quite often in solving adaptive controllers for control purpose.
You can also get that m,n are even by checking mod 4, in general for any exponential diophantine equation checking mod 3, mod 4 usually gives some interesting results.
Oops, checking mod 4 only gives you that m is even, mod 8 is still needed. My bad
Thank you, professor!
#6:31 why is m=4a+1 and n=4b+3 not a posibility?
Because if m = 4a+1, then 3^m = 3^(4a).3 = 3.81^a ≡ 3.1^a ≡ 3 (mod 8) for all natural numbers a.
Whereas 5^n = 5^(4b+3) = 5^(4b).125 = 125.625^b ≡ 5.1^b ≡ 5 (mod 8) for all natural numbers b.
So you can never have 3^(4a+1) equal to 5^(4b+3) for any natural numbers a, b.
It can only work if a and b are even, which is a necessary (but not sufficient) condition.
Yes, I see this now. Maybe I should change ny glasses. :-)
just learnt how problems are generated! very nice presentation.
Thank you for this inspiring video! It is good to see how to build some exercises.
New merchandising idea: shirts with sleeves that can be used as erasers!
This channel made me realize just how much math I don't understand. I don't think I've seen a single video in which I could have solved the problem on my own.
This looks like fun. No need to hunt number theory problems any more - just make my own 😁
Well posed!
but this only works if the equation is made to have even solutions. how about something like 2 + 5^m = 7^n how do we solve that.
Me encantó que propusieras este video enseñando a armar problemas. Muchas Gracias!
I loved that you proposed this video teaching how to set up problems. Thank you very much!
thank you.
9:52 *power
Nice video as always
In some equations , we need to check the remainder on division by sth . How does the person who makes the problem know which polynomials or numbers give us unique answer or contradiction ?
9:54, not a power of 3, but a multiple anyway.
I really hoped for a proof that exponentiation is Diophantine
Michael Penn, your videos are amazing!
Now's a good place to stop this comment.
Sweet
🙏🏼👍🏼👍🏼
Can anyone solve this for me:- Show that there are infinitely many pairs (a,b) of coprime integers (which may be negative, but not zero) such that x^2 + ax + b = 0 and x^2 + 2ax + b have integral roots.
Basically you try to find a relation between a and b(in particular try to express b as a function of a) so that the statement is true. I found that for odd a b(a) = - (a²-9)(a²-1)/16
For integer p the functions: x²+(2p+1)x-(p-1)p(p+1)(p+2)=0 has roots: x=-p²-2p and x=p²-1 whilst x²+2(2p+1)x-(p-1)p(p+1)(p+2)=0 has roots: x=-p²-3p-2 and x=p(p-1). Still need to show there are infinitely many ( 2p+1 , -(p-1)p(p+1)(p+2) ) which are co-prime.
It's nice to see another video with a meaningful title.
Have you run out of tasks? Then create your own one.
I don't know if it's because (to me) you look like Neil Patrick or because you are a great explainer but i love your videos! Thanks for making the boring interesting!
n = 1/2, m = 2/2
I miss olympiad probkemms
lectures 6 & 7 need to be added to the linear algebra playlist
In an other way: n satisfies the equation if n >or = m/2, because 3^x-5^(x/2)-56 is an increasing function that is zero only if x = 4, and also n < or = m -2, because 3^x-5^(x-1)-56 is negative so the only solution seems to be m = 4... the better way seems the other...
Hmmm is this 2.th comment ?
Здається,це легко, дякую 🇺🇦🇺🇦🇺🇦🇺🇦🇺🇦
Why did I click on this?
The scattered blade spindly hover because polo metabolically saw absent a quizzical withdrawal. spiritual, innate zoo
LOL silly