Shorter solution: Suppose that p is odd if r is also odd, then from B: q is odd, but then A is not integer. if q is odd then from C: r is odd, but then again A is not an integer. So when p is odd: r=q=2, from this A is not an integer. Hence we can assume that p=2, in this case: A=(2*q+4)/(q+r), if A=2 then r=2, from B we got q is even q=2, but then B=22/4 is not an integer. We left only A=1 (obviously 0
LEFT FOR THE VIEWER : Prove that if a and b are positive integers then there maybe at most a finite number of integers n, such that both an² + b and a(n+ 1)² + b are perfect squares. SOURCE : 93rd Eötvös-Kürschák Competition 1993
@@adityaekbote8498 I'm doing all by myself, no crawler. When I feel things are not balanced, for example when I have too much number theory problems, I'm looking for other topics like calculus or geometry. Same goes for countries and levels (junior, senior, olympiad)
Hello Dr. Penn The solution was interesting. But I think we still have to check if our found values for p, q, and r guarantee that C is an integer (in this case they do). Do we still need to check C regarding the given solution, or is that obvious?
@michael penn Great video! The claim that q and r are not even is a strong start, you probably started this problem thinking that all the three variables couldnt be even, right? I think it would be a cool ideia to make a video showing your thought process including the ideas that dont work! Just an idea! Have a nice day!
What about a video with a selection of recomentations of good or best books for the main fields of maths? At least the ones that helped you along the way. It would be great!
Small problem: While it is true that the values you presented work for B, you never showed that they work for A and C. You showed necessary conditions for them to work for A and C, but never showed that they were sufficient.
Agreed. For this kind of problem, I think that one always needs to verify that the found solution actually works, i.e. checking that A, B and C are integers with the proposed solution
I’m sure what he said in the end “so this is in fact the only triple that satisfies these conditions.” implies that we can easily check A and C. i.e. left to the listeners as exercise.
A: q+r even -> p even B: p+r even -> q+r even C: p+q even -> r+p even so using = in mod 2 to denote parity A: if q=r then p is even B: if p=r then q=r then p is even (from A) so all are even C: if p=q then p=r then p=q=r so all are even. By contradiction, (2,2,2) doesn't work, so we know p=r and p=q are both false, p+r and p+q are both odd, so r and q are odd and p is even.
I have a question, why you suppose that A=1, when the problem says that A,B and C must be integers (which means that A can be 0, -1, -2,...)? Maybe it doesn't change anything, but I'm not seeing it.
Was that a good place to stop? You've shown that if there is a so.option then it's (2,3,7). But is this a solution. I think you've shown that A and C are both integers but not B. Or did I miss something?
I hate the way you write lowercase q's so much. Especially the q in a's numerator. I couldn't tell what that said until you verbalized it, I thought it was some arcane mathematical symbol I was unfamiliar with, it definitely looked nothing like a q to me.
your style of "show whether these can be even or odd" argumentation almost feels like it's contrived for how much shorter it could have been if you had just said "let's show q and r cannot be 2" and so forth, further eliminating the need to repeat "the only even prime number is 2" as though being even isn't just defined as being a multiple of 2
14:19 q = {-5, -4, -3, 0, 3, 12} but the correct element is still negative so the solution isn't affected.
Shorter solution:
Suppose that p is odd
if r is also odd, then from B: q is odd, but then A is not integer.
if q is odd then from C: r is odd, but then again A is not an integer.
So when p is odd: r=q=2, from this A is not an integer.
Hence we can assume that p=2, in this case:
A=(2*q+4)/(q+r), if A=2 then r=2, from B we got q is even q=2, but then B=22/4 is not an integer.
We left only A=1 (obviously 0
15:04
Thank you, professor!
LEFT FOR THE VIEWER : Prove that if a and b are positive integers then there maybe at most a finite number of integers n, such that both an² + b and a(n+ 1)² + b are perfect squares.
SOURCE : 93rd Eötvös-Kürschák Competition 1993
How do you get so many problems from is there any crawler you use or you just go like today I feel Russian and then a random Russian olympiad question
@@adityaekbote8498 I'm doing all by myself, no crawler. When I feel things are not balanced, for example when I have too much number theory problems, I'm looking for other topics like calculus or geometry. Same goes for countries and levels (junior, senior, olympiad)
Hello Dr. Penn
The solution was interesting. But I think we still have to check if our found values for p, q, and r guarantee that C is an integer (in this case they do). Do we still need to check C regarding the given solution, or is that obvious?
No, you're correct he left it as an exercise to the viewer but it definitely needs to be checked
I know I'm not the only one curious about this, so I went and did the calculations.
A = 1
B = 8
C = 11
And that’s even better place to stop.
@michael penn
Great video!
The claim that q and r are not even is a strong start, you probably started this problem thinking that all the three variables couldnt be even, right?
I think it would be a cool ideia to make a video showing your thought process including the ideas that dont work!
Just an idea!
Have a nice day!
We made the same solutions. I am getting better at these thanks to you professor.
What about a video with a selection of recomentations of good or best books for the main fields of maths? At least the ones that helped you along the way. It would be great!
Small problem: While it is true that the values you presented work for B, you never showed that they work for A and C. You showed necessary conditions for them to work for A and C, but never showed that they were sufficient.
Agreed. For this kind of problem, I think that one always needs to verify that the found solution actually works, i.e. checking that A, B and C are integers with the proposed solution
He fixed A to be 1 in the previous step so that's fine, but yes he should have shown that C is an integer too.
I’m sure what he said in the end “so this is in fact the only triple that satisfies these conditions.” implies that we can easily check A and C. i.e. left to the listeners as exercise.
@@williamadams137 A is set to be equal to 1. We only need to check C I think
Technically A is already checked as p=2 and r=q+4 was sufficient for A to be an integer, namely 1
But I agree he should have checked for C
Nice problem. Inequalities are underrated, in my opinion. They are very useful
15:05 MichealPenn.exe has stopped working
A: q+r even -> p even
B: p+r even -> q+r even
C: p+q even -> r+p even
so using = in mod 2 to denote parity
A: if q=r then p is even
B: if p=r then q=r then p is even (from A) so all are even
C: if p=q then p=r then p=q=r so all are even.
By contradiction, (2,2,2) doesn't work, so we know p=r and p=q are both false, p+r and p+q are both odd, so r and q are odd and p is even.
Hey Mike, what's 3 minus 6?
Good observation lol
We can quickly discount p being odd by considering A, which requires exactly one of q and r is 2. Then from C that's not possible.
Hooo, it was also in the shortlist of a french contest, that's fun
I have a question, why you suppose that A=1, when the problem says that A,B and C must be integers (which means that A can be 0, -1, -2,...)?
Maybe it doesn't change anything, but I'm not seeing it.
@Thijs Van bosch thanks bro
3:24 -1 not 1
At the end, you should also check that C is in fact an integer, just so we know (2,3,7) is in fact a solution.
good thanks
Was that a good place to stop? You've shown that if there is a so.option then it's (2,3,7). But is this a solution. I think you've shown that A and C are both integers but not B. Or did I miss something?
Very nice
Can you solve the Turkish math exam questions? (It was very difficult)
You can delete your comments all you like, we all know what your country did.
Talaat Pasha had it coming.
@@fracaralhostupid you don't care
The word odd now sounds weird to me
am I here before Good Place To Stop? No................... At least it was another one of my suggestions.
Nice question!
I didn't get it how did you comment 1 month earlier 🤔🤔
I think he is eautistic.
@@Czmlol Ok...
Please add laugh track to bloopers
I wonder how many Math Olympiads there are in the World?
2,3,7 resp. by inspection and that’s a good place to stop.
I hate the way you write lowercase q's so much. Especially the q in a's numerator. I couldn't tell what that said until you verbalized it, I thought it was some arcane mathematical symbol I was unfamiliar with, it definitely looked nothing like a q to me.
your style of "show whether these can be even or odd" argumentation almost feels like it's contrived for how much shorter it could have been if you had just said "let's show q and r cannot be 2" and so forth, further eliminating the need to repeat "the only even prime number is 2" as though being even isn't just defined as being a multiple of 2
First