239 is the number of the best russian math school (Fields laureates Grigori Perelman and Stanislav Smirnov are from there). This problem is from their annual open math olympiad.
if a+b is a perfect square, one of the primes that satisfy this expression is a Mersenne prime where a is 2x the perfect number associated with that Mersenne prime and b - 1 = Mersenne Prime. For example a = 56, b = 8. Other primes work, for example b = 6, a = 30, therefore p = 5.
Quick minor point, if like me you define the Naturals as including 0, then the hypothesis still holds for b=0. i.e. a + 0 = 0(b - c) = 0 so a = 0, and therefore ab = 0 which is a perfect square.
North American standard pedagogy defines N as 1, 2, 3, ... } and W, the set of Whole Numbers, is N + {0}. British pedagogy and many number theoriticians have different preferences - but as a North American professor Prof. Penn is reasonable to conform to N.A. pedagogy.
@@pietergeerkens6324 I didn't say he's "unreasonable". I said "if like me you include 0...." But like you said, not everybody defines the Natural Numbers starting at 1. The ISO for example uses 0 in its definition, as do most people working in set theory. There is no official standard that North American mathematicians comport to 0 being excluded, though. (In fact, all my college professors started it at 0, probably based on their backgrounds in set theory and combinatorics.)
Yes! This is the way I like to think of it: For p >= 5, p^2 can be written as 24n + k, when n is a positive integer and k is a positive integer between 0 and 23. k must be 1,5,7,11,13,17,19, or 23, otherwise 24n + k would be divisible by some number other than 1 or p, making p not prime. Notice that 1*1 = 1, 5*5 = 25 = 24 + 1, 7*7 = 49 = 24(2) + 1, ... , 23 * 23 = 529 = 24(22) + 1. Since k must be 1,5,7,11,13,17,19, or 23, and since each one of these is one more than a multiple of 24, k must be one more than a multiple of 24. Then k = 24t + 1. k is less than 24, so t = 0. Then k = 1. We see that p^2 is one more than a multiple of 24. So c = p^2 - 1 = (24n + 1) - 1 = 24n. This is even easier to see if you're familiar with congruence classes modulo 24.
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Saint Petersburg Lyceum 239 is one of (if not the) best math school in Russia. Grigori Perelman went there for example.
One could say it produced the greatest math mind of the Millennium
239 is the number of the best russian math school (Fields laureates Grigori Perelman and Stanislav Smirnov are from there). This problem is from their annual open math olympiad.
239 is a number of the devil
this post was made by PML 30 gang
I think in 'Case 3' you could have gotten b=p+1 "easier" out of the equation p^2=m*(b-1), knowing that in 'Case 3' m=p.
Extremely true
7:50
Good place to stop
Loved your (Michael Penn)^9 so much, I'd look forward to every "Good Place To Stop" combined, although that's a crazy idea
@@t39an8r Thank you, I’m lacking of ideas for a new video but once I get a very good one, I’ll go very fast to do it
239 is a prime
I’m just amazed that a number theory professor didn’t know every prime
I figured that out in my head. It was easier than the problem he was doing.
@@idjles an intellegent man never do useless things
What a nice problem.
I usually shy away from number theory, but you are making me like it.
Thank you, professor!
Very elegantly done. Many thanks for this wonderful problem.
if a+b is a perfect square, one of the primes that satisfy this expression is a Mersenne prime where a is 2x the perfect number associated with that Mersenne prime and b - 1 = Mersenne Prime. For example a = 56, b = 8. Other primes work, for example b = 6, a = 30, therefore p = 5.
No, that it's wrong?
It's a prime, and a twin prime to boot.
1:16 so 0 is not a natural number ? i thought that all non negative numbers were natural
case 3, we have that m=p so m(b-1)=p^2 means that b-1 = p so b = p+1, a = p^2+p so a+b=p^2+2p+1 = (p+1)^2
Quick minor point, if like me you define the Naturals as including 0, then the hypothesis still holds for b=0. i.e. a + 0 = 0(b - c) = 0 so a = 0, and therefore ab = 0 which is a perfect square.
North American standard pedagogy defines N as 1, 2, 3, ... } and W, the set of Whole Numbers, is N + {0}.
British pedagogy and many number theoriticians have different preferences - but as a North American professor Prof. Penn is reasonable to conform to N.A. pedagogy.
@@pietergeerkens6324 I didn't say he's "unreasonable". I said "if like me you include 0...."
But like you said, not everybody defines the Natural Numbers starting at 1. The ISO for example uses 0 in its definition, as do most people working in set theory. There is no official standard that North American mathematicians comport to 0 being excluded, though. (In fact, all my college professors started it at 0, probably based on their backgrounds in set theory and combinatorics.)
I am feeling proud after solving this problem. Though it was a easy one . Number theory is my strong point.😊😊😊😎😎
Why find the value of a, b, c and p is a prime we get the ab and a+b are both perfect square?
for case 3, m=p and we have p=b-1 as well
a=bp=b(b-1)
a+b=b^2. done
3:27 shouldn't you also consider m=-1, m=-p, and m=-pp? It's trivial to rule them out, of course, because a/b = m > 0. But shouldn't you consider it?
The question states that we are working with natural numbers, so I don’t think we have to.
Fascinatingly contrived problem.
Elegant summation of number theory.
St.Petersburg
I like when he says "and that's a good place todo stop"
239 is indeed a special integer because pi/4 = 4 arctan 1/5 - arctan 1/239.
See en.wikipedia.org/wiki/Machin-like_formula
And sometimes they can be both perfect square like when a=b=50 and c=48
I actually did a problem! A nice problem, but I'll take it nonetheless lol
Very nice
Big Michael Penn..
Nice problem
oh, i failed to find out where it came from, but thank you, the problem was so useful😊😊😊
Sir can You discuss paper of ISI entrance held today
Random and unrelated, c seems to be a multiple of 24?
Yes! This is the way I like to think of it:
For p >= 5, p^2 can be written as 24n + k, when n is a positive integer and k is a positive integer between 0 and 23.
k must be 1,5,7,11,13,17,19, or 23, otherwise 24n + k would be divisible by some number other than 1 or p, making p not prime.
Notice that 1*1 = 1, 5*5 = 25 = 24 + 1, 7*7 = 49 = 24(2) + 1, ... , 23 * 23 = 529 = 24(22) + 1.
Since k must be 1,5,7,11,13,17,19, or 23, and since each one of these is one more than a multiple of 24, k must be one more than a multiple of 24.
Then k = 24t + 1. k is less than 24, so t = 0. Then k = 1.
We see that p^2 is one more than a multiple of 24.
So c = p^2 - 1 = (24n + 1) - 1 = 24n.
This is even easier to see if you're familiar with congruence classes modulo 24.
239 is a prime number.
Nice..
SOS! SOS! SOS! SOS! Derivative of (cos inverse x /x) using first principle. I got stuck in too many terms.
Noice I would love to know the source
Yes 239 is a prime as it gives -1 as remainder when divided by 6
That is a property of primes, but not only primes. You accidentally used the converse of a statement.
If it was that simple to know whether a number is prime or not then we wouldn't have so many theorems to predict the pattern of primes?
Very nice proof
"That's the nice place to stop",common sentence used in every video
Well done for spotting that
/s
So, 60 is a prime!!! 😁😁👍👍
You must be thinking of 57 😉
@@BlackTigerClaws Ok good, the perfect square is ab and a+b are both perfect squares for the evaluate value and the answer is not prime!!! 😁😁👍👍
@@BlackTigerClaws If Grothendieck says it's prime, it's ok for me.
OK
pee pee