In other words, when you are given some ugly quantity is a perfect square, Always try proving that this quantity is between two consecutive perfect squares. And your problem will be done!
Around 9:00 when he offers a^2+2a+1 as being greater or equal to a^2+a+2 this is only true of a>0. By this point though we could have observed that the formula m=2^(a+1)-3 gives a negative m when a is zero and so excluded this case.
My favorite proof of the "exponential outpaces polynomial" fact is the following: Let P(x) = sum a_i x_i be a polynomial and E(x) be some exponential function e^(kx). Then lim as x tends to infinity of E(x) / P(x) is infinity by L'Hospital. This is because the quotient is infinity / infinity, so applying L'Hospital we get an exponential over a polynomial of degree 1 minus what we started with. Continuing this process will leave a constant polynomial in the bottom and infinity in the top.
What happens if we allow m,n to be negative? EDIT: The fact that we always want a perfect square forces n to be positive. If we allow m to be negative, then it can be argued similar to how it was done in the video that m=-1 and n=1 is the only other possible solution over the integers.
5:48 I don't understand why the middle term must be a perfect square. I follow the proof up to that point, but this step confuses me. Edit: OH! It's part of the requirements on the left. I couldn't see because Michael was standing in front of it.
No need for calculus to prove that 2^a > (a+1)^2 for a ≥ 6. This can be done by induction: true for a = 6, and then 2^a doubles at each iteration while the factor for the square is ((a+2)/(a+1))^2 < 2.
Hey Michael, It'd be great if you tried some problems from the 2021 INMO(Indian National Mathematical Olympiad). Regards, Adarsha Chandra(Fan from India)
This is confusing, so I recommend switching the orders in which he made the assumptions. First note how the polynomial that bounds m is less that 2^(a+2), then substitute that in for m in the bottom to create an inequality and notice how it factors into (2^a + 2)²
Here is an extremely hard math problem you can try: Find all (a,b,c) which are natural numbers which satisfy both of these equations: 1/a=(1/b)+(1/c)+(1/abc) and (a^2)+(b^2)=c^2 dosen't seem like much but trust me it is a monster to prove.....
HOMEWORK : Let n be an even number which is divisible by a prime bigger than √n. Show that n and n^3 cannot be expressed in the form 1 + (2x + 1)(2x + 3), i.e., as one more than the product of two consecutive odd numbers, but that n^2 and n^4 can be so expressed. SOURCE : Suggested for the 25th Spanish Olympiad but that didn’t make the cut.
The conclusion is equivalent to the number being an even perfect square, so obviously n^2 and n^4 are, but n (and therefore n^3) can’t be because it can only have one copy of the large prime in its factorization.
1+(2x+1)(2x+3) is (2x+2)². Let p be the prime larger then √n that divides n. Then we have p²>n; Therefore p² does not divide n. Hence n is not a perfect square. Similarly you can prove that n³ is not a perfect square. N²,n⁴ are perfect squares therefore we can express them as (2x+2)² with x=(n/2)-1 and (n²/2)-1 respectively. Both of which must be natural.
One more way to solve; (n^2+1) is divisible by 2m So n^2+1 is even; consequently n^2 is odd. Therefore n is odd. n-1 is even. Let (n-1)/2=k ( eq.1). Where k is some natural no. n-1=2k. 2^(n-1)+m+4= 2^2k+m+4= (2k)^2+m+2^2. Now; 2^(n-1)+m+4 is a perfect square. So (2k)^2+m+2^2 is a perfect square. So 'm' must be equal to +2*2*2^k or -2*2*2^k. as it will give (2^k +/- 2)^2 since m is natural. We have m=2*2*2^k 2^k=m/4. 2= kth root of (m/4). Now k cannot be grater than 2 other m/4 will not be a natural no. and cannot be equal to 2. so k=1 or 2. putting (n-1)/2 =k= 1 or 2 (by eq.1) Therefore n=3 or n=5
This is a little inelegant; but, here goes. Assume there is some value of a such that 2a^2+2a+1=2. So, what this shows is that if the inequality holds for any a>=2, it also holds for all higher values of a. And, it just remains to check 0, 1, and 2, which is simple enough and returns 1
Выручили!) Мне как раз сегодня дали эту задачку, я порешала, но ничего не вышло( а это видео быдо на очереди к просмотру уже месяц, и вот как пригодилось🤩🤩
You seem to be incorrect here. (5,57) satisfies the first condition, namely (57^2+1)/(2*5) = 325 which is an integer. However, the second condition yields: 2^(57-1)+5+4 = 2^56+9 which is not a perfect square. It lies strictly between two consecutive perfect squares, namely (2^28)^2 and (2^28+1)^2, and therefore cannot be a perfect square itself. Did sqrt(2^56+9) perhaps look like an integer on a computer with insufficient precision?
In other words, when you are given some ugly quantity is a perfect square, Always try proving that this quantity is between two consecutive perfect squares.
And your problem will be done!
Yup! A typical bounding trip which I got to know 3 months back.
Lol, i dunno why, but we call it policeman's lemma😂
Nice. I like the style of leaving some of the easier bits as HW and avoiding a 20 minute video.
Well explained. I like it when the answer isn't just 0 or 1.
Around 9:00 when he offers a^2+2a+1 as being greater or equal to a^2+a+2 this is only true of a>0.
By this point though we could have observed that the formula m=2^(a+1)-3 gives a negative m when a is zero and so excluded this case.
My favorite proof of the "exponential outpaces polynomial" fact is the following:
Let P(x) = sum a_i x_i be a polynomial and E(x) be some exponential function e^(kx).
Then lim as x tends to infinity of E(x) / P(x) is infinity by L'Hospital.
This is because the quotient is infinity / infinity, so applying L'Hospital we get an exponential over a polynomial of degree 1 minus what we started with. Continuing this process will leave a constant polynomial in the bottom and infinity in the top.
Important caveats:
(1) this is directionally dependant. So if x is going to positive infinity, k>0 or else e^(kx)
Yes, but this does not give a quantitative bound.
What happens if we allow m,n to be negative?
EDIT: The fact that we always want a perfect square forces n to be positive. If we allow m to be negative, then it can be argued similar to how it was done in the video that m=-1 and n=1 is the only other possible solution over the integers.
5:48 I don't understand why the middle term must be a perfect square. I follow the proof up to that point, but this step confuses me.
Edit: OH! It's part of the requirements on the left. I couldn't see because Michael was standing in front of it.
We want it to be. It's not necessary. According to the problem statement.
No need for calculus to prove that 2^a > (a+1)^2 for a ≥ 6. This can be done by induction: true for a = 6, and then 2^a doubles at each iteration while the factor for the square is ((a+2)/(a+1))^2 < 2.
That's what I thought too.
8:58 a²+a+2
I was just warming up to studying Japanese, this sort of counts right?
Inequality proved by calculus could also be proved by induction more easily
Brilliant! Love this channel
I thought this was going to be one of the ones where all the possible values are like 2 and 3, but that 61 and 11 works is kind of surprising.
They probably wrote the question intentionally to suit to those numbers.
Hey Michael,
It'd be great if you tried some problems from the 2021 INMO(Indian National Mathematical Olympiad).
Regards,
Adarsha Chandra(Fan from India)
13:57 それはいいところで止まっている
That’s so good 😂
4:10 why is this possible? Why can we just put that bound on m
This is confusing, so I recommend switching the orders in which he made the assumptions. First note how the polynomial that bounds m is less that 2^(a+2), then substitute that in for m in the bottom to create an inequality and notice how it factors into (2^a + 2)²
Here is an extremely hard math problem you can try:
Find all (a,b,c) which are natural numbers which satisfy both of these equations:
1/a=(1/b)+(1/c)+(1/abc) and (a^2)+(b^2)=c^2
dosen't seem like much but trust me it is a monster to prove.....
Pythagorean triplets
HOMEWORK : Let n be an even number which is divisible by a prime bigger than √n. Show that n and n^3 cannot be expressed in the form 1 + (2x + 1)(2x + 3), i.e., as one more than the product of two consecutive odd numbers, but that n^2 and n^4 can be so expressed.
SOURCE : Suggested for the 25th Spanish Olympiad but that didn’t make the cut.
The conclusion is equivalent to the number being an even perfect square, so obviously n^2 and n^4 are, but n (and therefore n^3) can’t be because it can only have one copy of the large prime in its factorization.
Video was uploaded 1 week ago, then how is your comment 3 weeks old?
@@keyanaskar1981 Time travel
Nice brother, but wth.
1+(2x+1)(2x+3) is (2x+2)².
Let p be the prime larger then √n that divides n.
Then we have p²>n;
Therefore p² does not divide n.
Hence n is not a perfect square.
Similarly you can prove that n³ is not a perfect square.
N²,n⁴ are perfect squares therefore we can express them as (2x+2)² with x=(n/2)-1 and (n²/2)-1 respectively. Both of which must be natural.
One more way to solve;
(n^2+1) is divisible by 2m
So n^2+1 is even; consequently n^2 is odd. Therefore n is odd.
n-1 is even. Let (n-1)/2=k ( eq.1). Where k is some natural no.
n-1=2k. 2^(n-1)+m+4= 2^2k+m+4= (2k)^2+m+2^2.
Now; 2^(n-1)+m+4 is a perfect square. So (2k)^2+m+2^2 is a perfect square.
So 'm' must be equal to +2*2*2^k or -2*2*2^k.
as it will give (2^k +/- 2)^2
since m is natural.
We have m=2*2*2^k
2^k=m/4. 2= kth root of (m/4).
Now k cannot be grater than 2 other m/4 will not be a natural no. and cannot be equal to 2. so k=1 or 2.
putting (n-1)/2 =k= 1 or 2 (by eq.1)
Therefore n=3 or n=5
but this is false. 25+1/2^k+3 which is not an integer
Nice little sojourn!
I am a time traveller
@@aashsyed1277 liar! he is just smart
@@timetraveller2818 🤣😂🤪
@@amiasam3354 xd
@@timetraveller2818 🤪
The intervention prism in math is equal to zero. Not equal one of n
And one value.
I got one answer (m,n) = (1,3) by just observation
But second answer is quite surprising 😅😉😁👍
Thank you...
1:13 why did you already delete the m value??
Why 2^(2a)+m+4
Well, we don't need to check the perfect square condition at all, since m was constructed to be a perfect square.
2:15 why did you say that the denominator is equal or less than the numerator
Otherwise you can get fractions / decimals and according to the question you can get only natural numbers.
@@harivatsaparameshwaran4174 thank you very much
Its not hard to see how m
Nice video, this comment is for the algorithm
Harder one next time please!
this was really interesting
Nice video
Great !
can someone please explain to me how to prove 2a^2+2a+1
This is a little inelegant; but, here goes. Assume there is some value of a such that 2a^2+2a+1=2.
So, what this shows is that if the inequality holds for any a>=2, it also holds for all higher values of a. And, it just remains to check 0, 1, and 2, which is simple enough and returns 1
a mathematical induction!
It's very nice work .I love this challenge 👍👍👍
Выручили!) Мне как раз сегодня дали эту задачку, я порешала, но ничего не вышло( а это видео быдо на очереди к просмотру уже месяц, и вот как пригодилось🤩🤩
the couple m = 5, n = 57 seems to work but you will see below why it isn't , through informed responses
sqrt(2^(57-1) + 5 + 4) = 268,435,456.0000000167638063...
so m = 5 n = 57 doesn't work, 2^{n-1} + m + 4 isn't perfect square
@@erikpedersen2245 Indeed, my digital calculation software rounds this number to an integer...
You seem to be incorrect here. (5,57) satisfies the first condition, namely (57^2+1)/(2*5) = 325 which is an integer. However, the second condition yields: 2^(57-1)+5+4 = 2^56+9 which is not a perfect square. It lies strictly between two consecutive perfect squares, namely (2^28)^2 and (2^28+1)^2, and therefore cannot be a perfect square itself.
Did sqrt(2^56+9) perhaps look like an integer on a computer with insufficient precision?
He proved mathematically there's no other solution, Python won't help you...
please sir... can I get some theoretical maths thats not niche equation solutions?
Nice
Bella appiddaveru.
Weeb problem FTW! Great!
I follow math elite idea
Great video btw
😍😍😍
Nice