When is this a perfect square?
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- Опубліковано 28 вер 2024
- We solve a nice quadratic Diophantine equation.
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I really like these types of problems which donot require immense maths knowledge but still ineteresting to solve.
The goal of the Internet should be to help people that don't have immense math knowledge gain immense math knowledge by solving simple problems. Otherwise, the Internet is nothing more than a bunch of stupid viral videos and pictures of cats.
My solution is a bit different:
n²
Very good solution
This man is on another level of humanity
Could you solve
sps n,a is a natural number s.c
n(n+a)+2 is a perfect square
Please,i need your help
Without watching it, I got the following 4 answers (n,b):
(4,90)
(20727,21714)
(22747,23736)
(1020100,1021110)
Yea correct there are 4 solutions, i also got the same
@@sabitasahoo5388 I kinda found it funny that the smallest solution is 4, and the largest is well over a million
It looks like n=0 is obvious solution even not included in all natural numbers definitions and others one need calculator. WFA found 10 solutions in total when included negative values for n and m. Wikipedia:
Some definitions, including the standard ISO 80000-2, begin the natural numbers with 0, corresponding to the non-negative integers 0, 1, 2, 3, ... (sometimes collectively denoted by the symbol {\displaystyle \mathbb {N} _{0}}, to emphasize that zero is included), whereas others start with 1, corresponding to the positive integers 1, 2, 3, ... (sometimes collectively denoted by the symbol {\displaystyle \mathbb {N} _{1}}, {\displaystyle \mathbb {N} ^{+}}, or {\displaystyle \mathbb {N} ^{*}} for emphasizing that zero is excluded).
I hope to get better grab of this diophante style equations setting and solving.
Thanks for your posts. "Notice" that 2021 is 47*43 haha. In competitions, want to avoid lots of tedious trial and error for factors, can spot that it's 45^2-2^2. This is natural to consider since any non-prime n must have a prime factor less than or equal to its square root. Might as well check the difference at this point and say aha 2025-2021 is a square job done. No rabbit out of hat needed in this case. (By the way, checking for prime factors up to 13 has short cuts, so the trial and error only starts from 17.) More seriously, we would also only want to factorize when it's clear from the problem that it's necessary. It might be that for a random diophantine problem containing a number n, what you end up having to factorize is not n but say 2n+1. So laziness is the better strategy, until the precise need is known. Speaking of laziness, I wouldn't complete the square here: it doesn't achieve anything here, we can directly write n=a*k^2 and n+2021=a*l^2, where k&l are coprime. So a*(l+k)*(l-k)=2021 and our numbers stay much smaller, leading to 5 solutions n=0, 4, 20727, 22747, 1020100. I think from another video, you prefer that 0 is not a natural, but you should be aware this is just one interpretation (see wikipedia page of natural numbers, which says there are two conventions here) and it would be kinder if you always articulate that yours is just one of the two possible assumptions, to avoid any audience member slipping up in some future encounter with the naturals. Or better, use Z+ and talk about positive integers, if you insist on excluding 0 (although I disagree with that approach, as you can miss nice simple solutions, and creates a hole if you want to identify negative solutions too). Thanks again, and great to see the rock-climbing!
We were taught in school in multiple classes, and also it was written in multiple math textbooks (like multiple classes, I think class 6th and 7th) that natural numbers mean all the numbers from 1,2,3 and so on (also called counting numbers) and whole numbers were the numbers starting from 0,1,2,3 and so on (basically all natural numbers including 0)
@@anshumanagrawal346 Thanks for this. My point is only that the definition of naturals is not universal. But a new point is that in many problems (e.g. this one), the whole numbers are the right concept to start with, but are somehow overlooked because of a reflex to always work with the counting numbers.
Now days I think its de rigeur to check the factorization for the current year before sitting to write the exam. Use of the year number as a constant seems to be a standard way to customize problems for the one year of the current contest.
There is zero chance that I would notice that 2021 is 45^2-2^2, whereas I would be quite likely to notice that its square root is in the low 40s and it ends in 1, and give a try dividing it by 43 or 47.
Or you can use the Batman prime decomposition strategy : come prepared. If you are doing some sort of annual maths competition in year X, you can be sure some problems will involve the number X, so check beforehand X's prime decomposition and check if it's special in any way :D
Fun fact, The last perfect square year was 1936 (44²) the next one is 2025 (45²). This means most everybody on earth hasn't lived to see a perfect square!
Most of us will be lucky to see one, and it is pretty incredible to see two!
My mom was born in 1933. She is 90 now and I have every reason to believe she'll live to see 2025 ❤️
8:10
Incomplete lol
And that's a good place
@@srijanbhowmick9570 😂
Hw work n=4
H.W.: The rest of the answers are n=4 and n=20727.
Hi Michael,
I gave a go at this in the more general case where n^2 + pq*n = m^2 where p and q are non even primes (your case falls into it). This equation has 4 solutions of the form (2(p-1/2)(q-1/2) + p-1/2 + q-1/2)2, p(q-1/2)2, q(p-1/2)2 et (q-p/2)^2 which gives for p=43, q=47 n in {4, 20727, 22747, 1020100}
I need to watch the video again to try to understand why you stated at first that the number of solutions was only three :/
Anyway, I love your videos especially the calculus videos. Here's a nice one : int(0, 1) (ln(t))^n/1-t dt which has a nice closed form involving zeta and gamma.
another approach
n^2 + 2021n = m^2 => n^2 + 2021n - m^2 = 0 --(1)
since n is integer, discriminant(D) of (1) must be a square.
D = 2021^2 + 4m^2 = k^2 for some positive integer k
=> k^2 - 4m^2 = ( k + 2m) * (k - 2m) =2021^2
here we can get the values of k and m and also n.
n = 0, 21^2*47, 23^2*43, 2^2, and 1010^2 for non-negative solutions. For each of these m = -2021-n is also a negative solution
As a follow-up because Michael didn't use my method at all:
We can factor the expression as n(n+2021). If we let d be the gcd of n and n+2021, then n = da and n+2021 = db for some relatively prime integers a and b with d(b-a) = 2021. Then, our original expression is n(n+2021) = abd^2, which is a perfect square only if ab is.
Lemma (without proof, but just look prime by prime at the factors): If a and b are relatively prime positive integers, and ab is a perfect square, then a and b are both perfect squares.
So a = x^2 and b = y^2, and y^2 - x^2 = 2021/d. From this, we see that d must divide 2021 = 43*47, so d = 1, 43, 47, or 2021, leading respectively to the equations (y-x)(y+x) = 2021, 47, 43, or 1. If y+x=g and y-x=h, then x = (g-h)/2, so each pair of factors multiplying to either 2021, 47, 43, or 1 leads to a different value of x. So for d = 2021, we get x = (1-1)/2 = 0; for d = 47, we get x = (43-1)/2 = 21; for d = 43, we get x = (47-1)/2 = 23; and for d = 1, we get in one case x = (47-43)/2 = 2 (oops, gotta edit my original post!) and in the other x = (2021-1)/2 = 1010 (oops again, mental arithmetic is hard).
Finally, n = dx^2, so these lead to n = 2021*0^2 = 0, 47*21^2, 43*23^2, 1*2^2, and 1*1010^2.
If n doesn't have to be positive, then it's also possible for a = -x^2 and b = -y^2. This gives an alternate set of 5 solutions, though each of these solutions m corresponds to m = -n-2021 and m+2021 = -n for one of the original five
And that's a good place to stop this comment
0 is a solution! 0 is a square. The first thing I did was to write it as 2021n = m^2-n^2 = (m+n)(m-n). Are these 43 and 47? No, because that gives fractions. So double those. 90-4 and 90+4. Their product is 8100-16 = 8084 = 2021*4 and that gives 4*4 + 2021*4 = 8100 = 90^2, so 4 is a solution. There is one other solution 20727 and 21714.
Amazing !! i love it, Thank you for posting such great content.
Another solution factored this as:
n(n+2021)=m^2
If (n,n+2021)=d
Then d|2021 so we have four cases:
Case1 d=1then a^2-b^2=2021
Case 2 d=43 then a^2-b^2=47
Case 3 d=43 then a^2-b^2=43
Case 4 d=2021 then a^2-b^2=1
My approach: n^2+2021n
n(n+2021)
n=-2021
Plugging it in gets n^2+2021n=0 which is a perfect square
If we are just ignoring that n is a natural number, why not just use n=0?
@@BrutalBeast666 haha good point didnt see that
case 3: n=20727
case 4: n=4
Solution of the homework
From Morocco
Michael clearly stated that n is a NATURAL NUMBER by that he meant positive integers. N=0 should not be a solution.
? I argue that 0 is a natural number
@@jatloe watch 0:11 - 0:17
@@JankkPL o ok
@@The_Math_Enthusiast yes for sure
n= 4 and n=20727
4 & 20,727
Watching this a second time, I realized that you really can't ignore m completely. You need to make sure that its value is a natural number. It seems possible for instance that it could be fractional while still leaving n integral. Perhaps an argument could be made why that would never happen. Or you can double check each of the values of n you get.
It can't happen because n(n+2021) equals m^2, so if n(n+2021) is natural than so is m^2 and m.
4,20727,0
n shud be a natural no.
So it will not be 0.😁
Rest all are correct
@@ok-ez1xl 0 can be a natural number. Sometimes it is included in the natural numbers
@@ronludmer126 bruh
Natural numbers are the counting numbers, and in counting , we do not start from 0.
In whole numbers, 0 is included along with all the natural numbers.
Brute-forcing it in R
library(dplyr)
library(tibble)
numarr = tibble(
n = 1:2021^2,
result = n^2+2021*n,
sqresult = sqrt(result),
fl_sqresult = floor(sqresult),
resid = sqresult - fl_sqresult
)
) %>% filter(resid==0) %>% View()
I did the same thing, but in Python:
import math
for n in range(1000000000):
val = n**2+2021*n
root = math.sqrt(val)
above = math.ceil(root)
if(above == root):
print(n, val, root)
Hw work n=4
This was a bad place to stop! Lol
forget the problem 😜 I just want that hoodie
n=0 also works.
I mean how to get the factor of 2021is not easy in the first place. Any comnent ?
I think if you're doing a test then you would memorise the prime factors of that year and the two or three years either side. If you're at home doing it for fun then you can either use Google or the sieve of eratosthenes.
Umm u can first check whether it is a prime or not by checking its divisibility with the prime nos less than √2021 ≈ 45.
So after checking the last prime no. Before 45 , i.e. 43, u will get that 43|2021.
And that's a gud place to start solving 😂.
You don't memorise the prime factorisation of the current year? Weird.
you can also recognize that 2021 is a difference of squares:
2021 = 45^2 - 2^2
45^2 - 2^2 = (45-2)(45+2)
2021 = (43)(47)
@@leadnitrate2194 Ha HA
4 is a nice answer to this
Homework (and spoiler): The third is 4.
what about n=4 ? I just tried it in fact ...
Nope
@@tomctutor 4*4+2021*4 = 8100 = 90*90 .... What's wrong ?
@@gvomet1 Sorry you are right, can't imagine how I missed that even checking with my calculator, I raise my hand! シ
@@tomctutor Not an issue. As I told you, I often (always ?) try as the teatcher says : I try for the first figures (with excel) just to see what happens .....
Mayt he force of math be with you :-)
... to stop.
Scintillating problem
And that's a good place.
to stop
AND THAT'S A BAD PLACE!!! 😠😠😠😠😠😠😠
This channel is my good place.
TO START!!!😃😃😃😃😃😃😃
My approach to the problem;
n² + 2021n
=n(n + 2021)
=n(n + 47.43)
=n(n + (45+2)(45-2))
=n(n + 45² + 90 - 90 - 4)
=n(n + 45² - 4)
For n=4;
4(4 + 45² - 4)
=2²(45²)
=90²
Unfortunately, I see two problems in your approach.
1) you don't explain how you find n=4
2) you lack the 3 other solutions
But keep practicing :)
Math is cool
@@SogeYann Guessing a solution and checking if it works in the equation is perfectly valid. It can (and often is) used to prove that at least one solution exists. However, it *cannot* be used to find ALL solutions, unless you somehow manage to find the number of solutions beforehand and are able to guess all of them.
2021 has a very unique factorisation... At first I thought it was prime lol 😂😂😂
I never learned my 43 nor 47 times tables 😁
But it's a difference of squares, hence the factorization!
2021 = 45² − 2² = (45−2)(45+2)
@@samsonblack That's probably harder to spot than the prime factors. Although if you clock that 45^2 = 2025 it helps!
@@samsonblack I too did the same thing later
@@mcwulf25 well that's not hard to do either, (10a+5)² always ends in 25 and the other digits at the beginning are a(a+1)
Interesting and easy after watching XD
If n = 4, then n^2 + 2021n = 16 + 8084 = 8,100. The square root of 8,100 = 90, therefore when n = 4 you get 8100 [4^2 + 2021*4], a perfect square, since 90 * 90=8100. Maybe I missed something, but I didn't see that in your solutions. ???
Nice problem, that was what i needed to see
HOMEWORK : 7:58
Hw work n=4
hw: n=4 and n=20727
@ u missed the factorization 47*(43^2*47)
If n is not a natural (i.e., a positive integer), then n=0 is a solution as well.
H.w answer=20727,4
n^2+2021n
=n(n+2021)
Since 45*45=2025
So n=4
(4)(4+2021)
=90^2
4; 20'727; 22'747; 1'020'100.
Uma solução é n = 4
Another solution is n = 4
20,714.
Nice video!!
There is a nicer way to treat cases in the end.
You just let (2n + 2m + 2021) and (2n - 2m + 2021) be equal to arbitrary factors a and b of 2021^2. Then you can solve the system you get in terms of a and b, and adding you get
n = (a + b - 2*2021)/4
And all we need to do is just substituing a and b.
Yes, I've noticed (over a good number of his videos) that Michael is very strong on strategic thinking, but less so on tactical manouvering.
1/2019 also works as a solution
Ah, but that involves the rationals, and this is specifically a Diophantine equation - one where the only solutions we're interested in are integers.
I did it differently. One writes n² + 2021n = (n+k)² = n² + 2kn + k². Then n = k² / (2021−2k), which must be an integer. There are 3 cases. 1) If 43 | k, one writes k = 43m and n = 43m² / (47−2m), where gcd(m²,47−2m) = 1. Thus 47−2m | 43. Solutions: m = 2 and m = 23, i.e. n = 4 and n = 22747. 2) If 47 | k, one writes k = 47m and n = 47m² / (43−2m), where gcd(m²,43−2m) = 1. Thus 43−2m | 47. Solution: m = 21, i.e. n = 20727. 3) One has gcd(k²,2021−2k) = 1, and n is an integer only if 2021−2k = 1, i.e. k = 1010 and n = 1020100.
i get n = 4 and 21227
sorry..20727 (calculator typing error :-) )
That quantity is a multiple of 4 always cheking odd,even so what if you take mod 4? Then n is a multiple of 4 or n+1 is.
4 and 20,727
🥰🥰
1.020.100 ; 22.747 ; 20.727 ; 4
N=0 qed
n=-2021
This channel's a great place to stop
Case 3 : 2021² = 43²(47²)
2n - 2m + 2021 = 43²
2n + 2m + 2021 = 47²
Upon adding the equations
4n + 2(2021) = 47² + 43²
According to wolframalpha:
n =4
There is also a fourth case. Following your lead:
Case 4: 2021² = 43²×47(47)
2n - 2m + 2021 = 47
2n + 2m + 2021 = 43²×47
Upon adding the equations
4n + 2(2021) = 47 + 43²×47
n = 20,727
Right!
CASE 3 = 4
1/2019 works:
Start with two consecutive squares, aka n^2 plus (n+1)^2. The difference between them is equal to 2n+1, I calculated. So then if we set 2021n equal to 2n+1 you get 1/2019.
Edit: so then are there five solutions?
You first assume a natural number n, just as th äe problem in the video does. Obviously 1/2019 is not natural
@@johannesh7610 I totally missed him saying it has to be a natural number
This one was pretty easy to solve if you remembered that (far more challenging) one from October: ua-cam.com/video/n0RF03Rm2qw/v-deo.html
The problem seemed impossible to solve at the first glance.Amazingly you solve it easily .
I got the same answer for real numbers, but I think my way was simpler in this case.
n^2 + 2021n will be a perfect square whenever it can be written as the sum of an arithmetic series: 1 + 3 + 5 + 7 + 9 +....
I used this as the base assumption.
Another interesting way to look at it for all real numbers is to say n^2 + 2021n will be a perfect square whenever it may be written as (n+k)^2 for some real integer, k. We have n^2 + 2kn + k^2 = n^2 + 2021n
2kn + k^2 = 2021n
k^2 = 2021n - 2kn
k^2 = n(2021 - 2k)
(k^2)/(2021 - 2k) = n; k doesn't equal 2021/2
Thus, n^2 + 2021n will be a perfect square for all integer solutions of (k^2)/(2021 - 2k).
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If we try to factor 2021^2 as 2022(2021) we see that that means -2m must be equal to 2m, which means m is 0. If 0 is not considered a natural number, we can stop here for this attempt, even if it is, we will see that n^2+2022n = 0 can only be satisfied with negative n values, which are obviously not natural numbers and n has to be a natural no.
Who do that third one 😂😂😂😂😂😂😂
You missed one case where (2n+2m+2021) = 47^2
No, he didn't. He left the other cases as homework (listen at 7:57).
Try this : when is n+3 and 2020n+1 is both a perfect square for smallest n.
n=2022?
First step would probably be to find all values of n for which 2020n+1 is a perfect square (2020n=(m+1)(m-1)), then figure out which ones work for n+3 being a perfect square. Oddly enough, I can only come up with three values for n that work, due to the factorization of 2020: 20 (m+1=202, m-1=200), 2018, and 2022. And of them, only 2022 is three less than a perfect square.
n(2021 + n) is square so we can write n = x^2*d and 2021 + n = y^2*d we get 2021 = d(y-x)(x+y) we just have consider this equation
n(n+2021)=k^2
First, let's suppose n=m^2 is a perfect square, then n+2021 must be also, so n+2021=m^2+2021=(m+a)^2 for some a>0. From this we have 2021=43·47=a(2m+a). Checking all possible combinations of factors, we end up with m=1010,2 as only solutions. So n=1010^2=1020100 and n=4 are the only solutions if n is a perfect square.
Now let's rearrange the initial equation into 2021n=k^2-n^2, and 2021=43·47=k^2/n - n. This means n | k^2. Let's express n as n=a·(s)^2 where a has only prime powers of exponent 1. Then k^2=s^2 · a^2 · q^2, where q has the rest of k that doesn't belong to s or a. Then the equation is
43·47=a·q^2-a·s^2=a(q-s)(q+s). Since we've already dealt with n being a perfect square, we can impose a!=1, so we're left with
43=(q-s)(q+s) (for a=47) and 47=(q-s)(q+s) (for a=43). Since q,s>0 the only solutions are q-s=1 and q+s=47,43. Those solutions correspond to n=22747 and n=20727.
0:01 Good Place To Start