Problem Solving | Techniques from Number Theory

We look a few concepts and results from Number Theory that are commonly used in mathematics competitions. Solutions to two examples are also given, including one to the 1985 AIME.

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[Bodyknock]

7:46 When constructing that table of squares it’s actually helpful to, instead of list numbers 0 to 8, list the number -4, -3, -2, -1, 0, 1, 2, 3 and 4. Notice that the negative numbers in that list are congruent to 5, 6, 7 and 8 respectively, so you’re constructing the same equivalent list of residues. But, the key thing here is that obviously -3 squared equals 3 squared, -4 squared equals 4 squared, and so on. Thus you get a natural symmetry in this list of squares - the upper part of the list is just a reflection of the lower part.

It’s just a nice way to view the squares in modular arithmetic.

[goodplacetostop]

0:24 I hope whoever it was, that person is okay
28:20 All of a sudden, a video unlisted for months now goes public...

[ashrafyaseer]

If you read my comment, then I request you to make some videos like this on combinatorics

[littlefermat]

Nice video!
Number theory is definitely one of the most fun types in math Olympiad.
I actually see that the main subtopics in NT are (divisibility - mod - Diophantine equations- quadratic residues)
Of course there lots of other things as well...

[sharathpr]

Easy solution to problem 1:
The GCD must divide n² +100 as well as (n + 1)² + 100 ..
=> it also divides their difference (n + 1)² + 100 - (n² + 100) i.e. 2n + 1
Note that 2n + 1 is always smaller than n² + 100..
Therefore the max GCD is itself 2n + 1 such that it satisfies the condition
2n + 1 | n² + 100
=> 2n + 1 | 4n² + 400
=> 2n + 1 | (2n + 1)² - (4n + 2) + 400 + 1
=> 2n + 1 | 401
For maximum GCD, 2n + 1 = 401
n = 200 and max(GCD) = 401

[goodplacetostop]

HOMEWORK : Find all positive integers n which satisfy the following two conditions: (i) n has at least four different positive divisors;
(ii) for any divisors a and b of n satisfying 1 < a < b < n, the number b - a divides n.

SOURCE : 2010 Middle European Mathematical Olympiad

[ramakrishnasen]

This is like the summary of all your past number theory videos. "Now thats a good place to stop!"

[divyanshtripathi]

Thank you for this problem solving techniques in number theory.

[Zxv]

I always struggle to remember Fermat's theorem, but writing it as
a^p ≡ a mod p
It makes perfect sense and is borderline obvious. As you multiply a by itself it takes on a different value in the range 0-(p-1), and only after p multiplications does it return back to where it starts from.

[nl]

27:31 When I was minding my own business and suddenly my brain think a cringy moment I did 10 years ago

[level]

Straight up olympiad tutoring
Revolutionary stuff

[miserepoignee]

Regarding the last question, it is an open problem to find a general expression for an arbitrary number as a sum of three cubes, nor is it known whether every number congruent to something other than 4 or 5 mod 9 has a solution. Numberphile has a series of videos where they talk about these problems. While most numbers admit relatively tame solutions, there are a handful of numbers even under 100 that require numbers many orders of magnitude greater in their solutions.

[robertgerbicz]

Much easier way:
let d=gcd(a(n), a(n+1)) | a(n+1)-a(n)=2*n+1

then d |

and clearly for n=200 we have d=401, what we needed.

[KoshiPrime]

Your number theory videos are one of my favourites! 😊

[danielferdiansyah]

what a great video, it really helps for the beginner!

[reshmikuntichandra]

Thank you so so so so so much for this! I really needed it, and it'd be really helpful if you could make similar videos on the other topics of Math Contests (Geometry, Combinatorics, Functional Equations and

[maharanirani]

I'm kinda struggling for number theory problems, so I've been waiting for this kind of video cz I really need it for preparing contest. And I just can't believe that this video comes at the right moment when I really need it!

I have watched many videos of number theory problems from this channel.
Thank you very much for your help, Professor🙏.

[thekingadomas]

It's scary enough when there's 3 variables. The man literally intruduces like 7 variables :D

[davidemasi__]

What a video! However, I'd like to know better how to understand quickly I have to use 9 in the last exercise (if there is a trick)

[farissaadat]

17:30 Does anyone have a more full explanation of why x and y ought to be linear in n?