A nice math contest trick.

We present an important math contest trick involving reduction modulo n and perfect squares. A short problem is presented as an application.

Books I like:

Abstract Algebra:

Differential Forms:

Number Theory:

Calculus:

[joseluishablutzelaceijas]

Haven't seen the video yet, but I would say: Since k! ends in 0 for k > 4, the LHS will end in 3 for m > 3. But no square ends in 3, so the only possible solutions are the ones for m \in {1, 2, 3}, from which one sees that (m, n) \in {(1, 1), (3, 3)} are the only solutions.
Now I have seen the video... fine!
Wanted to say what many people have already said here: Thank you very much for the amazing videos and content!

[marcmaticas]

About the third follow up question that you proposed at the end of the video, I've tried all the values of m1 from 0 to 11 ({0, 1, 2, ..., 10, 11}) and I've concluded that the only solutions that exists with m1 being less than 12 are:
>> 0! =1^2
>> 0! + 1! + 2! = 2^2
>> 1! = 1^2
>> 1! + 2! + 3! = 3^2
>> 4! + 5! = 12^2

The method that I used is a generalization of the one used in the video:
i) We know by the Bertrand's postulate that n! for n>1 is never a perfect square, thus m2 > m1.
ii) Then we have to find a number M > m1 s.t. [(m1)! + (m1 + 1)! + (m1 + 2)! + ... + (M-1)! + M!] cannot be congruent to any square mod M+1
iii) Then, since for any m2 >= M you can express the factorial sum as [(m1)! + (m1 + 1)! + (m1 + 2)! + ... + (M-1)! + M!] + (M+1)*x where x is an integer, we conclude that the sum can only be a square when m2 < M.
iv) The last step is to compute all the values of m2 from m1+1 to M-1.

As an example I will prove that if m1 = 11, the factorial sum is never a square:
i) We notice that (11!+12!) ≡ 0 (mod 13), which can be a perfect square since 0 is a square mod 13.
ii) Then we notice that M+1 can't be 14 nor 15 nor 16 since 11! is a multiple of those numbers.
iii) For the next step we see that the squares mod 17 are {0, 1, 2, 4, 8, 9, 13, 15, 16}
iv) Finally we notice that (11! + 12! + 13! + 14! + 15! + 16!) ≡ 7 (mod 17). Since 7 is not a square mod 17 we just have to compute all the possibilities 11 < m2 < 16.
v) After checking we see that there does not exist any factorial sum starting with m1 = 11 that is a perfect square.

This method is ok for fixed values of m1, but if we want to see the full picture we need a more powerful method. But the technique from above lets us have some intuition of how rare the solutions are. I think that there are infinitely many solutions but they become ridiculously rare. This problem is also interesting because if we could show for what m1 and m2 the sum is a square, setting m1 = m2 we should get a proof of the Bertrand's postulate (assuming that we don't use it for this problem like I did above).

By the way, has anyone found any other solution? I specially loved the solution 4! + 5! = 24 + 120 = 144 = 12^2

[viratrobbie]

A nice observation about perfect squares is that they never end with 2, 3, 7, 8. 1! + 2!+3!+4! =33, and the subsquent factorials end with zeros: SO after 4, the sum always ends with 3. So you need to check till 4 only!

[cmielo]

I really like your videos. You deserve much more viewers. Keep doing great work.

[ramk]

Modular Arithmetic is a great tool for number theory problems.

Great explanation as always. Thank you.

[ramk]

For the follow-up question with starting index 0, I have a solution.

Please try yourself first if you haven't.

Since 0! + 1! + 2! + 3! = 10 = 2 (mod 4) and x^2 = 0, 1 (mod 4), m must be less than 4.

Checking the cases we get, (m, n) = (0, 1) and (2, 2).

[小江-ji]

I love your strong handwritting, it's a kind of confidence.

[andreben]

I absorb so many bits and pieces of olympiad level math with you it's amazing. I always had a problem with taking initiative with them and trying to do them myself. It's particularly nice and rewarding when I do work them out and watch your solution.

Thank you for all the hard and consistent work you put in.

[llumnate]

Not looking at the video, the sum of factorials from 5 onwards will have '3' as its last digit and squares never end up with 3 as a last digit.
Checking n=1..4, we get the answers (1, 1) and (3, 3)

[pokoknyaakuimut]

I love this channel. So many videos that help so many people in learning to solve math olympiad problem.

[kenthedawg]

I appreciate the follow up questions! It’s great to learn/relearn a trick but it makes it stick if I have to reinforce or expand upon it just after!

[sufchan]

I had the same approach, only that I used the fact that there is never a perfect square congruent to 3 mod 10.

the sum upto 4! is 33, and from 5! onwards there is at least one factor of 10, so the unit digit of all the "factorial sums" upto k! is 3 for n≥4, so k belongs to the set {1, 2, 3}.

[josephquinto]

Being a recent high school grad who is taking a gap year, your videos are very helpful in keeping my brain fresh. And the fact that I can follow along with you with only AP Calc under my belt shows how great of a teacher you are.

[andrewhaar]

In the case m=0 we just consider it to be the empty sum, which is defined to be 0, which is a perfect square. So I would argue that (0, 0) is a solution (if we are including 0 in the naturals).

[zakzaki]

Could you not say that anything above 4! willl be a multiple of 10, and since sigma4!=33, the sigma of anything above 4! will end in a 3. Since there are no square numbers that end in a 3, the only solutions are (k, n)={(1, 1), (3, 9)}

[bangaloremathematicalinsti]

You are a Great Inspiration to me...Love your videos

[tuffleader]

There are n possible residues modulo n, not n-1 residues.

[nnaammuuss]

Including the 0! term makes it easier; viz. the sum upto 3 is now 10=2 (mod 4). So, the more familiar _“every square is either of the form 4k or 4k+1“_ (just squaring 2r or 2r+1—iin fact gives 1 (mod 8) for odd numbers) leaves the only solutions to be (m, n) = (0, 1), (2, 2).

[axelperezmachado]

I really like your channel. You present very interesting topics that are not common to find in other more established math channels in a very understandable manner

[antoniopalacios]

The only solution to the first one is m=2, n=1. I get it the same way you do, Michael, but with p=7, where 0, 1, 2 and 4 are the only remainders that satisfy the condition (necessary but not sufficient) to obtain a perfect square. Thanks.