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@@wilfredrohlfing7738 every cube is 0, 1, or 6 modulo 7. Therefore, we have 3x3 =9 possible outcomes modulo 7. You could simulate this easily in Python to determine if your claim of 19/49 is correct

​Select 1st integer n: (let n==a be n³=a mod7) P​(n==0) = 1/7 P(n==1) = 3/7 P(n==6) = 3/7 Same probabilities for 2d integer m. Now probability of sum of cubes of two independent, randomly drawn integers is divisible by 7 becomes: P(n==0) × P(m==0) + P(n==1) × P(m==6) + P(n==6) × P(m==1) = 1/49 + 9/49 + 9/49 = 19/49 @@MyOneFiftiethOfADollar

The sample space/total outcome space is comprised 9 ordered pairs of the form (A^3,B^3) . This is because any integer cubed is congruent to 0,1,or 6 modulo 7. Only three ordered cubed pairs sum to zero modulo 7, namely (0,0), (1,6) and (6,1) Therefore the probability is 3/9=1/3. I made the same natural mistake you made of assuming there where 49 possible outcomes(which is true if we are summing x+y where x and y are random integers, but not x^3 + y^3 where only 9 outcomes exist modulo 7

​@@MyOneFiftiethOfADollarx³ and y³ are random integers and the same space of their actual sums totals 49. If you actually added all 49 combinations of 1³ thru 7³ you would see exactly 19 of those sums are divisible by 7. I've run 3 trials of 1000 selections on programmable calculator getting 0.378, 0.391, & 0.385 after summing cubes of two randomly generated integers from {1,2,3,4,5,6,7}

The question is to select a random integer, not just 1,2,3,4,5,6,7.I realize every integer is congruent to 1,2,3,4,5,6,7 modulo 7, but still prefer to try and simulate the exact question) I used excel and selected two random integers between 0 and 200. Summed their cubes and got 8/24=1/3.(8 multiples of 7) I got a variance from 1/3 when number of trials was less than 24. I will do larger number of trials later on today. 19/49 and 1/3 are close enough as decimals to where small numbers of trials won’t necessarily distinguish……

@@samdaman2510 that’s true, but do you think what you said would be accepted as a proof on an exam in a number theory or discrete math course?

It is always “easy” if one uses technology or a calculator or guess and test at a particular step.

True enough if you are talking about the linear algebra version of Cauchy Schwarz inequality. There is a real number formulation of CS which was used since sinx and cosx are real numbers. One is not required to always interpret CS in the vector sense.

@@piraka_mistika would you just show us all the steps if one chooses long division over modular arithmetic?

@ it’s 10 iterations of multiplication and subtraction

It is also quite error prone and tedious, but you do produce the quotient also which is something modular arithmetic does not give us.

Thanks, it looks like you did it essentially the same way taking advantage of the fact that 10^2 - -1 = 101 and (10^2)^n == (-1)^n mod 101

You may already know this, but this problem is really simple to solve. Any n-digit number which start with a digit k and the rest of the digits are 9s are solutions. So your conjecture that this is an "one-off" coincidence is false.

Thanks! Your binomial expansion idea appears to give the last four digits, namely 6447. Was cool how you split it up to yield tractable computations.

Both, this one was done when Covid had shut down physical classes at the school where I taught. So I was getting used to zoom, canvas and other distance learning online platforms. Now I try to do one video a day on whatever topic interests me at the moment. I’m not caught in the trap of trying to increase subscriptions or page views that so many are stuck in! There are so many math channels competing trying to get viewer’s attention so the channel owner can earn ad revenue. Michael Penn’s channel was the one that got me interested in the idea of creating high quantity videos to find other people who share similar math interests.

@@MyOneFiftiethOfADollar That sounds good, and I imagine it can be pretty rough trying to put yourself at the mercy of The Algorithm to make money!

Right, one is definitely subject to the whimsical algorithm which has no idea about the quality of the math, but favors the most profitable video from an advertiser perspective.

I may be wrong but read somewhere that an average quality math video with a lot of comments, can earn more than a high quality math video with few to no comments. Or some viral phrase like real men and Ramanugan.

@@MyOneFiftiethOfADollar Oh yeah, I'd totally believe that. Tracking metrics like number of comments is probably much much easier than somehow trying to evaluate the quality of the math being done in a video...

Thanks, I also liked the way complete the square was key in the solution process. We have lots of hummingbirds in our backyard. I am fascinated by them and the way they can scare off much bigger birds! I think the background image is kind of a fractal rendition of a female ghost, but not sure.

@@MyOneFiftiethOfADollar That's cool! I like hummingbirds too, though we don't have any in our backyard. A fractal rendering of a female ghost sounds interesting.

I'll take that as a sign that you understood the key concept, which is that the units group of a ring is cyclic. 😂

I will take your comment as a sign you enjoy showing off your "knowledge". Tell us more about this ring structure and its two operations and the cyclic nature of the units group. What are the units for the problem you are referencing?

@@MyOneFiftiethOfADollar Aren't some units groups noncyclic? Did you mean specifically the units groups of Z/nZ where n is prime?

@@MyOneFiftiethOfADollar No, I don't really enjoy showing off, I was just trying to clarify something and lock in my own understanding. As far as I understand it, and looking things up just now, "units" are the elements that have multiplicative inverses in rings of integers modulo n. And the "group of units" consists of those units together with the operation of multiplication, and is only cyclic when n = 1, 2, 4, p^k or 2p^k for some odd prime p and k is an integer >= 1. Does that match your understanding?

All the nonzero elements in Z/19Z are units since 19 is prime and every positive integer less than 19 is relatively prime to 19 and hence a unit, but I still don't know how that relates to your original comment in this thread. "units group of a ring is cyclic"

I have done quite a few sum of digits(sigma function) problems. Can you be more specific about the type of problem? The bounding is solely determined by the prime factorization form of the solution

@ what’s the title? I can’t found it

Proof that sqrt((x+y)/(x+y+z)) + sqrt((x+z)/(x+y+z)) + sqrt((y+z)/(x+y+z)) less than sqrt(6) th-cam.com/video/uSs_BgUe2G4/w-d-xo.html

Your choice of 6 and 5 is not a counterexample. Reason: (a+b) = 6 and (a-b) = 5 implies a =11/2 (Not an integer) Your "refutation" is the mathematical equivalent of the legal straw man as what you said is true, but not applicable to this problem. Be careful

​@@MyOneFiftiethOfADollar I did not understand the step before that : how can u say that ((a+b) = 2 and (a-b) = (1+2k)) or ((a+b) = (1+2k) and (a-b)=2) ?

I think what you are overlooking is the fact that (a+b)(a-b) is completely factored as is 2(1+2k). Your "straw man" was x=pq and choosing 6 and 5 which are factors of 30, but do not "fit" the form (a+b)(a-b) x= 2 + 4k leads to all of the aforementioned which results in a = (2k+3)/2 being a rational number that is not an integer Remember, we are trying to show that any integer in the form x= 2 + 4k cannot be expressed as the difference of two square integers. The video did that independent of your concerns over alternate irrelevant integer factorizations

@@MyOneFiftiethOfADollar I'm sorry but I still don't get it : how can we say that (a+b)(a-b) is completely factored as is 2(1+2k) ? Does that mean that all (a+b), (a-b) and (2k+1) are primes ?

x=(a+b)(a-b) AND x=2(1+2k) Take your example of 30 = 2(1+2k)=6*5 = 2(1 + 2*7) = (a+b)(a-b) This implies a+b = 2 and a-b=15 OR a+b = 15 and a-b=2 In either case after summing, we see 2a =17 or a=17/2(NOT an integer) If this doesn't help you, consider finding a number theory tutor. You are fixating on things that are not relevant to the problem statement. Good luck

Could you take the time to show us the steps in your "long division" of a Seven Digit integer with a single variable digit? I knew that was theoretical approach, but it seemed tedious/difficult. Thanks

@@MyOneFiftiethOfADollar I only did long division between integers without any variable digits (first 3 lines in my previous comment). When I wrote it I didn't think about how similar 0 (zero) and D (dee) look in 1760283 vs. 176D283. The only Ds in my previous comment are in the final (4th) line. 176D283 = 1760283 + D * 1000 = 5 + D * (-1) (mod 13) Hope that helps.

@ it does help! Your way is the most direct way to solve problem. What I did in video depends on a divisibility test that I proved in another video and then some modular arithmetic.

The way I did it depends on an ad hoc divisibility test for the integer 13 and then modular arithmetic. Your way uses a reliable classic approach called long division and modular arithmetic. Long division becomes more difficult as the size of the integer increases, but if we switched the problem to say divisibility by 29, long division is still a better strategy since coming up with divisibility test for 29 could be unwieldy!

There is a number theory text book that strongly advocates parity as a problem solving approach over a broad range of problem categories. I don't remember the author's name but will enter it here if I remember. Parity is essentially the verbose form of modulo 2. What you shared seems right and an example of a wordy proof that some prefer over terse notation and symbols.

@@arsenypogosov7206 since 0-1 = -1==3 modulo 4

@@arsenypogosov7206 glad to know that != is the same as “not equal” 😀

@@SrisailamNavuluri How do you know 1691001273 is divisible by 13?

@@Phanatomicool yes that is also a proof. The video proved it directly as a consequence of the integer equalling 4k + 2

I had seen this stated equivalently as the sum of the first n odd integers is n^2. It’s a pretty standard induction proof probably with a cool geometric interpretation. Quite interesting that one can use it as a proof by exclusion in this case.

Also, just noticed that x^2==0,1 modulo 4 is easier to prove directly by just checking 0,1,2,3 which are the possible remainders when any integer is divided by 4.

@@AlgebraicAnalysis yes, the binomial theorem is often “mr reliable” for these problem types. The video took advantage of the simple truth that 46 must divide x-1 because the given tells us 46^51 divides x-1. It’s a case of two different algebraic identities yielding same result.

Thanks for the proof you did not watch video. Consider doing videos on your channel so you have to deal with half-wit comments such as yours

You wrote sqrt(2)/14 = sqrt(1/99) which is false as an equality. Perhaps you did not notice we are solving an inequality, not an equality. If last fraction is 35/36, then we have (1/2)(3/4)(5/6)……(35/36) < 1/6 = 1/(sqrt(36)) Hope this helps. In the middle of doing a video for the general case of this problem type.

1/99 < 1/98 leads to stronger result you referenced. The end of video showed entire string of fractions < 1/sqrt(98) = sqrt(2)/14

@@krisbrandenberger544 right, I thought it was neat how the floor function combination “signaled” all the divisors of the function input

If you are asking, are there an infinite number of n such that mu(n) is not equal to zero then yes. This is straightforward from definition of mu(n) since product of n = product of distinct prime numbers results in mu(n)=+/- 1

Thanks much for your professional and well informed refutation of a thumbnail statement which is erroneous by counterexample. The counterexample was revealed in video, but not the thumbnail which irritated a few, but you can’t please all the people all the time. It is common mistake with undergrads and professors alike to want their proof to be correct to the point of overlooking the statements you mentioned in your comment.

You are missing something. Geometric series is part of the derivation of the sigma function. You reinvented the wheel. I started with the formula for the sigma function as a given. You are like many, eager to impress and show off your knowledge without taking the time to understand the video. You did not understand the claim well enough to ascertain whether it was correct or not. Create some videos on your own channel and be prepare for people like yourself to leave comments similar to the one you composed here.

@@graf_paper for the general product of fractions you mentioned, the upper bound is 1/sqrt(n). Yes, it is “tight” in the sense the ratio (1/(n+1))/(1/n) approaches 1 as n approaches infinity.

Yes, one is divisible by 2^2 and the other by 3^2. The bigger challenge may be to show there are an infinite number of these type of proofs. 😀

Thanks much. Your way is more "first principles" since it does not rely on the cardinality of the prime numbers being infinite. Did you take number theory or more of a hobby for you?

Also, try proving u(n) + u(n+1) + u(n+2) = 0 also has an infinite number of solutions.

@@MyOneFiftiethOfADollar I found that 48, 49, and 50 are the first three consecutive non-square-free numbers. 48 = 2⁴ · 3 49 = 7² 50 = 2 · 5² You can generate infinitely many such triples in the following manner : Since the prime divisors of 48, 49, 50 with powers greater then 1 of these three numbers are 2 ,5, 7: we can build infinitly many triples of the form: 2² | { 48 + k(2×5×7)² } 7² | { 49 + k(2×5×7)² } 5² | { 50 + k(2×5×7)² } And get another sequence of three consecutive non-square free numbers (k∈ℕ). I'll be honest, my first thought is to write out and solve a system of congruences: n ≡ 0 (mod 4) n + 1 ≡ 0 (mod 9) n + 2 ≡ 0 (mod 25) Ha, that was a really fun problem!! Thank you so much. Now I am super curious about what can be proven about n-conservative non-square-free numbers. Did you prove it in a different way?

@@MyOneFiftiethOfADollar I am a math teacher, tutor, and a math team coach. Math is definitely a hobby and have recently started working through textbooks more seriously. I have found TH-cam to be an extraordinary place to learn and love the math community here.

I hadn't done it yet. Just learned it from you! AND yes n-consecutive of the form u(k) = +/- 1 is the interesting natural generalization. These are all problem types from Charles Eynden's elementary number theory text book

Thanks again! I believe there is a class of arithmetic functions that are multiplicative without the condition of relatively prime inputs being imposed. Called totally multiplicative or something? Quite straight forward to see why both the number of divisors function called tau by many is multiplicative. Perhaps a little more difficult to show sum of the divisors sigma is multiplicative. Not so sure about the Totient function/Euler Phi function

@MyOneFiftiethOfADollar I have encountered the proof that σ(ab) = σ(a)σ(b) in the Euclid-Euler theorem which relates perfect numbers to Mersenne Primes. (an even number is perfect if and only if it has the form 2^(p−1)(2ᵖ − 1), where 2ᵖ - 1 is a prime number) Eulers proof uses the multiplicative property of σ(x) and it was very memorable the first time I saw it. I actually haven't thought about the class if multiplicative arithmetic functions before and now am pretty interested!!

@@MyOneFiftiethOfADollar Oh that is interesting, I haven’t really considered the class of multiplicative functions before. I have seen the prof that σ(ab) = σ(a)σ(b) in the proof of the Euclid-Euler Theorem that an even number is perfect if and only if it has the form 2ᵖ⁻¹(2ᵖ − 1), where 2ᵖ − 1 is a prime number. The proof of the multiplicativity of σ(x) and φ(m) are both wonderful bits of mathematics and i have the Micheal Penn videos of their proofs saved.

@@graf_paper Michael Penn’s channel was my first math love during pandemic. He is extraordinarily gifted at producing high quality content. He holds down a full time face to face teaching position as well. Not surprised at all you mentioned him! If you can prove there are no ODD perfect numbers, you will become an overnight sensation. Still unproven as far As I know. The form you mentioned generates new even perfect numbers every time they waste the electricity to find another prime of the form 2^k - 1 And of course i don’t really believe it is a waste of energy 😄

Thrilled about your interest in this topic. The Mobius function is theoretically important when finding inverses of arithmetic functions. You may be more familiar with that than me. Currently studying it....... The notion of square free integers is a big deal and the Mobius function measures that in a manner of speaking. An integer is either the product of distinct primes OR is divisible by at least one squared prime. All integers are a member of one of those two sets. Simple but still fascinating.

@@RexxSchneider thx for writing back! k must be 4 since k(k+1) is the product of consecutive integers and is also = 4(4n^2 + 1) which cannot be equal to the product of consecutive integers unless n=1. I probably didn’t say or write enough during the video to convince viewers k has to be 4………I still have some ennui because 4(4n^2 + 1) logically does not have to be product of consecutive integers but think the uniqueness of the prime factorization/parity might play into claim k=4

You just proved you are afflicted with attention deficit disorder. The video proved what you “merely asserted”. Also it’s not clear whether you can read and interpret plain English based on your careless and thoughtless commentary. How dreadful it must have been for those poor University of Kentucky professors to give your parents a break from you. .

I would hope for a more mature response from both parties involved, especially if you're both scholars. Can't we just accept that there are mathematical concepts that are simply fun to explore? Everything we do in life doesn't have to posses exceptional purpose. Also, kindly leave the neuro-divergent community alone with your harsh comments about ADHD. It's clear little empathetic consideration was taken before that response. Being in a STEM field, there's a good chance you're unknowingly attacking your colleagues as the field is popular with those having ADHD, Autism, or both.

You would or you do hope? Keep hoping. Your comment constitutes an act of immaturity by mislabeling an action as an “attack” Another over sensitive petulant who is probably ADD himself. In your safe space world it’s never OK to criticize anything unless you are the one imposing your personal sense of right and wrong.

I hope your situation improves. Cheers mate

I am guessing a guy like you hopes everyone’s “situation” will improve.

That is true

The statement is absurd because it does not hold for all prime k as the counterexample in the video shows for k=11. It is not “absurd” for the reason you stated since there are an infinite number of primes 😬

@@davidcawthorne7115 The point of the problem is to find the mistake in the “proof”. I believe 2^11 - 1 = 2047=23*89 is the first counterexample that refutes the thumbnail. The converse of the thumbnail is true, namely if 2^k - 1 is prime, then k is prime. The thumbnail is misleading because it does not hold for all prime k, but its converse does. Thx for viewing.

These numbers are known as Mersenne numbers and whenever number is prime it is known as Mersenne prime.Smallest Mersenne composite is 2047. Largest Mersenne prime is M(136,279,841) i.e. 2^136,279,841 -1 Confirmed on October 21, 2024 It has 41,024,320 digits and it is 52nd Mersenne prime. If M is Mersenne prime then M(M+1)/2 is a perfect number for M = 3 , 7 and 31 perfect numbers are 6 , 28 and 496 respectively.

@@raghvendrasingh1289 right, and for every Mersenne prime there is a perfect number of the form (2^n - 1)(2^(n-1)) where 2^n - 1 is prime. I think Mersenne numbers became of huge interest because there is an even perfect number corresponding to each Mersenne number. The interest in perfect numbers historically preceded the interest in Mersenne numbers I believe.

Thanks, you are the first to find the flaw in the proof. That must mean you are a reasonable and realistic person. The proof that 2=1 is based on an “unreasonable and ridiculous” division by zero on last step of the “proof”, yet many can’t discover that including teachers who are seeing it for the first time. You get an A for correctly noting the factoring formula I used is not the only possible factorization. See if you can prove the converse of the thumbnail which is 2^k - 1 prime implies k is prime

The “proof” started with assuming k is prime, therefore k=kx1 is only factorization of k. The correct factoring formula I used “proved” 2^k - 1 is prime , but as you pointed out that is not the only possible factorization as the counterexample 2^11 - 1 = 2047 = 23x89 shows. In the eighteenth century, pre calculators, a few number theory types, thought 2047 was prime!! Less talented people today may scoff at that since technology immediately shows 2047 is composite, but had you lived in the 18th century you may have thought the same erroneous thing.

@@MyOneFiftiethOfADollar Technological advantage aside, I was wondering what the basis was for concluding that this was the only factorization? The proof basically amounted to taking a complicated route to reach the conclusion n*1 = n (which was obviously unnecessary filler for such a fundamental result), from which we conclude "we've found a factorization of n, therefore this is the only one". Not even 17th century mathematicians, nor a 9th grader, would come to that conclusion. If you subtract the number theory filler, your proof amounts to: n*1 = n; this is the only factorization of n, thus n is prime. I understand that the criteria for a ridiculous assumption is subjective, but this is the biggest leap in logic I've ever seen in a proof, including the proof of 2 = 1, and it completely ruins the point for most people. Who is your target audience?

@@freepimaths9698 the factorization I used is from a modular arithmetic identity. A priori, there is no reason to believe it would be the only possible factorization. That was the error in my “proof”. The counter example of 2^11 - 1 shows there have to be other factorizations than aforementioned modular arithmetic identity. The flaw in the “proof” was assuming this was the only factorization. If it was, then my “proof” is correct starting with assuming k is prime as in the video. k=kx1 is not “unnecessary fodder”. The thumbnail statement assumes k is prime and attempts to show 2^k - 1 is prime which it does based on the modular arithmetic identity factoring formula. The “proof” fails because of the assumption that is the only possible factorization.

​​​@@MyOneFiftiethOfADollar Why would anyone assume that the identity you used was the only possible factorization? That is not at all implied, and for that reason, the identity that you used to derive 2^k-1 = (2^k-1)*1 contributes no additional information or importance in your proof, hence why I called it filler. People know that n*1 = n without using it, and it's not as if your formula's factorization being unique is even a believable assumption for somebody to just make unpromted, so nobody gains from you using it. You seem to be under the impression that I was calling the expression "n*1 = n" filler though, which was the opposite of my point. You also seem to keep irrelevantly re-explaining things we both understand, like how 2^11 - 1 is a counter example, as if that's related to how your attempt at a fake proof is a completely useless teaching tool, which was my point this entire time. It's based on your students having a completely unprompted and strange assumption that the factorization using your modular arithmetic formula was unique.

@j_sum1 you only proved you viewed the thumbnail and not the video which gave a counter example to the “proof”. Will you list 5 out of those “billions” ?

@@MyOneFiftiethOfADollar So why give a deceptive thumbnail then?

@@j_sum1 I agree it was deceptive, but it is difficult to cram what I was trying to convey onto small thumbnail. The description for the video italicized proof to indicate the counterexample 2^11 -1 = 2047=23x89 refutes the pseudo proof I presented. A common practice in the teaching of mathematics is to give a proof that is flawed and have students find the flaw.

oh crap i spent like 40 minutes trying to prove it

The converse of the thumbnail is true, namely if 2^k - 1 is prime, then k is prime. Try to prove that. Thx for viewing.

@@alphs4184 if you want to call appealing to the geometric series and binary “pure brute force”, then OK 😀 Geometric series is used frequently in number theory, e.g. the sum of the divisors of an integer formula. I used to code in assembly and have fond memories of binary and hexadecimal! Thx much for your point of view.

Your way is better! 2^2013 + 13 = (2^3)^671 + 13 == 1 + 13 = 14 == 0 mod 7 I didn’t notice that since FLT gives 2^6 == 1 mod 7 Nice find! Thanks

Thx, nerd humor at its finest😀 2 cents = dollar/50

Yes, (5/-1) = -5 is an integer, so -2 is a solution. Did I not include -2 as a solution in the video?

ok, i uderstand now thank you for the help

Knowing to rewrite the problem in the way you described was the challenge. (The upper limit of the summation minus the lower limit of the summation) plus one is the key to producing A^2 + B^2 = 261.

If you want to go down a rabbit hole, A001481 in the Encyclopaedia of Integer Sequences is the list of numbers that are expressible as a sum of two squares of whole numbers.

Right, I am aware of OEIS, but did not want to copy from that. Another viewer commented that 15,6 and -15,-6 are the only two integer solutions based on exhaustively checking all possibilities after establishing an upper bound for the possible squares.

Thanks, Did you notice that 14^2 + 8^2 + 1 = 261 (within a whisker of being a solution :) )?! If you have time will you check the recent one I did on Triangular numbers? Would be interested to see if you agree with some of the claims I made.

Got accepted into where?

Arml

Do you mean you qualified to compete in the ARML competition in Chicago?

@@rls5907 Yes, you are the only person who took the time to characterize the problem as “contrived”. Would you share with us what you mean by contrived?

@MyOneFiftiethOfADollar I think that you describe exactly what I mean 60 seconds before the end of the video. At that time, you state that it’s not clear exactly what the purpose of this problem is. For me, it comes across as a fairly contrived problem (I.e. artificially created for the sake of it - rather than to show some interesting result). In the video you yourself say that many people will not find this problem interesting. To be clear, I enjoyed the video, watched it through to the end, but was surprised that I’m the only person to comment.

The notion of interesting is almost as vague as the notion of contrived. I wouldn’t know a priori if even a single triangular number could be expressed as the sum of the squares of consecutive odd integers. Would you? The video showed there is only one such instance of this occurring. The problem is from an older number theory book by sierpinski titled 250 problems and was of interest to many in the 70s. There was a slight mistake in the way the problem was stated which was corrected in the video.

I’ll check out the book. Thanks for the reference.

There is nice free PDF version of 250 problems with solutions if you just search Sierpinski on the Web. It was problem 224 and you will see the error in problem wording after reading the solution. Are you a math teacher?