I love how the video is trying to explain it to 3 year old and comments are trying to use every number theory theorem to solve it.

For all of this, a = b means a = b mod 5. I just don't want to have to write mod 5 everywhere. By Fermat's Little Theorem, we know that: m^5 = m Multiplying both sides by m^4 m^9 = m^5 We already know that m^5 = m, so m^9 = m. Multiplying both sides by m^8 m^17 = m^9 We already know that m^9 = m, so m^17 = m Nothing here is reliant on the value of m, so all of this is true for n as well. So, we can rewrite the statement m^17 n - m n^17 as mn - mn, or 0

my life is now complete

It's easy to be wise after the event, at least for me! If we consider: n.m^x - m.n^x This divisibility holds for x = 5, 9, 17, 33, ..., 1 + 2^n Looking forward the next friday!

Incredibly clear and thorough. The content is accessible and engaging with your step-by-step explanation, I really like it. Thank you for such informative videos❤

Nice! I have a shorter proof. Same method for divisibility by 2. Now consider divisibility by 5. If m or n = 0 mod 5, we're done. Now consider m = 1, 2 , 3 or 4 mod 5. m^4 is either 1, 16, 81 or 256, all = 1 mod 5. So m^4 - n^4 is 0 mod 5. CQFD. This is shorter because I studied 5 cases instead of 25 😊

For the divisibility by five, we can factor m^{17}n-mn^{17} into mn(n^8+m^8)(n^8-m^8). By Fermats Little Theorem we know that for a prime p and all integers a that are no multiple of p, that a^{p-1} = 1 (mod p). (if m or n is a multiple of p the statement follows directly). We can write m^8-n^8= m^4*m*4-n^4*n^4 and reducing modulo 5 we get 1-1 =0. Therefore the whole term is divisible by 5

This is how I’d solve this using logic. We only need to prove the last digit is equal to zero. Any integer raised to the power 5 has the same last digit as the original integer. This is also true if it’s raised to the power of 9,13,17 etc.then the last digit of M^17.N will be equal to the Last digit of N^17.M . So subtract them one from the other it will equal Zero.

I have an ugly solution, but it’s quick and doesn’t require paper. The statement is basically equivalent to saying that m^17n and mn^17 have the same ones digit, since if they do m^17n - mn^17 is divisible by 10. Well as it turns out, since all the possible units digits (0-9) when raised to the power of 17 retain their units digit(for example 2^17 = 131072), the units digit of m^17n is the same as the units digit of mn and the units digit of mn^17 is the same as the units digit of mn. Thus, m^17n and mn^17 must have the same units digit and m^17n - mn^17 is divisible by 10. Done.

Given that we only needed the 1st few factors, m^5n-mn^5 would have been sufficient.

Since you only used up to m^2 + n^2 for divisibility of 5, isn't m^5n - mn^5 also divisible by 10?

Really, REALLY dig the idea of this modulo table for digits 0-9. Done it for Boolean and Truth tables, dunno why I never considered it for modulo. That's really cool, and so now of course my mind wants to take it to geometry and sets...

I figured out the divisibility by 2 the exact way you did. for divisibility by 5 - fermat's little theorem states that a^(p-1) ≡ 1 mod p. therefore m^16 - n^ 16 = (m^4)^4 - (n^4)^4 ≡ 1^4 - 1^4 = 0

phi(10)=4 and 17 mod 4 = 1 thus x^17=x mod 10. Hence m^17 * n - m * n^ 17 = m * n - m * n = 0 mod 10

Nice one. Since we just needed to go up to squares, we can say that any expression of the form: mn^5 - nm^5 is also divisible by 10

It is well known that m^5==m (mod 10). Therefore, m^17*n - m*n^17 (mod 10) == (m^5)^3*m^2*n - (n^5)^3*n^2*m == m^5*n - n^5*m == mn - nm == 0 (mod 10).

Thanks for this video. I tried to do the divisibility by 5 myself by checking the different remainders. Later that day I wondered if the same thing works for all primes. I vaguely recalled seeing something like that, so I looked it up, and yep it was Fermat's Little Theorem. I'm never going to forget that theorem now.

Goodness. All you need to know is that if x between 1 to 10, then x^5 and x have the same unit's digit: ie x^5 = x mod 10. Of course, it's clearly true for all x. now modulo 10: m^17 * n =( m^5)^3 * m^2 * n = m^3 * m^2 * n = m^5 * n = m * n exchanging m, n and taking the difference you get the result.

I did it using modular arithmetic, an integer to power of 17 is congruent to itself mod(10). So m^17 is congruent to m mod(10) and n^17 is congruent to n mod(10). Multiplying these by n and m respectively and subtracting shows that nm^17-mn^17 is congruent to 0 mod(10).

If either is even, we have an even number by m*n. We also have addition in one of the factors, so this becomes even if both are odd. So divisible by 2. 5 has the residue classes 0, 1, 2, 3, and 4. If either belongs to 0, mn as the factor leads the whole thing to be zero. We also know addition/subtraction and multiplication are valid operations for modulo. So, we can rewrite 4 as -1 and 3 as -2 by the addition aspect. 1 2 -2 -1 1 2 -2 -1 The whole main diagonal is 0 by m - n since if m = n, this = 0. The opposite signs become = 0 by m + n. And by m^2 + n^2, the rest also = 0, for any sign of 1s and 2s together.

Very elegant.

For generalization, if x=p*q, where p and q are primes, then [a^(phi(x)+1) = a] mod x. In our case, x=10=2*5 and phi(10) = 4. Then (m^17 * n = m^5 * m^5 * m^5 * m^2 * n = m * m * m * m^2 * n = m^5 * n = m*n) mod 10 Same thing with (n^17 * m = n*m) mod 10 So we have (mn - nm = 0) mod 10

Divisibility by 2 can be easily checked but for 5 theres a really clean method where you show 1^4=1 2^4=1 3^4=1 4^2=1 So a^16=1 (mod 5) Meaning a^17=a and mn^17-nm^17 reduces to mn-nm=0

If x = 1, 2, 3, 4 mod 5, it's fairly easy to find out that x^16 = 1 mod 5. for example with x = 3. x = 3 mod 5, x^2 = 4 mod 5, x^4 = 1 mod 5, x^16 = 1 mod 5. So for any situation such that m, n e 0 mod 5, m^16 - n^16 = 0 mod 5.

Very nice. Much more concise and elegant than my workings.

very nice, I factorised only to (m2+n2)(m2-n2) this is a liitle bit more complicated maybe but if we superpose the two 4x4 matrices (m or n multiple of 5 is trivial of course) of the two factors of this product we get such a beautiful symetric on the x and y axis superposition of zeroes and nonzeroes which is just so elegant.

Beautiful use of number theory! I don't think I'd have ever gotten there, but I at least noticed the DOTS reduction all the way down. I just hadn't figured it out from that point

Top tier math. Thank you

Or simply use the amazing fact: x⁵=x (mod 10) Considering powers mod 10, we have 0,1,5,6 doesn't change 9 has period 2 2,3,4,7,8 have period 4 Done.

For divisibility by 5, the factor m^4-n^5 is divisible by 5 by fermat's little theorem, if m and n are coprime to 5 else the factor mn is divisible by five.

Since 2^2=4=-1 mod 5, then m^4 = 1 mod 5 for every m such that m != 0 mod 5. Hence m^16 = 1 or 0 mod 5.

Simply put... Ends with 1 -> 1 Ends with 2 -> 2 4 6 8 Ends with 3 -> 3 9 7 1 Ends with 4 -> 4 6 Ends with 5 -> 5 Ends with 6 -> 6 Ends with 7 -> 7 9 3 1 Ends with 8 -> 8 4 2 6 Ends with 9 -> 9 Ends with 0 -> 0 (Dividable by 10) These are cycles that finishes itself at n^(x+4), and starts again at n^(x+5) 17 fulfills this criteria, so let's pick randoms and test it... xx7*x1^17-x1*xx7^17 = (Some digits)7-(Some digits)7 = Something dividable by 10 (Some digits)2*(Some digits)9^17-(Some digits)2^17*(Some digits)2 = (Some digits)9*2-(Some digits)9*2 = (Some digits)8-(Some digits)8 = (Some digits)0 True for every 4n+1

For me this was engrossing. Thanks.

The interesting thing is that you could've just stopped your factorization at mn(m^4-n^4)(m^4+m^4)(m^8+m^8), For all k not divisible by 5, k^4 is 1 mod 5 (1^4=1=0*5+1, 2^4=16=3*5+1, 3^4=81=16*5+1, 4^4=256=51*5+1). Thus, if both m and n are not divisible by 5, then m^4-n^4 is equivalent to 1-1 mod 5, which is 0 mod 5.

If neither m,n get's you the 5, I thought about the relationship between m and n to force you to go to m^2 +n^2. If m = 3n then (3n)(n)(3n+n)(3n-n) doesn't get you the 5 but (9n^2 + n^2) = 10n^2 does. 24 and 8 is an example.

Okay, that's even spookier than the 24 people in a room with a > 50% chance of sharing a birth month + day. But i^i being an element of the Reals is, of course, obvious, and I'm happy with that. :)

It's also true of mn^5-m^5n. You factor out mn to get mn(m^4-n^4) and by fermat's little theorem if m and n aren't zero then that second part is 1-1 (mod 5) and 1-1=0

thats indeed a very beautiful proof

For 2, exponents don't matter. For 5, reduce exponents by mod 4.

If we are using modulo arithmetic anyway then it would be easier to teach how things work modulo 10

This one was very pleasant. Thank you.

So really, this works for ((m^5)*n-m*(n^5)) as a base case - the original equation makes it look a little flashier. Pretty clever stuff.

This is very cool

Using basic rules of module comparison: if m^17 * n - m * n^17 ⁝ 10 then: m^17 * n ≡ m * n^17 (mod 10) using theorem: if a ≡ b (mod m) and c ≡ d (mod m) then ac ≡ bd (mod m) we get: m^17 ≡ m (mod 10) Originally saw a top comment that proves this statement using Fermat's Little Theorem, but why would you need that if you have the basic rules of number theory that every 8th grader knows?

Since you didn’t use the quartic and octic(?) factors then you could find smaller expressions with this same property, namely m^9*n - m*n^9 and m^5*n - m*n^5. I suppose the latter isn’t as interesting to discover always produces a multiple of ten given the exponent of five.

You could also say that when m, n ≠ 0 mod 5, m^4 = n^4 = 1 mod 5 Then if either m or n = 0 mod 5, then mn(m^16 - n^16) = 0 mod 5 And if neither m nor n is 0 mod 5, then mn(m^16 - n^16) = mn(1^4 - 1^4) = 0 mod 5

Very cool! If I was in a flexing mood I might have run the table from -2 to 2 for even more symmetry and fun

If you use (-2) and (-1) as representatives mod 5 instead of 3 and 4, then you can make your life easier.

Good. Note also that m, n both not divisible by 5 implies m⁴=1 (mod 5) => m^16 = 1 (mod 5). Similarly, n^16 = 1 (mod 5). Therefore, their difference is 0 (mod 5). 👍

Mesmerized by how Dr. Barker always closes brackets with an upward stroke

Nice mechanical proof.

It's a one-liner. k^(4 j + 1) mod 10 ≡ k mod 10. Then m^17 n mod 10 ≡ m n^17 mod 10 ≡ m n mod 10 ⇒ difference ≡ 0 mod 10 QED.

F(m,n) = n m^17 - m n^17 = mn (m-n) (m+n) (m^2+n^2) (m^4+n^4) (m^8 + n^8). If mn is odd then both m and n are odd, so m+n and m-n are both even. Otherwise mn is even. So, mn (m+n) (m-n) is always even. So F(m,n) is always even. Now it remains to prove that at least one of the factors of F(m,n) is a multiple of 5 even when m and n are both not multiples of 5. m and n (mod 5) are either +/- (1 or 2). If equal then m - n is multiple of 5, If opposite then m + n is a multiple of 5. So, they have to be unequal (one 1 and the other 2). So, (m^2 mod 5) and (n^2 mod 5) would add to 1 + 4 = 5 (a multiple of 5). So, (m^2 + n^2) is a multiple of 5. In all cases, mn (m-n) (m+n) (m^2+n^2) is a multiple of both 2 and 5, hence a multiple of 10. So, the entire F(m,n) is a multiple of 10. Hence proved.

Excellent!!!

الله يحفظك

nice, but grinding proof. 'Divisible' in this sense means: when 'divided by 10 gives an integer' (since 1/10 = 0.1 so 1 is divisible by 10). Exception: m=n OR m*n=0 gives 0 and . . . nope, not divisible by 10.

alternatively for 5, if one of m or n is a multiple of 5 then done, else fermat little finishes

Nice.

very simple very quick very impressive nice stuff :O

thank you so much!!!

m^(2^k+1)*n - m*n^(2^k+1) is divisible by the product of all primes of the form 2^r+1 less than or equal to 2^k+1. Those are called Fermat Primes, and the only known such primes are 2, 3, 5, 17, 257 and 65537. m^2*n - m*n^2 are divisible by 2. m^3*n - m*n^3 by 2*3 = 6. m^5*n - m*n^5 by 2*3*5 = 30. m^17*n - m*n^17 by 2*3*5*17 = 510. m^257*n - m*n^257 by 2*3*5*17*257 = 131070. Last one is m^65537*n - m*n^65537 is divisible by 2*3*5*17*257*65537 = 8589934590.

With luck and more power to you.

i checked every ones place digit and they all equal themselves when taken to the 17th power, so a*b-b*a=0 for the ones place digit

Loved it 👌👌👌

No offense, but this answer wasn't satisfying. I liked the resolution, but you didn't need all those factors. So I have a different question: what's the GCD of that number? Is it possible to figure that out?

Satisfying indeed.

Disappointed we didn't get to use that (m^8 + n^8) term

So 17 was a red hereing. You could have done any number 1 more than a power of 2, 5 or greater.

A nice proof!

Fab as always

Satisfying indeed!

So could you have just done mod 10 from the start instead of doing 2 and 5?

Too much lived in Moebius Strip

cool😮

Now prove it's also divisible by 17

17=1(mod 4) so x^17=x(mod 10) so m^17n-mn^17=mn-mn=0(mod 10)

Cant the problem be to the power of 5 also

2^n -1 /q is integer for odd q and n is a(q-1/2) where a(x) is A002326 and a(0) is 1, a(1)=2 etc.

But what if m and n are both equal to 1? Surely 0 isn't divisible by 10?

not really satisfying when u had to check all the mod 5s and not give like a general solution to slice it all up

wish it was more theoretical

Can i get a shout out

😮

Strange way to write n

by the same logic you can have mn5-m5n : 10 (and it works) - why do you use 17?

Small typo on the whiteboard: N, not Z

It is not satisfying, it is working out 25 cases of n and m modulo 5