the largest number that divides the difference of quartic primes

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[@MichaelPennMath]

Try @Blinkist for 7-days AND get 25% off a Premium Subscription using my link. Invite a friend with Blinkist Connect. That's 25% off a 2-for-1 deal: blinkist.com/michaelpenn

[@leif1075]

How or why would.anyine discover this and why would anyone care?

[@RexxSchneider]

Three facts: any number n not divisible by 3 when raised to the fourth power is congruent to 1 (mod 3); any number n not divisible by 5 when raised to the fourth power is congruent to 1 (mod 5); any odd number n when raised to the fourth power is congruent to 1 (mod 16). Obviously if p and q are primes greater than 5, they fulfil those conditions, but any odd numbers p and q that aren't divisible by 3 or 5 will also have 240 | (p^4 - q^4). An analogous result applies to eighth powers: 480 | (p^8 - p^8) since the eighth power of any odd number is congruent to 1 (mod 32).

[@gniedu]

To prove p^4 and q^4 congruent to 1 mod 16 there's a quicker route. All primes bigger than 11 (bigger than 2 for that matter) are odd. Write them as 2n+1 and expand. That will give (2n+1)^4 = 16n^4 + 32n^3 + 24n^2 + 8n + 1 Reducing mod 16 the first two terms go to 0, and we're left with 8n(3n+1) + 1. If n is odd, 3n+1 is even; if n is even, 3n+1 is odd. Either way, n(3n+1) is even and, therefore, 8n(3n+1) is congruent to 0 mod 16. So, all that remains is only 1 mod 16. So (2n+1)^4 is always congruent to 1 mod 16

[@mwelinder]

Indeed. And the mod-5 and mod-3 cases follow from p=6k+-1 and p not divisible by 5. Just calculate (6k+-1)^4 mod 15 for k=0..4 and notice you always get 1, except for the cases that make 6k+-1 a multiple of 5. Conclusion: this has very little to do with primes. And there is no need to exclude 7 -- it works just fine.

[@TheEternalVortex42]

It's only quicker if you are good at algebra ;)

[@khoozu7802]

Small typo error, it should be 32n^3 not 16n^3

[@gniedu]

@@khoozu7802 indeed, I've fixed it. Thanks

[@stanleydodds9]

@@TheEternalVortex42 Well, not really. If you're good at algebra, then you already know the structure of the multiplicative groups of integers modulo powers of 2, namely that (Z/2^nZ)* is isomorphic to C_2 x C_2^(n-2) for n >= 3. This immediately tells you the required fact that p^4 and q^4 are always congruent to 1 mod 16.

[@JohanRimez]

Solving these kind of Olympiad questions is fantastic in itself. However, somebody or a team had to invent and test them in the first place. Respect.

[@MichaelPennMath]

Foreshadowing is a literary device used to indicate something important will happen in the future. ;) -Stephanie MP Editor

[@noahtaul]

@@MichaelPennMath Thanks Lemony Snicket!

[@Utesfan100]

Usually such problems have the strictest restriction possible. Since 11^4-7^4=51*240, this works for p>q>=7.

[@HagenvonEitzen]

Allowing smaller primes makes it easier to show that the magic number cannot be too big

[@erazmtylko4773]

To prove 240 divides p^4-q^4 I've factored the latter out: (p^2+q^2)(p-q)(p+q) p and q can be +/- 1 mod4, substituting combinations of 1 and -1 into the product we see it's divisible by four twice - is a multiple of 16. p and q can be +/- 1 mod3, thus either (p-q) or (p+q) is 0 mod 3, therefore the product is also multiple of 3. p and q can be +/- 1 or +/- 2 mod5. Whichever way we combine +/- 1 and +/- 2 in the product, one of the terms will be multiple of 5. Therefore p^4-q^4 is divisible by 3*5*16=240.

[@Rbmukthegreat]

Small mistake at around 10:46. Instead of saying p^4-q^4 cong 1 mod 5, it should say p^4 cong q^4 cong 1 mod 5 (the minus should be replaced with a cong). He writes p^4 - q^4 cong 1 mod 5 in the next line

[@eiseks3410]

Next video: what is special about the number 420?

[@neilgerace355]

Groovy!

[@TheEternalVortex42]

It's the smallest number divisible by 1, 2, 3, 4, 5, 6, and 7.

[@cameronbigley7483]

@@TheEternalVortex42 I believe the term is subfactorial. Would be neat to see a video on those numbers.

[@cH3rtzb3rg]

This is actually true for p,q >= 7. And (excluding the part that 240 is the greatest number) it is equivalent to p^4 = 1 (mod 240) for all primes >= 7. Also, the proof that 15 | p^4 - q^4 could be done "manually" similar to the last part of your proof, by showing that p^4 = 1 (mod 3) and p^4 = 1 (mod 5) -- which is true for all numbers which are neither multiples of 3 or 5: p in {+-1} (mod 3) --> p² = 1 = p^4 (mod 3). p in {+-1, +-2} (mod 5) --> p² in {1,-1} (mod 5) --> p^4 = 1 (mod 5).

[@wolfmanjacksaid]

The video quality is getting better and better.

[@Phylaetra]

So - 240 is the largest for p>= 11; what about p>=13? 17? Does this number grow very quickly? What about smaller primes (i.e. p^4-q^4 for p,q

[@arthurfibich112]

I would give the hypothesis that since 240=2*(2*3*4*5), 5=11//2, 1440 would divide everything for the 13 case

[@Phylaetra]

@@arthurfibich112 Well... Except it fails in the 17^4 - 13^4 case given in the video (=240 * 229, and 229 is not divisible by 6...) I suspect that, if there is another, larger, number that it will take a little more work to find than just extending a pattern...

[@pepefrogic3034]

Lets see, we essentially need that number to the 4th power is equal to 1 mod that number. That is true for 16 but not for 32, so no increase i even factors, and for odd primes p-1 needs to be a factor of 4, so only 5 and 3 will work. So no, you do not gain anything there by increasing numbers. Now for 7 i do not understand why it does not work, clarly 7 to the power 4 is 1 mod 5 and also 1 mod 3 and also 1 mod 16, so he could start from 7. If you want more than 240 you have to increase the power

[@stratehorthy3351]

To show that p^4 - q^4 is divisible by 16=2^4, one can use the famous Lifting The Exponent lemma (LTE) ! Indeed, since p,q are both odd, and since 2 divides p-q, it follows that v_2(p^4-q^4) = v_2(p-q) + v_2(p+q) + v_2(4) - 1 = v_2(p-q) + v_2(p+q) + 1. Since p is either q or -q mod 4, it follows that one of p-q or p+q is divisible by 4, and hence that v_2(p^4-q^4) >= 2+1+1 = 4 and thus that 2^4 divides p^4 - q^4 ! Remark : The p-adic valuation is a function defined from N to Z+, such that v_p(x) is the biggest power of the prime p in x's unique prime factorization. For example, v_5(2^3 * 5^2) = 2, and v_7(1001) = 1, and so on. There are many nice results related to this function, for example LTE as used above, which can give us an idea on v_p(a^n - b^n). There is also Legendre's Formula for v_p(n!) (Ever heard of a problem asking you how many zeroes are at the end of a given factorial ? You merely have to compute it's v_2 and v_5 and take the smallest of the two !). It would be really awesome to see more problems related to the prime exponent of a number in its unique factorization, because it can really lead to beautiful solutions and ideas (For example, you can show that a sequence (a_n) of positive integers is eventually constant if all the sequences v_p(a_n) are eventually constant for all primes p, with certain extra conditions to avoid weird cases like the sequence of all primes). This can help solve a lot of olympiad-type problems, like IMO P5 2018 for example! Here's a little suggestion for a problem : Take three positive integers a,b,c such that a/b + b/c + c/a is an integer, and show that cbrt(abc) is also an integer, where cbrt(x) is the cube-root of x. Another suggestion, more creative, beautiful and astonishing would be Putnam 2017 B3 ! en.wikipedia.org/wiki/Lifting-the-exponent_lemma en.wikipedia.org/wiki/Legendre%27s_formula en.wikipedia.org/wiki/P-adic_valuation

[@Maths_3.1415]

You are amazing dude 😲

[@skalderman]

Whats v ?

[@stratehorthy3351]

@@skalderman It's a function, denoted by v_p(x), spitting the power of the prime p in x's factorization. Check the third Wikipedia link for more info!

[@MyOneFiftiethOfADollar]

Please do your own videos on your own channel. What a waste of excellent mathematics in a difficult to read comment section. I will be your first subscriber!

[@matteodaniello2594]

Shouldn't be possible to relax the conditions to primes bigger or equal to 7? After all we only need them to be different from 2, 3, and 5 in order to apply Fermat/Euler

[@charlottedarroch]

You can! If the pair (p,q) has q = 7 and p > 7, it's still the case that 240|(p^4-q^4). Not sure why Michael stated the conditions with a bound of 11. As you say, the only obstructions in the proof are that in utilising Euler's theorem (possibly with Carmichael's extension in the mod 16 case) you require that your base be coprime to the modulus. So we need only that the primes p,q be coprime to 3,5,16, so not including prime factors 2,3,5.

[@johnmartorana196]

Not a real mathematician here, but I tried this one... I started working on this a completely different way. I factored p^4-q^4 = (p-q)(p+q)(p^2+q^2). Then argued from the fact that p and q are +/-1 mod 3 so then p= +/-q mod 3. So either p-q=0mod3 or p+q=0mod3. Same argument with mod 4 (along with the other two factors being even) gives me 16 divides p^4-q^4. (5 being a factor was similar, but required a check-a-bunch-of-cases sort of argument that was probably a bit too much effort)

[@WILDXHjordi]

From p, q = +/-1 mod 3, it follows that p^4=q^4=1 mod 3, and hence the result

[@mikeoakes2]

Here's an alternative and I think easier route:- p and q are each congruent to 1,7,11,13,17,19,23 or 29 mod 30; so p^2 and q^2 are each congruent to 1 or 19 mod 30; so p^4 and q^4 are each congruent to 1 mod 30, so 30 divides p^4-q^4. Also, 16 divides p^4-q^4 as per your video. But lcm(16,30)=240, so 240 divides p^4-q^4. QED.

[@MasterHigure]

I am pretty certain the thumbnail for this video is wrong, as it seems to say p^4 + q^4. Also, for divisible by 16, I would personally have gone with p^4 - q^4 = (p^2 + q^2)(p^2 - q^2) The first factor is even, and the second factor is always divisible by 8 (by elementary mod 8 considerations).

[@MichaelPennMath]

thanks for catching that. it's been fixed. -Stephanie MP Editor

[@HagenvonEitzen]

4:20 Perhaps faster: By Bezout, there are integers u and v such that 58 u + 287 v = 1. Now id d divides both 58 * 240 and 287 * 240, it also divides 58 u *240 + 287 v*240 = 240.

[@evreatic3438]

GENERALIZATION: If p and q are not divisible by 2, 3, 5 then 240 divides p^4 - q^4. Examples: 7^4 − 1^4 = 240 * 10 79^4 − 77^4 = 240 * 15821

[@arthurfibich112]

Question: can this be generalized? I mean, 240=2*(2*3*4*5), 5 being the rounded half of 11... Hypothesis: 2*(2*3*4*5*6)=1440 divides the same for primes above 13

[@RexxSchneider]

Since 7^4 = 2401 ≡ 1 (mod 240), we have 240 | (p^4 - 7^4), so the 11 is a red herring. The result is true for all primes >= 7. In fact, it's true for all odd numbers not divisible by 3 or 5, not just the primes, e.g. 49. I'm pretty certain that any odd number to the power 2^n is congruent to 1 mod 2^(n+2). So we can say that (15*2^(n+2)) | (p^n - q^n) where p and q are not divisible by 3 or 5. For example, 480 | (p^8 - q^8). Sadly, your hypothesis fails at the first hurdle since 19^4 - 17^4 = 46800, which is equal to 1440*32 + 720.

[@Simio_Da_Tundra]

Another method for proving p^4-q^4 is a multiple of 16: Write p^4-q^4=(p-q)(p+q)(p^2+q^2). Since p, q>2, they are both odd, so p+q, p-q and p^2+q^2 are multiples of 2. Furthermore, either p-q or p+q is a multiple of 4. Indeed, if p-q is not a multiple of 4. then p=1(mod 4) and q=3(mod 4) or p=3(mod 4) and q=1(mod 4), and in both cases p+q=0(mod 4). Therefore, p^4-q^4 is a multiple of 2*4*2=16.

[@cernejr]

All odd numbers raised to 4th power give reminder 1 when divided by 16, not only primes. The proof is super simple. Sometimes learning advanced math damages our common sense, it has happened to me too.

[@pepefrogic3034]

Do you mean to the 4th power

[@cernejr]

@@pepefrogic3034 Yes. I have edited my comment to make it more clear/accurate.

[@Anonymous-zp4hb]

My instinct to factorize when I see a difference of squares is insatiable: p^4 - q^4 = (p-q) (p+q) (pp+qq) It's sufficient to : Show that all the factors are even, exactly one is a multiple of 4 and there exists at least one case where none of them are multiples of 8 Show that one of the factors is always a multiple of 3 Show that one of the factors is always a multiple of 5 Show that besides 15, no odd numbers greater than 5 can be common. ( hint: use p==2, q==1 (mod n) ) Individually, each point is quite easy to show.

[@martinnyberg9295]

Green t-shirt! 😊👍🏻

[@emanuellandeholm5657]

I got the divisor 8 from factoring p4 - q4 as (p2 - q2)(p2 + q2) = (p - q)(p + q)(p2 + q2). All even => 2^3 | p4 - q4 In fact, rewriting p2 + q2 as p2 + (p - 2n)^2, we see that we have p2 - p2 - 4pn + 4n = 4n(-p +1) = 0 mod 4, so there's the final divisor is 2*2*4 = 16.

[@Mosux2007]

240: the lowest number of beans that yields an excessively gassy bean soup.

[@teeweezeven]

Why do we exclude 7? Everything works out when q=7 right?

[@karlstrecker6002]

I agree, 7 works fine too. Maybe Mike isn‘t a friend of 7?

[@barbietripping]

p^4 - q^4 = (p-q)(p+q)(p^2+q^2) 2 | (p-q) 2 | (p+q) 4 | (p^2+q^2) Either 3 | (p-q) or 3 | (p+q) If p and q have units digits {1,9}, or {3,7}, then 5|(p+q) If p and q have the same units digit, then 5|(p-q) If else, p and q have units digits {1,3}, {1,7}, {3,9}, or {7,9}, then 5|(p^2+q^2) Note that In mod5: 1^2=1 3^2=4 7^2=4 9^2=1 Thanks for a fun problem

[@noahtaul]

I think the fastest way to prove that p^4-q^4 is divisible by 16 is by writing p*=(p-1)/2, q*=(q-1)/2, and then (p^4-q^4)/16 = 24((p* choose 4) - (q* choose 4)) + 48((p* choose 3) - (q* choose 3)) + 29((p* choose 2) - (q* choose 2)) + 5(p* - q*) which is clearly an integer.

[@shishkabob984]

10:33 in other words, 15. that was hilarious

[@bobbobson6867]

A sphere can touch 240 other spheres in 8 dimensions. 😁

[@evionlast]

Ad breaks feels like listening to radio doesn't it?

[@rvaisberg]

Didn't understand what Michael is trying to say at 13:19. The statement "There are no primitive roots mod n" is certainly false for arbitrary n. For example 3 is a primitive root mod 14.

[@matteodaniello2594]

It actually meant "for n being a power of 2", more specifically any power of 2 bigger than 4 doesn't have primitive roots

[@Tehom1]

It puzzled me too. One usually speaks of "primitive nth roots" for some particular n and he didn't mention what n is. It seems like n=8 was intended, since he goes on to rule out the other divisors of 8.

[@rvaisberg]

@@matteodaniello2594 Yeah that must've been it

[@wolfmanjacksaid]

Around 10:10 he writes that p^4 - q^4 is congruent to 1mod5 then immediately says it's congruent to 0mod5. I'm confused there

[@uffe997]

It should have been a congruent sign instead of minus sign.

[@Phylaetra]

A couple of notes - (1) at 10:11 you are writing and saying that "p^4-q^4 is congruent to 1 mod 5", but I think you meant to say (and write) both p^4 and q^4 are congruent to 1 modulo 5. (2) the argument requires p and q to both be greater than 3, and then greater than 5, but there is no requirement that p be greater than 11, that requirement doesn't really show up, and appears not to be necessary (note - 7^4-5^4 is not divisible by 5, and so not 240; but 11^4-7^4 = 240 * 51 and 13^4-7^4 = 240*109(. (3) I took a different path, first noting that we can factor p^4 - q^4 = (p^2 - q^2) (p^2 + q^2). For p>3, p^2-q^2 is always divisible by 3, for p>5 if p^2=q^2 mod5 then p^2-q^2 is 0 mod 5, but if p^2 != q^2 mod 5, then p^2+q^2 is 0 mod 5! Yay. Similarly, it's easy to see that p^2-q^2 is always congruent to 0 mod 8 (for all odd numbers, so all odd primes too), and then p^2+q^2 is obviously always even, so their product always has a factor of 8*2 = 16. Again - our only requirement is that p & q both be greater than 5, so I am not sure why the problem required them to be >=11...

[@erikheddergott5514]

What about 7? From 7 to 11 it is 61*240 - 10*240 = 51*240. Why is 7 not included?

[@evreatic3438]

All p and q not divisible by 2, 3, 5 (even non primes) satisfy this condition, e.g. 1, 7, 49, 77, 91, etc.

[@erikheddergott5514]

@@evreatic3438 You are right, so I wonder why the 7 got excluded. It is claimed that every Double- Digit Prime fulfills the Demand. I really would like to know why it came to that „Exclusion“. Numerology? Or other occult Reasons?

[@evreatic3438]

​@@erikheddergott5514 Since this was a math olympiad problem, I guess they may have intentionally excluded 7 to make it a bit harder to solve, to "muddy the waters" so to say... but I really don't know. Occult reasons could be equally good of an explanation :)

[@erikheddergott5514]

@@evreatic3438 As Muddy Waters sang: „Got my Mojo working, it just don’t work on 7.

[@ke9tv]

Didn't you fail to prove the second half of the problem? (Show that there exist p,q where 32 does not divide p**4-q**4, that there exist p, q where 9 does not divide p**4-q**4, show that there exist p,q where 25 does not divide p**4-q**4, and show that for all primes r>7, there are p, q such that r does not divide p**4-q**4. Euler's extension to FlT makes this straightforward, but you also promised a "down market" solution.) ETA: Not as hard as I thought - a quick search can show that there are two pairs p, q for which p**4-q**4/240 are distinct primes.

[@charlottedarroch]

In fact, you don't need two pairs (p,q) with (p^4-q^4)/240 are distinct primes. You merely need pairs (p,q) and (r,s) with (p^4-q^4)/240 and (r^4-s^4)/240 are coprime in order to show that 240 is the largest number dividing the differences of fourth powers. And Michael gave the example of the pairs (13,11) and (17,11) with this property, so indeed he did prove that 240 is the largest number with the desired property.

[@easports2618]

Oh wow, I didn’t know that

[@MrMctastics]

and they said coolmathgames was a waste of time smh

[@nikoladjuric9904]

(2k+1)⁴=16k⁴+32k³+24k²+8k+1 =16(k⁴+2k³+k²)+8k(k+1)+1 So (2k+1)⁴=1 (mod 16) p⁴=q⁴=1 (mod 16) And that's the good place to stop.

[@ianmathwiz7]

Wonder if there are numbers with similar properties, except they divide p^n-q^n for some other values of n.

[@Anonymous-zp4hb]

If n is a power of 2, then a larger number than 240 definitely divides p^n - q^n for all n > 4. Example: p^8 - q^8 = (p^4 - q^4) (p^4 + q^4) we already know that 240 divides the left factor. And the factor on the right is always even. So 480 must divide all p^8 - q^8

[@RexxSchneider]

It turns out that odd numbers to the 8th power are all congruent to 1 (mod 32) and to the 16th power are all congruent to 1 (mod 64). If you consider an odd number to the power 8 as (1+2n)^8, that expands to: 1 + 8*2n + 28*4n^2 + 56*8n^3 + 70*16n^4 + other terms in higher powers of 2n. So you can see that is congruent to: 1 + 16n + 16n^2 + 0 + 0 + ... (mod 32) ≡ 1 + 16n(1+n). Since either n or n+1 is even, the second term is divisible by 16*2, therefore 1 + 16n(1+n) ≡ 1 (mod 32). That shows that an odd number to the 8th power is congruent to 1 (mod 32). If you consider a similar expansion of an odd number to the 16th power mod 64, you get: 1 + 16*2n + 120*4n^2 + 560*8n^3 + 1820*16n^4 + 4368*32n^5 + 8008*64n^6 + ... (mod 64) which is congruent to: 1+ 32n + 32n^2 + 0 + 0 + 0 + 0 + ... ≡ 1 + 32n(1+n). The analogous argument shows that to be congruent to 1 mod 64. I'm pretty certain that any odd number to the power 2^m is congruent to 1 mod 2^(m+2), but I can't say for sure what is the greatest number that divides the difference between those powers. I'll have a think about that.

[@jamesfortune243]

In a humorous parody of Nikki Haley, have a T-shirt that says, 'Into my primes', instead of 'IN MY PRIME'.

[@Bemajster]

That's preety nice!

[@morgane3501]

actually what i found was that for any (p,q) relative integers, where 2,3, and 5 do not divide either p or q, the statement is true. to do that i decomposed the polynomial p^4-q^4 into (p-q)(p+q)(p^2+q^2). we can easily see that (p^2+q^2) is even, but we also find that either (p-q) is divisible by 4 or (p+q) is divisible by 4. if p and q have the same parity modulo 4 then (p-q) is divisible by 4, and if they have different parity then their sum has a parity of 1 + 3 = 4, so (p+q) is divisible by 4. So in the end (p-q)(p+q) is divisible by 8, and p^4-q^4 is divisble by 16. even if we use the prime numbers because "numbers that aren't divisible by 2, 3, or 5" isn't as nice and might too easily lead to the solution, the limit should have ben set to 7 and not 11.

[@MCMCFan1]

tyler1 reference?

[@JavierSalcedoC]

I'm really liking the AI generated thumbnails but the best thumbnail is still the puppy

[@Uranyus36]

if we can work with modulus with squaring, can we do the same for square-rooting? like if p = 1 (mod n), then p^2 = 1 (mod n). can we do the step p^4 = 1 (mod 16) => p^2 = 1 (mod 16) by just square-rooting both sides?

[@calvincrady]

No. Some counterexamples: p² ≡ 15 ≡ -1 (mod 16) ⇒ p⁴ ≡ 1 (mod 16) p² ≡ 7 (mod 16) ⇒ p⁴ ≡ 49 ≡ 1 (mod 16) Even for the real numbers this wouldn't work, since when taking the square root of both sides of an equation you could flip the signs: (-a)² = a², so b² = c² does not imply b = c (this is equivalent to the first counterexample above). But the second counterexample doesn't have an equivalent in the real numbers.

[@calvincrady]

If, however, our modulus was a prime number q, then there would be no counterexamples of the second type: Let n be some integer s.t. n² ≡ 1 (mod q). n² - 1 ≡ 0 (mod q) ⇒ q divides n² - 1. n² - 1 = (n - 1)(n + 1), so q divides (n - 1)(n + 1). Because q is a prime, q must therefore divide either n - 1, n + 1, or both. That is, at least one of the following is true: q divides n - 1 ⇒ n - 1 ≡ 0 (mod q) ⇒ n ≡ 1 (mod q) q divides n + 1 ⇒ n + 1 ≡ 0 (mod q) ⇒ n ≡ -1 (mod q) Therefore n ≡ ±1 (mod q).

[@Uranyus36]

Oh, i see. I understand now. Thanks for your explanations!

[@RexxSchneider]

@@calvincrady But p² is never congruent to 7 (mod 16), or equivalently 7 is not a quadratic residue mod 16. The same applies to 15. The only quadratic residues mod 16 are { 0, 1, 4, 9 }, all of which when squared again are congruent to 0 or 1 (mod 16), showing that the only quartic residues mod 16 are { 0, 1 }. Your examples need to be if you consider p² ≡ 1 (mod 16) ⇒ p⁴ ≡ 1 (mod 16) and p² ≡ 9 (mod 16) ⇒ p⁴ ≡ 81 ≡ 1 (mod 16).

[@calvincrady]

@@RexxSchneider Hmm, fair point. I suppose if I wanted to show that you can't simply take the square root of both sides of a congruence relation mod n, it would've been more rigorous to use p ≡ 7 (mod 16) ⇒ p² ≡ 49 ≡ 1 (mod 16) as a counterexample 😅

[@przemekmajewski1]

Where have you shown a problem for p = 7? I'd say nowhere, and also I'd say, it works for prime p > 5. Have I missed sth? Also going into primitive roots with this solve lacks finesse and doesn't show the real magic behind importance of primitive roots. Great you've shown how to do it from scratch. Also again, 7 did not pose any problem at all anywhere.

[@byronwatkins2565]

At 10:35, how can p^4 - q^4 be congruent to both 0 and 1 (mod 5)?

[@luisaleman9512]

He made a mistake. It should be p^4 ≅ q^4 ≅ 1 (mod 5) which implies that p^4 - q^4 ≅ 0 (mod 5)

[@fatbudyn]

needless

[@kendebusk2540]

This is one of those things that only 1 in a million would have "discovered". I find these nuggets fascinating even if there's not much worth in the "real" world.

[@HagenvonEitzen]

And here I thought 420 is special ...

[@The1RandomFool]

What's up with the "green new deal" shirt?

[@petersievert6830]

I'm not from the US, but that seems to be a not very subtle political statement, right?

[@MyOneFiftiethOfADollar]

Easy, from the prime factorization, 240 is the smallest number divisible by 15 and 16. 🙄 isn’t that special?!

[@tomholroyd7519]

p^5 - q^5?

[@RexxSchneider]

There's nothing I can find as nice for the quintics. Although any number to the fifth power is only congruent to ±1 or 0 (mod 11). However, I conjecture that all primes greater than 7, when raised to the sixth power, are congruent to 1 (mod 6, 7, 8, 9).

[@someperson9052]

Stay hard

[@tomholroyd7519]

wut

[@stephenhamer8192]

Little brother: for every prime, p > 3, p^2 - 1 is divisible by 24

[@easports2618]

240 Fr Fr built different 🐐

[@matematicacommarcospaulo]

It seems that you have been changing the image of your videos from one day to another... Sometime I think that there's a new video, but not

[@MichaelPennMath]

That’s just A/B testing. -Stephanie MP Editor

[@matematicacommarcospaulo]

@@MichaelPennMath what does A/b mean? Something like the old LP records?

[@schweinmachtbree1013]

The term 'universal property' is not appropriate here; universal properties of an object _u_ take the form "for all _x_ such that ... there exists a unique _y_ such that P( _u, x, y_ )". That is, 'universal property' is really an abbreviation of 'universal unique-existential property' - another term for 'universal property' is _'universal mapping property'_ , because such properties of the form "for all _x_ there exists a unique _y_ such that P( _u, x, y_ )" can be framed in the form "for all _x_ there exists a unique mapping from _u_ to _x_ / from _x_ to _u_ "

[@easports2618]

So if I do, 97^4-47^4 it’ll give me some number times 240? Damn that’s crazy

[@forcelifeforce]

97^4 - 47^4 = 240(348,540)

[@easports2618]

@@forcelifeforce 240 goateed bro ong maybe we could use this to verify if a number is a prime, any number n^4- largest current known prime number=240(Q)

[@andrycraft69]

​@@easports2618 I don't think it would work that way. We proved that all primes ≥ 11 satisfy the equation p⁴-q⁴ = 240*k, but we haven't proved that all solutions to this equation are primes, which is what you'd want for it to be a primality test.

[@easports2618]

@@andrycraft69 oooooo, very very true I didn’t think of that, nice observation

[@easports2618]

@@andrycraft69 wait, the theorem is, *all primes > or =11 satisfy the eqn, p^4-q^4=240*k, so after taking the current biggest prime number, can’t we just keep substituting numbers after that? Every number that does satisfy it then, should be a prime right?

[@pepefrogic3034]

What a sloppy presentation. He screwed up the "no primitive root mod n", should be no primitive root 2^n with n greater than 3 probably. Of course itbis not true for all n, primes, 4, 9, all have primitive roots. Should have stated that result which numbers have primitive roots, this is extremely frustrating why mention it and wrong. Yet he explains the other things yo extremely low level, and mentions some "course". Lazy and disrespectful to post this on youtube,

[@pierreabbat6157]

You can start with 7, since 7⁴=2401. But not 2, 3, or 5, since they are factors of 240.