when is the sum of squares a power of a prime??

[@goodplacetostop2973]

16:41 Happy New Year everyone! May 2022 be full of high energies and greater focus for you to study and make your dreams a reality.

[@kevinmartin7760]

This problem expands into an irritatingly large number of cases...

[@sleepycritical6950]

That's maths. I've been very sloppy with my proofs because of this.

[@suhail_69]

Thank you for putting together such amazing videos. Happy new year Mike!

[@ivankaznacheyeu4798]

Checking of n=14 case is incorrect, according to previous calculations product (n-1)(2n^2+5n+6) is divisible by 6 for any natural n.

[@davidgould9431]

You're right. (n-1)(2n² + 5n + 6) = (13)(2∙196 + 70 + 6) = (13)(468) = 2²∙3²∙13² which is divisible by 6 but clearly isn't a power of a prime.

[@superhightemperature2941]

Happy New Year everybody!

[@Ahmed-Youcef1959]

happy new year everybody

[@andrewkarsten5268]

The associated prime with the n=7 case is 139. Since at n=4, sum=29, we have sum=29+25=54 for n=5, sum=54+36=90 for n=6, and lastly sum=90+49=139 for the n=7 case, and 139 is a prime.

[@jakeb6977]

n=2 is the easy solution since it means p and k are equal to 2 as well

[@noumanegaou3227]

We have a good idea how to start 2022. You publishe a wormap problem and we send our solution And you're showing one of the right solutions in the last video.

[@alainbarnier1995]

Hard this one ! Happy new year :)

[@krisbrandenberger544]

For the n=14 case, the left hand side should be 6084 and the right hand side should be 36 × 13².

[@particleonazock2246]

Yes, summations are oddly the only thing I can solve. I feel so euphoric.

[@tinafeyalien]

Not everyday you hear a maths professor repeatedly referring to factors of a prime! He means prime power.

[@natepolidoro4565]

n=7 gives prime 139

[@minwithoutintroduction]

16:43

[@HagenvonEitzen]

One can exploit n-1 < 2n^2+… earlier and write n-1 = ap^x, 2n^2+… = bp^y with ab=6, x+y=k>0. In fact, we even have 2n^2+… > (n-1)^2. So bp^y>a^2p^(2x). If 2x>y, we must have p

[@siyuanhuo7301]

7 works, 27 fails, 40 fails, 79 fails

[@HagenvonEitzen]

One can stumble over the factor (n-1) also by noting that n=1 makes the dum the empty sum.

[@lox7182]

I find the fact that you use sub-subcases pretty cool

[@Hyakurin_]

If gcd(...)=1 one could also note that p^k does not divide n-1, since p^k>1+1+...+1=n-1, so p^k must divide the other factor (since they are relatively prime), this implies that 2n^2+5n+6=qp^k, therefore (n-1)q=6, that is n-1 divides 6.

[@MrRyanroberson1]

Another neat example of case-averse logic

[@梁偉康-d9k]

Vjvv

[@yahyaelkarkouri3117]

(7,139)

[@lyrimetacurl0]

Here the sum of squares is excluding m=1 for some reason? When 1 is included you get an interesting result that sum of squares is only a power of 2 when n=24 (if n>1). Not tried it for other powers.

[@threstytorres4306]

58SECONDS LATE EDIT: IF N=7, THE SUM FROM M=2 UP TO 7 OF M² IS 139, WHICH IS A PRIME NUMBER IF N=27, THE SUM FROM M=2 UP TO 27 OF M² IS 6929, WHICH IS NOT THE SOLUTION SINCE 6 ISN'T DIVISIBLE ON 6929 IF N=40

[@vikaskalsariya9425]

Shawty chill

[@threstytorres4306]

@@vikaskalsariya9425 I kinda like capitalized

[@vikaskalsariya9425]

@@threstytorres4306 It's just a bit harder to read

[@joaovictormelo6442]

How do we know that k needs to be 1?

[@n8cantor]

Far too many cases! I found it easier to note that p cannot divide both (n-1) and (2n^2+5n+6) unless p=13 because if n ≡ 1 mod p [i.e. p divides n-1], then 2n^2+5n+6 ≡ 13 mod p, which is only congruent to 0 if p = 13. If p divides only one of them, then we have the cases n = 2, 3, 4, and 7 like in the video. (And as shown in the video, p^k cannot divide n-1 because then 2n^2+5n+6 1 then the bracketed term ≡ 1 mod 13, and so does not divide 13^k, so must divide 6, but it is far too large to do so. If x = 1, then the bracketed term is still too large to divide solely the 6, and so it must have a factor of 13. The term reduces to 9a+1≡ 0 mod 13, and solving for a gives a ≡ 10. But a can only equal 1, 2, 3, or 6 and thus there are no solutions.

[@tomholroyd7519]

What's on your t-shirt?

[@chandranichaki9580]

Watch the latest video of wishing happy new year in the language of Mathematics kzbin.info/www/bejne/oqK5dpqMmsZnesk 😊

[@charlesglidden557]

Case 14 is 6084 not 5317

[@ashokhaldar9296]

Wow,easy to solve

[@trueriver1950]

Answer: always. I think a more interesting question might have been to ask when is it an *integer* power of a prime.

[@yoavbd123]

Not related to this vid, but I recall a video that showed that if in a sequence the limit of an+1 - an = 0 , then the limit an/ n is 0 Does anyone remember that vid and can share a link? Thanks!

[@ssttaann11]

You can set b_n = n (strictly increasing and approaching +\infty). Then b_{n+1}-b_n = 1.

[@nemoumbra0]

@@ssttaann11 What do you mean? Is your answer related to the comment you've answered to?

[@schweinmachtbree1013]

@@nemoumbra0 with a_n arbitrary and b_n = n you can apply the Stolz-Cesaro theorem which is the discrete version of L'Hopital's rule - it states that if the limit of (a_{n+1}-a_n)/(b_{n+1}-b_n) exists then the limit of a_n/b_n also exists and has the same value (provided that b_n satisfies a certain condition). Michael has done a video on this theorem, but probably OP wants a different video on this special case (b_n = n).

[@chandranichaki9580]

Watch the latest video of wishing happy new year in the language of Mathematics kzbin.info/www/bejne/oqK5dpqMmsZnesk 😊😊

[@TheEternalVortex42]

Simple proof: Take ε > 0 then for n ≥ N we have a_{n+1} - a_n < ε Now simply take a sum of both sides from N to n ∑ a_n - a_{n-1} < ∑ ε This sum telescopes, so a_n - a_{N-1} < (n - N + 1) ε < n ε Now just divide by n a_n / n < ε + a_{N-1}/n But the last term clearly ⭢ 0 since a_{N-1} is a constant. So we are done. (I didn't write it out, but you also need the other inequality for -ε but it's exactly the same).

[@leif1075]

Wyy would anyone think to rewrite the sum of squares as n minus 1 times something..surely you can solve by keeping it in its original form?

[@BiscuitZombies]

test

[@daneeko]

whaaaaaat

[@dougforkes563]

Let n=x+1, substitute to get x(2x^2 + 9x + 13) = 6p^k i.e. x = ap^c and 2x^2 + 9x + 13 = bp^d where ab = 6 and c + d = k if c = 0 then x = 1, 2, 3, or 6 and can be tested by hand. 1, 2, and 3 all succeed by cumulatively summing 4, 9, and 16. Plugging x=6 into formula gives 6(72 + 54 + 13) = 6*139 and 139 is clearly relatively prime to 2, 3, 5, 7, and 11 and thus prime. First four cases all succeed. if c = 1 or more, then substitute x = ap^c into 2x^2 + 9x + 13 to get 2a^2 p^2c + 9ap^c + 13 which has remainder of 13 when divided by p. Thus p must be 13. Also c cannot exceed 1 or the above expression has a factor with a remainder of 1 when divided by 13, Thus there may be four more cases: p=13, c=1, and a=1, 2, 3, or 6. We require that 9ap + p be divisible by p^2 ie 9a + 1 be divisible by p (where p is 13) But that is not the case for a = 1, 2, 3, 6 so these cases fail.