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.
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.
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
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.
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.
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
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.
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.
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!
@@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).
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).
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.
Thank you for putting together such amazing videos. Happy new year Mike!
This problem expands into an irritatingly large number of cases...
That's maths. I've been very sloppy with my proofs because of this.
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.
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.
happy new year everybody
Happy New Year everybody!
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.
n=2 is the easy solution since it means p and k are equal to 2 as well
Hard this one ! Happy new year :)
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.
For the n=14 case, the left hand side should be 6084 and the right hand side should be 36 × 13².
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
Yes, summations are oddly the only thing I can solve. I feel so euphoric.
One can stumble over the factor (n-1) also by noting that n=1 makes the dum the empty sum.
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.
Another neat example of case-averse logic
Vjvv
(7,139)
I find the fact that you use sub-subcases pretty cool
n=7 gives prime 139
Not everyday you hear a maths professor repeatedly referring to factors of a prime! He means prime power.
How do we know that k needs to be 1?
16:43
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.
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
Shawty chill
@@vikaskalsariya9425 I kinda like capitalized
@@threstytorres4306 It's just a bit harder to read
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.
7 works, 27 fails, 40 fails, 79 fails
Wow,easy to solve
What's on your t-shirt?
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html 😊
Answer: always.
I think a more interesting question might have been to ask when is it an *integer* power of a prime.
Case 14 is 6084
not 5317
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.
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!
You can set b_n = n (strictly increasing and approaching +\infty). Then b_{n+1}-b_n = 1.
@@ssttaann11 What do you mean? Is your answer related to the comment you've answered to?
@@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).
Watch the latest video of wishing happy new year in the language of Mathematics
th-cam.com/video/lqVEgVea4Dg/w-d-xo.html 😊😊
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).
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?
test
whaaaaaat