1 plus the product of four consecutive integers

[@saba6601]

Prime Newton, you are a brilliant Mathematician.Regards Dr Sabapathy (Mathematician, Singapore 🇸🇬)

[@bvenable78]

I love your grin at the end of each solution. Thanks for sharing your joy with us. 💙

[@BartBuzz]

Now that's pretty neat! The "difference of squares" is how I taught my granddaughter how to compute calculations like (60^2 - 59^2) in her head.

[@Akenfelds1]

I saw the problem, I tried to prove it. I came close but never made it. Then I watched the full video and sat there with my mouth wide open. Beautifually done!

[@perpetgholl5742]

very few smart channel on youtube. you're one of them. thanks for sharing.

[@SidneiMV]

1 + (k² + 3k + 2)(k² + 3k) k² + 3k = u 1 + u(u + 2) = u² + 2u + 1 = (u + 1)² (u + 1)² is a perfect square so (k² + 3k + 1)² is also a perfect square

[@saba6601]

@SidneiMV Very nice approach 👍 Regards Dr Sabapathy (Mathematician, Singapore 🇸🇬)

[@knupug]

Really liked you solution. Very elegant.

[@ahmedrafea8542]

As always, you continue to amaze me with your talent, great choice of problems, and wonderful style of explaining things. Thanks very much and keep up the wonderful work.

[@NoirKnight-uy2ce]

Although I am not taking the class required for that, I enjoyed the lesson I learned from here. NEVER STOP LEARNING!

[@keithhigh7773]

Brit here. Excellent, thanks. I managed to follow your workings, even when you took a shortcut! And you are the first person to explain the Capital pi in a manner l could follow. I shall recommend your channel to my grandchildren. Not sure how food technology and mathematics go together, but at the end of the day, everything can be explained by maths.

[@PrimeNewtons]

It all works out in the end.

[@chrisbarrington108]

Thank you. Nice presentation. You have a very reassuring way of teaching.… A more general version of the same problem is: Prove that the product of four positive integers in arithmetic progression with common difference d plus a constant (d^4) is a perfect square.

[@satyapalsingh4429]

You are a great mathematician .Keep it up your wonderful work !!!

[@PrimeNewtons]

You're very kind

[@aconite_claw]

Your lullabies like voice and mathematical acumen make maths story like. Quite appreciated. I like it a lot.

[@anandarunakumar6819]

Nice problem. After seeing the symmetric form of polynomial, I was tempted to use Pascal's triangle to simplify. The expression k^4+6k^3+11k^2+6k+1 = k^4+4k^3+6k^2+4k+1 + 2(k)(k^2+2k+1)+ k^2 , which reduces to a perfect square: ((k+1)^2+k)^2. Thanks for this problem, discovered a nice relation of Pascal's triangle of alternate rows. Take 3 alternate rows of Pascal's triangle, the sum of coefficients of 1st and third row, and add twice the sum of middle row to get a perfect square!

[@jpl569]

Brilliant ! Funny enough, I found a solution slightly longer, grouping the k in a different way…: X = 1 + k (k+3) (k+1) (k+2) = 1 + [(k + 3/2)^2 - 9/4] [(k + 3/2)^2 - 1/4] Fortunately, this leads to : X = 1 + (k^2 + 3k) (k^2 + 3k + 2), and the rest is the same. Strange to see that keeping the symmetry around k + 3/2 is less efficient than your grouping ! Thanks for your interesting videos ! 🙂

[@riccardofroz]

That was a pretty elegant solution. I got a more convoluted one as I expanded the whole thing: 1+k(k+1)(k+2)(k+3) 1+(k^2+k)(k^2+5k+6) 1+k^4+5k^3+6k^2+k^3+5k^2+6k k^4+6k^3+11k^2+6k+1 Extract a quartic binomial: (k^4+4k^3+6k^2+4k+1)+2k^3+5k^2+2k (k+1)^4+2k^3+5k^2+2k (k+1)^4+k(2k^2+5k+2) Extract a square binomial from the right term: (k+1)^4+2k(k^2+2k+1)+k^2 (k+1)^4+2k(k+1)^2+k^2 The previous is a square binomial with first term (k+1)^2 and second term k. ((k+1)^2+k)^2

[@giuseppemalaguti435]

Complimenti,ho avuto la stessa idea

[@HomePersonal-yc9rx]

What an amazing smile and intro ❤🎉

[@twlson49]

Excellent explanation. Kudos to you.

[@raymondseligman7003]

I have asked a number of times and I don’t want to be a pain in the butt but do you teach, where did you get your incredible mathematical knowledge, and where did you go to school? Your videos are absolutely brilliant. Every time one pops up I think there’s no way I could do it or understand it but by the time you’re done I do understand it. My compliments.

[@MrConverse]

0:49, maybe I’m being a little picky here but you can’t have a product of terms. Things that are added together are terms. Things that are multiplied together are factors. So we have a product of factors here. Hope it helps. Nice vid!

[@PrimeNewtons]

You're correct

[@henryubah5031]

This is brilliant

[@RoyPierce-fb8mt]

>>> from math import factorial as fc, sqrt as rt >>> 1 + fc(4) 25 >>> 1 + fc(8) / fc(4) 1681.0 >>> rt(1681) 41.0 >>>

[@vaibhav_._.]

Hello sir ! Please make one more video on solving questions of JEE ADVANCE. . Pleasee 🌸

[@robertpearce8394]

Neat

[@ernestdecsi5913]

Ez valóban egyszerű és nagyszerű!

[@fahadbinferoz6758]

When are you going to upload the previous video correction video?

[@PrimeNewtons]

Tomorrow

[@xbz24]

Where do I buy those chalkboard

[@maths01n]

Learnt ❤

[@makehimobsessedwithyou6412]

you are so clever

[@mathunt1130]

There might be a simplistic way. You know that the expression will be a quartic in k,and moreover a square of a quadratic, so write the quadratic as ak^2+bk+c. Now square and compare coefficients to find the quadratic. It's not as elegant as the solution provided.

[@frreinov]

Great. I tried it by myself but just couldn't come up with the way you went.

[@ricardoguzman5014]

Elegant manipulation. My solution is as follows: show k(k+1)(k+2)(k+3)+1 is a square. let k+1=a and let k+2=b. Then it can be rewritten as such: (a-1)a x b(b+1) +1= ab(a-1)(b+1)+1= ab(ab+a-b-1)+1; now notice the parenthesis (ab+a-b-1); a is 1 less than b, so a-b in the parenthesis is -1, thus it becomes: ab(ab-1-1)+1= ab(ab-2) +1 = a²b² -2ab +1, which factors as (ab-1)²; thus k(k+1)(k+2)(k+3) +1 is a square.

[@AtharvaKhadilkar-f4j]

Can you solve JEE advanced pyqs

[@domanicmarcus2176]

As a first-year undergraduate, I am taking proof writing for my major (Pure Mathematics). Can you please make videos on all types of proof writing? Thank you

[@sauzerfenicedinanto]

If n is a natural number then the product can be written as n(n+1)(n+2)(n+3)=[n(n+3)][(n+1)(n+2)]=[n^2+3n][n^2+3n+2]=[(n^2+3n+1)-1][(n^2+3n+1)+1]= (n^2+3n+1)^2-1 if we add 1 the we obtain always a perfect square, precisely (n^2+3n+1)^2.

[@MrGeorge1896]

Perf.Sq. = 1 + k (k + 1) (k + 2) (k + 3) = k⁴ + 6k³ + 11k² + 6k + 1 = k² [ ( k² + 1/k² ) + 6 ( k + 1/k ) + 11 ] = k² [ ( k + 1/k )² + 6 ( k + 1/k ) + 9 ] = k² [ ( k + 1/k ) + 3 ] ² = [ k² + 1 + 3k ] ²

[@raghvendrasingh1289]

❤ Generalization - if p , q , r and s are in AP then pqrs+d^4 is always a perfect square where d is common difference proof - let p = a-3c , q = a-c , r = a+c , s = a+3c then d = 2c pqrs+d^4 = (a^-9c^2)(a^2-c^2)+16c^4 = a^4- 10a^2c^2+25c^4 = (a^-5c^2)^2

[@StudyOnly-nn1xb]

K³-k(k+2)=k⁴+2k³-k²+2k (k²+k)²-2(k²+k)+1-1 (K²+k+1)²-1 +1 (K²+k+1)²

[@HrishikeshRaj-qv6lr]

Indians assemble here 🇮🇳❤

[@graf_paper]

1+ n(n+1)(n+2)(n+3) = (n + (n+1)²)² Very nice!!!

[@robertveith6383]

Show the connecting steps in between the first line and the last line.

[@Meowcat-kr6tc]

@@robertveith6383 They're just noting the general idea the problem suggests

[@mehdizangiabadi-iw6tn]

1≠i

[@Kakarot-kr]

Sir pls do this Let f(x)=lim n infty n^ n (x+n)(x+ n 2 )***(x+ n n ) n!(x ^ 2 + n ^ 2)(x ^ 2 + (n ^ 2)/4) ***(x^ 2 + n^ 2 n^ 2 ) ^ x n , for all x > 0 Then (A) f(1/2) >= f(1) (B) f(1/3) = (f' * (2))/(f(2)) Jee advanced 2016 shift 2

[@Bertin-q3y]

-9

[@josepherhardt164]

OMG. Somebody pinch me. My 70-something-year-old butt just figured this out using 50-year-old math. Maybe I should buy a lottery ticket. ||:-/

[@0lympy]

Your square symbol isn't perfect :(

[@mircoceccarelli6689]

👍 1 + k(k+1)(k+2)(k+3) = (k^2 + 3k + 1 )^2 = [ ( k + 1 )( k + 2 ) - 1 ]^2 = m^2 k € Z , m € Z 👍😁👋

[@marioenriquegarduno9676]

El Memín penguin de las matemáticas 👍