a fresh start to 2025

[@claireli88]

2025 is an interesting number: It is a perfect square number, 45²=2025 It is the sum of 3-squares, 40²+20²+5²=2025 It is the sum of cubes of all the single digits from 1 to 9, 1³+2³+3³+4³+5³+6³+7³+8³+9³=2025

[@bjornfeuerbacher5514]

1³+2³+3³+4³+5³+6³+7³+8³+9³=2025 follows from the fact that 1+2+3+4+5+6+7+8+9=45 and 45²=2025. ;)

[@ricardoguzman5014]

It is also the sum of consecutive integers from -10 to 64. -10 + -9 + -8 + -7 + -6 + -5 + -4 + -3 + -2 + -1 + 0 + 1 + 2 + ....+ 62 + 63 + 64 = 2025.

[@antoniopedrofalcaolopesmor6095]

​By induction I could easily prove (1+2+3+...+n)^2 = 1^3+2^3+3^3+...+n^3 I'm just wondering how this can be proved without induction. If you know the proof without induction, pls share. Thank you!

[@claireli88]

@@antoniopedrofalcaolopesmor6095 It can also be proved by using the method of differences or telescoping method as shown below: Using the expansion (k+1)⁴= k⁴+4k³+6k²+4k+1 (k+1)⁴-k⁴= 4k³+6k²+4k+1 ∑[(k+1)⁴-k⁴]= 4∑k³+6∑k²+4∑k+∑1 {All summations are from k=1 to n} Using method of differences on the LHS, (n+1)⁴-1= 4∑k³+6n(n+1)(2n+1)/6+4n(n+1)/2+n Simplifying, 4∑k³= n²(n+1)² ∑k³= [n(n+1)/2]² ∴∑k³= (∑k)² {Summation from k=1 to n}

[@bjornfeuerbacher5514]

@@antoniopedrofalcaolopesmor6095 Another method (quite complicated, but doable) would be using generating functions. Michael Penn once made a video on that.

[@Rai_Te]

Preview image is misleading.

[@bartekguz9371]

Asking just to be sure: Are we hiding the “two complex cubic roots of -1”by “introducing the external real root”? Lovely shenanigans👌

[@blackpenredpen]

2024+1=2025 Boom!

[@drpeyam]

Wooooow!!!

[@rinkumonigogoi5480]

Big fan of both of you

[@md2perpe]

It can be noted that y/x cannot be real but must be complex.

[@xinpingdonohoe3978]

That's true. x+1/x has a local maximum at -1, a local minimum at 1, and a discontinuity at 0. At -1 it's -2, and at 1 it's 2. The real image is then (-∞,-2]U[2,∞).

[@yurenchu]

Happy New Year to you too! Three remarks: - The problem in the thumbnail is a different one. - At 1:08 - 1:51 , you forgot to point out that t ≠ -1 and therefore it's safe to multiply both sides by (t+1) because we're not multiplying by 0 . - 2:18 The fact that 2025 is a multiple of 3 is not really a lifesaver; we could proceed in a similar fashion if the exponent wasn't a multiple of 3 . For example: (x/y)²⁰²⁶ + (y/x)²⁰²⁶ = = (t)²⁰²⁶ + (1/t)²⁰²⁶ = t * (t³)⁶⁷⁵ + (1/t) * (t³)⁻⁶⁷⁵ .... note: t³ = -1 ... = t * (-1)⁶⁷⁵ + (1/t) * (-1)⁻⁶⁷⁵ = t * (-1) + (1/t) * (-1) = (-1) * (t + 1/t) ... note: t + 1/t = 1 ... = (-1) * (1) = -1 (x/y)²⁰²⁴ + (y/x)²⁰²⁴ = = (t)²⁰²⁴ + (1/t)²⁰²⁴ = (1/t) * (t³)⁶⁷⁵ + t * (t³)⁻⁶⁷⁵ .... note: t³ = -1 ... = (1/t) * (-1)⁶⁷⁵ + t * (-1)⁻⁶⁷⁵ = (1/t) * (-1) + t * (-1) = (-1) * (1/t + t) = (-1) * (t + 1/t) ... note: t + 1/t = 1 ... = (-1) * (1) = -1

[@sensei9767]

Multiplying by zero would have been fine. The rest is true, though

[@Suryanshcontactpro]

happy new year ! bless you !

[@cheeseparis1]

Happy 45². Thanks for your videos and your energy.

[@9wyn]

Happy New Year Dr Peyam 🙏🏽

[@stevekaszycki8629]

Maybe a little two negative for a start to the year. On second thought, just right. Thanks for the New Years video.

[@mr.soundguy968]

This is how I solved it: I defined z_n := (x/y)^n + (y/x)^n. It is not hard to show that z_{n+2} = z_{n+1} - z_n for all integers n >= 1 (one can use, e.g., induction on n). Using z_1=1 and z_2 = -1, we quickly obtain a sequence z_1, z_2, z_3, ... that is recurring with period 6. Notice that 2025%6 = 3, so z_{2025} = z_3 = z_2 - z_1 = -1 - 1 = -2. Not only does this give us z_{2025} but it explicitly gives us z_n for all integers n >= 1.

[@slavinojunepri7648]

Excellent

[@antoniocampos9721]

Very clever solution!

[@ricardoguzman5014]

20+25=45, 45²=2025. You probably know this already, but here is something interesting about it. This won't happen again until 3025, in 1000 years, and 1000 is a perfect power, 10³. After that, it won't happen until 9801, which is in 7776 years, and that is also a perfect power, 6⁵. We are so priveleged to live in this exciting time. Happy 2025.

[@antoniopedrofalcaolopesmor6095]

Wondering whether this ever happened in the past 😅

[@antoniopedrofalcaolopesmor6095]

Happened just once before in year 81

[@antoniopedrofalcaolopesmor6095]

After 9801 it will happen next around year 500 000, almost half a million years later

[@antoniopedrofalcaolopesmor6095]

You have a list of such numbers if you google "On-Line Encyclopedia of Integer Sequences" and once in there search for "Numbers which when chopped into two parts with equal length, added and squared result in the same number"

[@Happy_Abe]

Happy New Year!🥳

[@nimaalz4513]

are you from Tabriz ?

[@nuklearwanze]

You really cant do what you did around the 1:40 mark - you just introduced a new root...

[@leozam326]

I think the argument holds: you can conclude that all the solutions of t^2 - t + 1 = 0 are such that t^3 = -1 (because t^3 + 1 = (t^2 - t +1)(t + 1)) and that’s the fact that is used in the end. He is not saying that t = -1 is a solution of the first equation

[@ArpanDasgupta-q4n]

HAPPY NEW YEAR

[@muriloporfirio7853]

The answer is wrong because there is no real solution for t²-t+1=0 t=-1 is an artifact from multiplying by (t+1) The way I would go for solving the problem is finding S_n = t^n + t^-n, as a function of S_i, being 1

[@muriloporfirio7853]

Just calculating is way faster, however and would give us: S_1 = 1 t = cos(2π/3) +- i sin(2π/3) S_n = t^n + t^-n = cos(2πn/3) + i sin(2πn/3) + cos(-2πn/3) + i sin(-2πn/3) S_n = 2cos(2πn/3) If n is a multiple of 3, then S_n = 2 Otherwise, S_n = -1

[@blakedylanmusic]

Dr Peyam never claimed t = -1, he only claimed t^3 = -1. 1:54 2:39

[@patipateeke]

Happy 3*675!!

[@xinpingdonohoe3978]

That's it? 3×675? Not (20+25)²? Not (1+2+3+4+5+6+7+8+9)²? Not 1³+2³+3³+4³+5³+6³+7³+8³+9³?

[@proguyz78]

2*10^3 + Integral e^-x x^4 [0,∞] - e^iπ

[@ArpanDasgupta-q4n]

first