A nice problem from the 2012 math olympiad

[@mathflipped]

Olympiad problems involving integers have always been my favorite. Great video, Michael!

[@udic01]

3:14 and you also need check for n=3... Because the inequality holds for n>3 and not for n>=3.

[@dyld921]

Not really, when n=3, the >= right before that turns into a strict >, i.e. (n+1)^n > (n+1)^2 >= 5n+1

[@charlied.4683]

Never seen analysing mod n^2 for these exponential vs polynomial equations before but it is super cool!

[@aqeel6842]

Definitely. There's always a new trick out there, no matter how good you are at math

[@bardamu9662]

Excellent example of the power using modulo p theory.

[@xavi835]

Patxi is an amazing climber, he had recovery from a hard accident and he still climbing in a hard level. I'm climber and mathemsthician as well! (And from Spain)

[@m11071997r]

The first image is the Alhambra of Granada. Saludos!

[@jkid1134]

@@m11071997r never heard a thumbnail called foreplay but you're kinda onto something

[@m11071997r]

@@jkid1134I don't know what do you mean. Who said anything about foreplay? Jajajaja

[@MrDestroys]

It truly is a nice problem

[@speedsterh]

Not overly technical, solution is easy to follow

[@batapanos2027]

Let me suggest the way i came up with.... First bring 1 to the other side of the equation and factor ...on LHS..((n+1)-1)( (n+1)^(n-1) + (n+1)^(n-2) +...+1) and see that the first factor is just n so we divide by n and we are left with ..... (n+1)^(n-1) +....+(n+1) = 2n^m-1 +2.. Then by taking modulo n we get that. (1+1+1+1+..+1)..(n-1 times) is congruent to 2 mod n , supposing m is greater than or equal to 2... So we have n-1 congurent to 2 mod n => n is conguent to 3 mon n , which further implies that n divides 3.... The rest is easy..

[@forcelifeforce]

Did you mean ... = 2n^(m - 1) + 2 ...?

[@themibo899]

if anyone is interested, the thumbnail shows the “Alhambra” in the spanish town of Granada

[@uszkaybalazs]

If we include 0 in the naturals (which is how I've been taught), we get the solution n=0, m€N

[@Pavgran]

And also n=2, m=0

[@andy-kg5fb]

Can you do a video on "friendly" binary sequences, which are sequences where each digit has atleast one 1 next to it.

[@advaykumar9726]

How much marks are you getting in ioqm?

[@Vikipedia11]

Appreciating the shoutout to Eminem at 0:29.

[@donofmath248]

Nice problem! I did something equivalent - subtracted 1 from both sides, factored the LHS as a difference of nth powers, which gives a factor of (n+1-1)=n, divided both sides by n, and then worked mod n to get that -1 is congruent to 2 mod n.

[@goodplacetostop2973]

9:20 Es un buen lugar para detenerse

[@Chris44351]

9:20 Und das ist eine gute Stelle zum Anhalten.

[@darkmask4767]

E questo è un buon posto per smettere

[@Neodynium.the_permanent_magnet]

C'est l'endroit idéal pour s'arrêter.

[@rialtho_the_magnificent]

9:20 dat is een goed moment om te stoppen

[@liliepepe65]

Um bom lugar para parar

[@alexfekken7599]

7:31: and that is a good place to stop 🙂

[@sahiltamang2352]

Now Homework:- Solving over non-negative integers.

[@pedropicapiedra4851]

Michael, we see you from spain too

[@ConManAU]

I like this problem, not just because of the neat solution, but because it’s one of those rare number theory Olympiad problems that isn’t solved by finding an obvious solution and then proving that it’s unique.

[@hernando-d]

Patxi has a baske- spanish name.

[@sahiltamang2352]

Post in your comments

[@Macisordi]

N=0 is a solution, i think in Europe we consider it a natural number!

[@charlessmith1931]

Interesting! I guess that's why it is better to say 'positive integers' vs. 'non-negative integers'. In the US, we are taught that the natural numbers are how we naturally count, i.e. 1, 2, 3, ... and that the whole numbers also include zero (which looks like a hole).

[@SpartaSpartan117]

Math training plan for today: a muerte

[@SuperYoonHo]

"He is from Spain and it's cool;)"

[@ayoubabid213]

Easy problem . Its a junior level Well here is my solution By the binomial formula (n+1)^n =1+n^2 +n^3 *(n-1)/2 .....+n^n So (n+1)^n =1 mod n^2 Lets suppose that m>=2 So 2n^m +3n+1 = 3n+1 mod n^3 So 1=3n +1 mod n^2 1=3n mod n^2 contradiction if n>1 It sufficed now to study the case when m=1 to find the solution Qed

[@goodplacetostart4606]

Early