A Nice Equation of Powers | Turkish National Maths Olympiad 2014

[@sadececansu9]

Thanks for the video!!!! Greetings from Adana,Türkiye🤩🤩

[@mcwulf25]

Similar solution. We can reduce options further because the product is odd so GCD must be 3 or 1. In your (1) we know there are no solutions because the LHS is a multiple of 3 and the RHS a multiple of 7. So with GCD=3, we know the factors must be 3*7^z and 3^(y-1). (We can discard x-2=3 by substituting in the quadratic factor, and it is not a multiple of 7).

[@rudipaganelli3398]

What a nice solution involving a lot of tricks! The way you justify every step is really fine for me. Thanks for sharing.

[@leif1075]

He doesn't justify why c has to be 1or.less though..

[@itzjanosbtw1509]

​@@leif1075bcs for c>1 the square is congruent to 6(mod 9) which is a contradiction

[@keishakwok4333]

For the c=0 case, another way after realising the 7^d part, may be: c=0, so 3 doesn't divide (x+1)^2, so doesn't divide (x+1), so doesn't divide (x-2), so a = 0, x = 3, but then x^2+2x+4 = 19 which isn't a power of 7, so empty set

[@vacuumcarexpo]

This is an intriguing question!

[@SuperYoonHo]

Amazing!!!

[@mobiusmathematics6901]

Hey do you have any advice for people who want to get better at solving these kinds of equations? I’d love to feature some on my channel

[@e39emfive]

How about taking mod 3 of x. The answer is x ≡2. Substituting x=3k+2 into (x-2)(x^2+2×+4) you get 3k(...). Since k shoud be either 3 or 7 , checking with those into(x^2+2x+4),(of course with substitution with x=3k+2) gives you the answer.

[@kimloantranthi6335]

why can't k be a power of 3.7 such as 3^a.7 or 3.7^b

[@e39emfive]

@@kimloantranthi6335 how about substituting k=21n(n=multiples of 3 and 7 ) into x^2+2x+4 , which is 3k^2+6k+4.

[@kimloantranthi6335]

@@e39emfive raising 21 to the second power ( when you plug k=21n into 3k^2) doesn't sound too ideal, but you can try

[@leif1075]

I don't see why anyone would.think to rewrite x squared plus 2x plus 4 as x plus 1 squared plus 3..and I any case I don't see why that tells you c has to be1..it could be more if x us 1 could be a larger multiple of 3 than 9..

[@mcwulf25]

You get the same answer if you write it as the more obvious (x-2)^2 + 6x and GCD(x-2,6x) = GCD(x-2,6x-6(x-2)) = GCD(x-2,12).

[@tridibeshsamantroy9837]

Sir can you please upload some double integral problems that were asked in PUTNAM

[@SimsHacks]

He does high school level problems. No calculus

[@tridibeshsamantroy9837]

@@SimsHacks ok bro

[@filipristovski88]

Can you maybe solve more problems that relates to sequences and requrence relations. Nice solution btw

[@otakurocklee]

Very nice!

[@disoriented3971]

I'm not clear on how you get a gcd of 12. Try plugging in various values for x in both "x-2" and "x^2 + 2x + 4" and take their quotient... it's not going to have a remainder of 12

[@TheArizus]

The gcd divides 12 and no divisor of 7 divides 12 except 1. However when x is odd and x=1(mod3) they actually do have a gcd of 1 sometimes e.g x=7 so...

[@mcwulf25]

It will have a remainder of 12 in the solution though. I show another way of getting GCD=12 in another comment.

[@mathcanbeeasy]

For the first case, at 4:35, why mode 6 and all that explanations? Is obvious that LHS is multiple of 3, but 7^b is not multiple of 3. 🙂 And maybe is better to prove more accurately than c is at most 1 in (x+1)^2+3=3^c*7^b. I will complete you. Indeed, if that expression is divisible with 3 than (x+1)^2 is multiple of 9 So, (x+1)^2+3=9*p^2+3=3*(3p^2+1) And 3*p^2+1 cannot be multiple of 3 because is always the next number after a multiple of 3. Therefore the expression is divisible at most one time with 3. Very good the gcd (x -2, x ^2+2x+4). Is the key to the entire solution. Congrats for another nice problem.

[@mcwulf25]

I agree - mod 3 or just spotting it's a multiple of 3 is enough to say there are no solutions.

[@bait6652]

Wow never knew to find gcd of var-fx, wsh i saw these in my younger days....how to know its gcd with 12

[@replicaacliper]

Polynomial division

[@willbishop1355]

Explanation of this one was pretty convoluted. Maybe slow down and write more next time.

[@CglravgHRjsksgS]

You are doing everything very quickly and that's annoying... Why c is bounded?

[@user-ry6ey8gq3t]

J

[@daoudandiaye4636]

🤝

[@MrLidless]

This should be solved in a quarter of the time. The methods used here are kindergarten, and I doubt that’s the target audience. 1/5. Has done, can do, and will do better.