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).
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
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.
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..
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
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...
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.
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.
![[ 2020년 61회 IMO ] 2번 문제](http://i.ytimg.com/vi/LAVdKk3HRGI/mqdefault.jpg)
Thanks for the video!!!! Greetings from Adana,Türkiye🤩🤩
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).
This is an intriguing question!
Amazing!!!
What a nice solution involving a lot of tricks! The way you justify every step is really fine for me. Thanks for sharing.
He doesn't justify why c has to be 1or.less though..
@@leif1075bcs for c>1 the square is congruent to 6(mod 9) which is a contradiction
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
Very nice!
Can you maybe solve more problems that relates to sequences and requrence relations. Nice solution btw
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
Sir can you please upload some double integral problems that were asked in PUTNAM
He does high school level problems. No calculus
@@SimsHacks ok bro
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.
why can't k be a power of 3.7 such as 3^a.7 or 3.7^b
@@kimloantranthi6335 how about substituting k=21n(n=multiples of 3 and 7 ) into x^2+2x+4 , which is 3k^2+6k+4.
@@e39emfive raising 21 to the second power ( when you plug k=21n into 3k^2) doesn't sound too ideal, but you can try
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..
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).
Wow never knew to find gcd of var-fx, wsh i saw these in my younger days....how to know its gcd with 12
Polynomial division
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
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...
12 is the remainder when you do polynomial long division
It will have a remainder of 12 in the solution though.
I show another way of getting GCD=12 in another comment.
Explanation of this one was pretty convoluted. Maybe slow down and write more next time.
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.
I agree - mod 3 or just spotting it's a multiple of 3 is enough to say there are no solutions.
🤝
J
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.
You are doing everything very quickly and that's annoying... Why c is bounded?