It seemed to me that the synthetic division method was equivalent to, but more elaborate than, appeal to the Factor Theorem. Having established that (x - 1) divides (x^2018 + 1), and since (x - 1) also obviously divides (x^2018 - 1) by the Factor Theorem, then (x - 1) divides their difference, which is 2. I wish I could have found the solution myself but, when you got to that stage the next step seemed obvious as above, whereas the division method made it seem more awkward. Great problem though!
Technically the synthetic division method is more precisely equivalent to using the Remainder Theorem to show the remainder is 2. But that would mean the remainder is both 2 and 0. It's easier to explain intelligibly using the Factor Theorem instead.
I wanted to know if this solution was correct. I originally posted this solution on the challenge video. The solution goes like: Let x = 2n + 1 and y = 2m + 1, by rearranging the original equation we can get (y^2 + 1)*(x-1) = 2(x^2018 + 1), Substituting for x and y we get the following equation ((2m + 1)^2 + 1)(2n + 1 -1) = 2 ((2n + 1)^2018 + 1), now simplifying this we get (4m^2 + 4m + 2)(n) = (2n+1)^2018 + 1. Now on RHS using binomial expansion we get RHS = (1 + C(2018,1)*(2n) + C(2018,2)*(2n)^2 + ... + (2n)^2018) + 1, this RHS can be re-written as RHS = 2 + 2nk for some nutural number k.Clearly if n > 1 then RHS gives remainder 2 when divided by 2n and if n = 1 then RHS is divisible by 2n. But we have LHS = (4m^2 + 4m + 2)(n) = (2m^2 + 2m + 1)(2n), clearly LHS is divisible by 2n. So from our conclusion about RHS we must have n = 1. If n = 1 this means x = 2n + 1 = 3, substituting the value back in the original equation we get 3y^2 + 3 = 2*3^2018 + y^2 + 3, which simplifies to y^2 = 3^2018 which means y = 3^1009 so log_x y = 1009.
@@ripansharma5259 Still less than the hard work done in uploading the videos with solution. Thanks to LetsSolveMathProblems, want more diophantine equation question :D
We can notice that as x-1 divides x^2018 +1 and x-1 also divides x^2018-1 (sum of geometric sequence) x-1 is 1 or 2. But since x is odd then x-1 =2 , x=3
I studied modular Arithmetic last year out of curiosity. Thanks for revealing its importance.
Nice video and interesting solutions.
I have never heard of synthetic long division of polynomials before, and I appreciate learning something new 👍
It seemed to me that the synthetic division method was equivalent to, but more elaborate than, appeal to the Factor Theorem. Having established that (x - 1) divides (x^2018 + 1), and since (x - 1) also obviously divides (x^2018 - 1) by the Factor Theorem, then (x - 1) divides their difference, which is 2.
I wish I could have found the solution myself but, when you got to that stage the next step seemed obvious as above, whereas the division method made it seem more awkward.
Great problem though!
That is an excellent insight. It does make the solution almost instantaneous once we reach the part you mentioned. =)
Technically the synthetic division method is more precisely equivalent to using the Remainder Theorem to show the remainder is 2. But that would mean the remainder is both 2 and 0. It's easier to explain intelligibly using the Factor Theorem instead.
Modular arithmetic is a usual suspect for many Diophantine equations but I wouldn’t have thought to try polynomial division. Very slick!
I wanted to know if this solution was correct. I originally posted this solution on the challenge video. The solution goes like:
Let x = 2n + 1 and y = 2m + 1, by rearranging the original equation we can get (y^2 + 1)*(x-1) = 2(x^2018 + 1), Substituting for x and y we get the following equation ((2m + 1)^2 + 1)(2n + 1 -1) = 2 ((2n + 1)^2018 + 1), now simplifying this we get (4m^2 + 4m + 2)(n) = (2n+1)^2018 + 1. Now on RHS using binomial expansion we get RHS = (1 + C(2018,1)*(2n) + C(2018,2)*(2n)^2 + ... + (2n)^2018) + 1, this RHS can be re-written as RHS = 2 + 2nk for some nutural number k.Clearly if n > 1 then RHS gives remainder 2 when divided by 2n and if n = 1 then RHS is divisible by 2n. But we have LHS = (4m^2 + 4m + 2)(n) = (2m^2 + 2m + 1)(2n), clearly LHS is divisible by 2n. So from our conclusion about RHS we must have n = 1. If n = 1 this means x = 2n + 1 = 3, substituting the value back in the original equation we get 3y^2 + 3 = 2*3^2018 + y^2 + 3, which simplifies to y^2 = 3^2018 which means y = 3^1009 so log_x y = 1009.
👍👍👍..I also solved it in the same way
That's a lot of hard work done in typing the entire solution
@@ripansharma5259 Still less than the hard work done in uploading the videos with solution. Thanks to LetsSolveMathProblems, want more diophantine equation question :D
Both solutions are elegant
I would have never thought to do polynomial division to that. A great problem though!
I imagine how incredible would be a number theory question with integer imaginary numbers. I think it is yet a big field to study
hell yeah
Sir good problem with excellent approach... Please continue ur uploading
We can notice that as x-1 divides x^2018 +1 and x-1 also divides x^2018-1 (sum of geometric sequence) x-1 is 1 or 2. But since x is odd then x-1 =2 , x=3
In what class is modular arithmetic covered?
Shitting class
I don't know what accent this is. Where are you from?
Sounds a bit east asian because he slurs his R, but idk this is the first time I've watch his videos.
Pretty sure it is an italian accent
@@IceSCream4u stfu lol
pure asian english accent
@@IceSCream4u wtf🤣🤣🤣
Respected sir,plz solve this question-find the value of sine^-1(sine20),where 20 is in radians.Answer is pi-20.
너무 대단....
Just wowwww....!!!!
A written solution for this question:
drive.google.com/file/d/1Qyz0zdczuJdhOLpwgfzd5dyLrGplruHu/view?usp=sharing
I have an integral question for you
Integrate the function :
(1-x²)/(x⁴+3x²+1)
It is an indefinite integral
Make denominator (x^2 +1)^2 + x^2 divide both sides by x^2 and u sub 1-1/x^2
Sunrit Roy Karmakar he didn’t ask u
@@newkid9807 why are you salty did you waste a day trying to solve this
Sunrit Roy Karmakar nope, you are just talking when you wernt supposed to porn man
Sounds easy form a x-1/x substitution