Solution 62: Cleverly Manipulating a Diophantine Equation with x^2018

Let's prove that there is a unique solution to an initially complicated-looking Diophantine equation.

Congratulations to Gabriel N., Allaizn, Theodore Leebrant, Bruno Alejandro Andrades, Nicholas Patel, Beshoy Nabil, Jaleb, adandap, Luis Alba Sarria, and Alex Edwards for successfully solving this math challenge question! Gabriel N. was the first person to solve the question.

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[carcinogenicthalidomide]

I studied modular Arithmetic last year out of curiosity. Thanks for revealing its importance.

[erock]

Modular arithmetic is a usual suspect for many Diophantine equations but I wouldn’t have thought to try polynomial division. Very slick!

[amitotc]

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.

[xCorvusx]

Nice video and interesting solutions.
I have never heard of synthetic long division of polynomials before, and I appreciate learning something new 👍

[benburdick]

I would have never thought to do polynomial division to that. A great problem though!

[supratimsantra]

Sir good problem with excellent approach... Please continue ur uploading

[srpenguinbr]

I imagine how incredible would be a number theory question with integer imaginary numbers. I think it is yet a big field to study

[AmitKumar-yejq]

Respected sir, plz solve this question-find the value of sine^-1(sine20), where 20 is in radians.Answer is pi-20.

[zzz]

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

[ianbrown]

In what class is modular arithmetic covered?

[shivammalluri]

I have an integral question for you
Integrate the function :
(1-x²)/(x⁴+3x²+1)

It is an indefinite integral

[joeaverage]

I don't know what accent this is. Where are you from?