Understand the problem

Find all positive integers $n$ such that equation $$3a^2-b^2=2018^n$$has a solution in integers $a$ and $b$.

Start with hints

Source of the problem

Belarus MO 2018 Problem 10.5

Topic
Number Theory
Difficulty Level
5/10
Suggested Book
An Introduction to Number Theory
Do you really need a hint? Try it first!

Let’s check for n = 1. Observe that a = 27, b = 13 gives a solutions for n = 1. What about higher degrees? Can we use this information?
Does it work for n = 2? Let’s prove something general! Prove that for a, b to have solutions, n must be odd.
If n is even, Take $\pmod{3}$ to see that $-b^2\equiv 1\pmod{3}$, which has no integer solutions in $b$. Hence, n must be odd.
Well now take n odd. Say $n=2m+1$ for some positive integer $m$. Then, the solution $(a,b)=(27\times 2018^m, 13\times 2018^m)$ exists and works.

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