Prove that there are no integer solutions to $ c^2 = 2018^a + 2018^b $ A friend of mine gave me this puzzle and I want to solve it, turns out its harder than I expected. I tried to prove this by contradiction, so let's just assume there is an integer solution. The first thing I noticed was this: $ 2018 = 2 \cdot 1009$, 2 and 1009 are both primes, also c must be an even number. 
My first idea was to rewrite a little:
$ c^2 = 2018^a + 2018^b \implies 2018^a = c^2 - 2018^b = (c+\sqrt{2018}^b)(c-\sqrt{2018}^b) $, but as it turns out, $(c+\sqrt{2018}^b)$ and $(c-\sqrt{2018}^b)$ don't necessarily have to be irrational, for example $ 4 = \sqrt{16} = \sqrt{2} \cdot \sqrt{8}$ is the product of two irrationals, but it's rational, so this idea doesn't really work. 
I never tried such problems so I don't really know what to look for and what to try. I would be glad if someone could give me a small hint, thanks in advance!  

Hints:
1) Assuming w.l.o.g $a\le b$, rewrite the equation as $c^2=2018^a(1+2018^{b-a})$, then prove that $a$ can't be odd (hint: $c^2$ contains each prime factor in an even power).
2) This leads to an equation with one less variable $d^2=1+2018^e$.
3) Rewrite this as $(d+1)(d-1)=2018^e$, so this number contains only the prime factors $2$ and $1009$, and in equal power. Discuss how they can distribute between $d+1$ and $d-1$.
4) Then show that this distribution is impossible due to both factors having vastly different sizes for $e > 0$.
