03-01-12 NT

Let 2ⁿ + 12ⁿ + 2011ⁿ = x²
→ x² = (-1)ⁿ + 1mod(3)
since all perfect squares are zero or 1 modulo 3 this means 'n' must be odd. n=1 is obviously a solution.

lets consider there exists some solution n≥2 ,
2ⁿ + 12ⁿ ≡ x² - 2011ⁿ
from this we can see that x must be odd ( since LHS is even )

so substituting x = 2k+1 and dividing by 4 on both the sides gives ,
2^n-2 + 3ⁿ4^n-1 = k²+k + 14(1-2011ⁿ)
→ 1- 2011ⁿ should be of the form 4p
→ 1-2011ⁿ ≡ 0mod(4)
→ (-1)ⁿ ≡ 1 mod(4) this is possible only when n is even.
that is contradiction!!
Hence n=1 is the only solution