71Hi Admin Sir! :)
1.Find all positive integers n, for which "2n+12n+2011n" is a perfect square.
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4 Answers
1057first of all n cannot be a multiple of 4 or of the form 4k+3(where k is any whole no)
i deduce this result on the basis of unit's digit which will be 3 in the first case and 7 in the second case....
1Let 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 ,
2n-2 + 3n4n-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