More fun in 2016, Part 3
Let be a monic polynomial whose non-leading coefficients are either 1 or . If has no real roots, what is the maximal number of coefficients the polynomial can have?
The answer is 1008.
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1 solution
We claim that the answer is 1 0 0 8 , in which a satisfying example is having negative coefficients on the odd exponents, e.g. x 2 0 1 6 − x 2 0 1 5 + x 2 0 1 4 − x 2 0 1 3 + . . . + 1 . This polynomial has no real roots because it can also be written as x + 1 x 2 0 1 7 + 1 , whose roots we know are complex.
If f ( x ) has more than 1 0 0 8 negative one coefficients, then more than half of the coefficients are negative. This means f ( 1 ) = # of positive coefficients-# of negative coefficients < 0 . Since f ( x ) has a positive leading coefficient, it is intuitive to assume that there exists a > 1 such that f ( a ) > 0 . Since f ( 1 ) < 0 < f ( a ) , there must be a c ∈ ( 1 , a ) satisfying f ( c ) = 0 by the intermediate value theorem, meaning f ( x ) must have a real root, a contradication.