Question

Find the number of real roots x, for the following equation.

(x - 1) (x - 3) (x - 5) (x - 7) ...... (x - 2015) (x - 2017) = (x - 2) (x - 4) (x - 6) (x - 8) .... (x - 2014) (x - 2016)

It may not need to solve a polynomial of degree 2017... Simpler solutions exist..

Answer 1

On the LHS there are (2017+1)/2 = 1009 (odd number) terms. On the RHS there are 1008 (even number) terms.
 
  We know that if a polynomial f(x) is positive at x = a and is negative at x = b, then f(x) = 0 at some x in between a and b. So that x is a real root.

Let  f(x) = LHS - RHS 
 = (x-1)(x-3)(x-5)...(x-2015)(x-2017) - (x-2)(x-4)(x-6)...(x-2014)(x-2016)
f(0) = (-1)(-3)(-5)...(-2015)(-2017) - (-2)(-4)(-6)..(-2014)(-2016)
      =  - 1*3*5*...* 2015*2017 - 2*4*6*...*2014*2016  = negative.
f(1) = 0  -  (-1)(-3)(-5)...(-2014)(-2015)    = negative
f(2) = 1*(-1)(-3)(-5)....(-2013)(-2015) - 0 = positive      ∵1008 terms are -ve
f(3) = 0 -  1 (-1)(-3)(-5)...(-2011)(-2013) = positive      ∵1007 terms are -ve
f(4) = 3*1* (-1)(-3)(-5)....(-2011)(-2013) - 0 = negative    ∵1007 terms -ve
f(5) = 0 - 3*1*(-1)(-3)(-5) ...(-2009)(-2011) = negative    ∵1006 terms -ve
f(6) = 5*3*1*(-1)(-3)(-5)...(-2009)(-2011) - 0 = positive   ∵ 1006 terms -ve
:  :
:  :
f(2013) = 0 - 2011*2009 *2007*...*1*(-1)(-3) = negative 
f(2014) = 2013*2011*...*1*(-1)(-3) - 0  = positive
f(2015) = 0 -  2013*2011*...*1*(-1) =  positive
f(2016) = 2015*2013*..*1*(-1) -  0  = negative
f(2017) = 0 - 2015*2013 *... *3*1  = negative
f(2018) = 2017*2015*2013*..3*1 -  2016*2014*..*4*2 = positive

We see that as x increases by 2, we find that f(x) changes sign from positive to negative or  from negative to positive.  

Thus there are  1009  real positive roots.