What is the number of positive solutions to (x1000 + 1)(1 + x2 + x4 + … + x998) = 1000x999?

What is the number of positive solutions to (x1000 + 1)(1 + x2 + x4 + … + x998) = 1000x999?

Question of the Day (30-Mar-12) option 2...for x=1. (x^1000 + 1)(1 + x^2 + x^4 + . + x^998) = 1000.x^999 -----(1) let S=1 + x^2 + x^4 + . + x^998. hence,S*x^2=x^2 + x^4 + . + x^1000. Now, S.x^2 - S=x^1000 - 1. means S(x^2 - 1)=x^1000 - 1. hence,S=(x^1000 - 1)/(x^2 - 1). Now, Equation (1) will become, (x^1000 + 1)*(x^1000 - 1)/(x^2 - 1)=1000*x^999. ===> (x^2000 - 1)=(x^2 - 1)(1000*x^999) ===> x^1001*x^999 -1 = 1000*x^1001 - 1000*x^999. ===> x^1001*x^999 - 1000*x^1001 =1 - 1000*x^999. ===> x^1001(x^999 - 1000)=(1 - 1000*x^999)------(2) in equation (2),we can see for all +ve values of X greater than 1, (x^999 - 1000) this will give +ve value. x^1001 will be +ve.Hence,x^1001(x^999 - 1000) will be +ve. But,for same x values (1 - 1000*x^999) will give -ve values.So,for x>1,above L.H.S won't be equal to R.H.S. Similarly,for all -ve values of x less than -1,x^1001(x^999 - 1000) will be far more than (1 - 1000*x^999). Hence,L.H.S won't will be equal to R.H.S for x less than -1. NOw,for x=1,both L.H.S will be same and equal -999. for x=-1,both L.H.S will be same and equal to 1001. Hence,solution for the equation will be x=1 and -1. But,question is asking us for only +ve solution,hence x can be only 1 not -1.