Math Olympiad | A Nice Algebra Math Problem 

Yes, I agree. First a=x+1/x then a=x, b=1/x. Which is which?

X^7+1/(X^7)=843 x=3/2±Sqrt[5]/2=1.5±0.5Sqrt[2]=(1 1/2)±Sqrt[2]/2

Nice solution but here is an alternative: Define Z(n) = x^n +1/(x^n) => Z(0)=2, Z(1)=x+1/x, these have nice recursion formula: Z(n+1)=Z(n)*Z(1)-Z(n-1) (also Z(2n)=Z(n)^2-2). From this Z(7)=Z(1) * (Z(1)^6-7*Z(1)^4+14*Z(1)^2-7) = 3*281 =843=> Z(1)=3; Use recursion from [Z(1)....Z(7)]=[3,7,18,47,123,322,843]. Hence Z(3)+Z(5) = 18+123 = 141

Once you let x+ 1/x =a and then you put x=a and 1/x=b How?ok .no problem.its just for understanding the formula.