Let (1 + x - √(x2 - 6x + 1) )/4 = ∑1∞ anxn. Show that all an are positive integers.
Solution
We have 1 + x - √(x2 - 6x + 1) ) = 4 ∑ an xn, and hence 1 + x + √(x2 - 6x + 1) ) = 2x + 2 - 4 ∑ an xn. Multiplying, 1 + 2x + x2 - x2 + 6x - 1 = 8 x = 8(a1x + a2x2 + ... ) (1 + (1 - 2a1)x - 2a2x2 - 2a3x3 - ... ). Hence:
a1 = 1 a2 + a1(1 - 2a1) = 0 a3 + a2(1 - 2a1) - 2a1a2 = 0 a4 + a3(1 - 2a1) - 2a2a2 - 2a1a3 = 0 ...The nth equation has the form an + combination of a1, ... , an-1 with integer coefficients, so by a simple induction, an is integral.
© John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002