### 9th Putnam 1949

**Problem B4**

Let (1 + x - √(x^{2} - 6x + 1) )/4 = ∑_{1}^{∞} a_{n}x^{n}. Show that all a_{n} are positive integers.

**Solution**

We have 1 + x - √(x^{2} - 6x + 1) ) = 4 ∑ a_{n} x^{n}, and hence 1 + x + √(x^{2} - 6x + 1) ) = 2x + 2 - 4 ∑ a_{n} x^{n}. Multiplying, 1 + 2x + x^{2} - x^{2} + 6x - 1 = 8 x = 8(a_{1}x + a_{2}x^{2} + ... ) (1 + (1 - 2a_{1})x - 2a_{2}x^{2} - 2a_{3}x^{3} - ... ). Hence:

a_{1} = 1
a_{2} + a_{1}(1 - 2a_{1}) = 0
a_{3} + a_{2}(1 - 2a_{1}) - 2a_{1}a_{2} = 0
a_{4} + a_{3}(1 - 2a_{1}) - 2a_{2}a_{2} - 2a_{1}a_{3} = 0
...

The nth equation has the form a_{n} + combination of a_{1}, ... , a_{n-1} with integer coefficients, so by a simple induction, a_{n} is integral.

9th Putnam 1949

© John Scholes

jscholes@kalva.demon.co.uk

11 Mar 2002