Solution to puzzle 134: Sum of reciprocal roots

If the equation x⁴ − x³ + x + 1 = 0 has roots a, b, c, d, show that 1/a + 1/b is a root of x⁶ + 3x⁵ + 3x⁴ + x³ − 5x² − 5x − 2 = 0.

Firstly, note that, since x = 0 is not a root,

a and b are roots of x⁴ − x³ + x + 1 = 0

1/a and 1/b are roots of (1/x)⁴ − (1/x)³ + (1/x) + 1 = 0.
Or, equivalently, multiplying by x⁴, 1/a and 1/b are roots of x⁴ + x³ − x + 1 = 0. (1)

Now let a' = 1/a, b' = 1/b, c' = 1/c, and d' = 1/d be the roots of equation (1). Using Viète's formulas, we can write down

a' + b' + c' + d' = −1,
a'b' + a'c' + a'd' + b'c' + b'd' + c'd' = 0,
a'b'c' + a'b'd' + a'c'd' + b'c'd' = 1,
a'b'c'd' = 1.

Now we express the above relations in terms of s = a' + b', t = c' + d', p = a'b', and q = c'd'. We seek the equation satisfied by s = a' + b' = 1/a + 1/b. Thus

s + t = −1 t = −1 − s,
st + p + q = 0, (2)
sq + tp = 1, (3)
pq = 1 q = 1/p.

Now substitute for t and q in (2) and (3):

−s(1 + s) + p + 1/p = 0, (4)
s/p − p(1 + s) = 1. (5)

Multiplying (4) by s, subtracting from (5), and simplifying, we obtain
p = (s³ + s² − 1)/(2s + 1). (6)

Now multiply (4) by p and substitute for p from (6):
−s(1 + s)(s³ + s² − 1)/(2s + 1) + (s³ + s² − 1)²/(2s + 1)² + 1 = 0.

Multiplying by −(2s + 1)², and simplifying, we obtain
s⁶ + 3s⁵ + 3s⁴ + s³ − 5s² − 5s − 2 = 0.

Therefore, if the equation x⁴ − x³ + x + 1 = 0 has roots a, b, c, and d, 1/a + 1/b is a root of x⁶ + 3x⁵ + 3x⁴ + x³ − 5x² − 5x − 2 = 0.

Source: Traditional