# 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.