# Solution to puzzle 130: Reciprocal polynomial?

Let p be a polynomial of degree n with complex coefficients.  Is there a value of n such that the equations

• p(1) = 1/1,
• p(2) = 1/2,
...
• p(n) = 1/n,
• p(n + 1) = 1/(n + 1),
• p(n + 2) = 1/(n + 2),

can be satisfied simultaneously.

It is easy to see that, for n = 1, the unique linear polynomial passing through (1, 1) and (2, 1/2) will pass through (3, 0), not (3, 1/3).
For n = 2, the unique quadratic passing through (1, 1), (2, 1/2), and (3, 1/3) is p(x) = (x2 − 6x +11)/6, which passes through (4, 1/2).
We suspect that even for higher values of n there is no polynomial p such that all of the above equations are satisfied simultaneously.

We will assume that a polynomial of degree n satisfying the above equations exists, and derive a contradiction.

Consider the polynomial defined by q(x) = x(p(x)) − 1.
Clearly, for x = 1, 2, ... , n + 2,  p(x) = 1/x q(x) = 0.  That is, q has (at least) n + 2 roots.
But, since p is a polynomial of degree n, q is a polynomial of degree n + 1, and hence by the Fundamental Theorem of Algebra, q has n + 1 roots, counting multiplicity.
This is a contradiction.

Therefore, for no value of n can a polynomial of degree n satisfy the above equations simultaneously.

## Remarks

We can show that, if p(x) = 1/x for x = 1, 2, ... , n + 1, then p(n + 2) = 0 or 2/(n + 2), according as n is odd or even, respectively.

Since q(1) = q(2) = ... = q(n + 1) = 0, by the Polynomial Factor Theorem q(x) has n + 1 linear factors: (x − 1), (x − 2), ... , (x − n − 1).
As q is of degree n + 1, it follows that q(x) = K(x − 1)(x − 2)...(x − n − 1), for some constant K.

Now consider q(0) = K(−1)n+1(n + 1)!.
But we also have q(0) = 0×p(0) − 1 = −1.
Hence K = (−1)n/(n + 1)!, and q(n + 2) = [(−1)n/(n + 1)!] (n + 1)! = (−1)n.
Since q(n + 2) = (n + 2)(p(n + 2)) − 1, we conclude that p(n + 2) = (1 + (−1)n)/(n + 2).

Source: Inspired by Polynomial Puzzle