# 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) = (x^{2} − 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

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