TITLE: Proof the Fibonacci numbers are not a polynomial.
QUESTION [36 upvotes]: I was asked a while ago to prove there is no polynomial $P$ in $\mathbb R$ such that $P(i)=f_i$ for all $i\geq0$. I tried to get a proof as slick as possible and here's what I got.
Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0$ . Construct the polynomials $p_1(x)=a_n(x-1)^n+a_{n-1}(x-1)^{n-1}+\dots +a_1(x-1)+a_0$ and $p_2(x)=a_n(x-2)^n+a_{n-1}(x-2)^{n-1}+\dots +a_1(x-2)+a_0$.
Then the polynomial $p-(p_1+p_2)$ has infinitely many roots since $p(i)=p_1(i)+p_2(i)$ for $i$ an integer $\ge 1$ because of the Fibonacci recurrence. Hence we have $p_1+p_2=p$ since a  non-zero polynomial of degree $n$ can have at most $n$ zeros.
However $p_1+p_2$ has leading coefficient $2a_n$. So they cannot be equal.
What do you think of my proof. Can you find other proofs? Simpler proofs? harder proofs? The less standard the better. I appreciate proofs which might look excessive in the amount of theory they use, although only if they finish quickly.
Thank you very much.
Regards

REPLY [2 votes]: The simplest I can think of, in broad strokes: Fibonacci numbers are roughly exponential, and an exponential function grows faster than any polynomial, so they can't be represented by a polynomial.  QED.
We can formalize this to pretty much any level desired.  Let's start with the first assertion, that Fibonacci numbers are roughly exponential.  Fibonacci numbers are defined by F(n) = F(n - 1) + F(n - 2), plus some initial conditions (here, F(0) = 0, F(1) = 1, but similar sequences with different initial conditions work just as well).  Dividing through (since F(n) is positive for n > 0, we'll assume an appropriate n), F(n)/F(n - 1) = 1 + F(n - 2)/F(n - 1), so if we define r(n) = F(n)/F(n - 1), r(n) = 1 + 1/r(n - 1).  Let r(k) = ø + e for some e and k.  Then r(k + 1) = 1 + 1/(ø + e).  1/(ø + e) = (1/ø)/(1 + e/ø), which, when |e/ø| < 1, is a geometric series, (1/ø)(1 - e/ø + (e/ø)^2 - ...).  Going back to r(k + 1), that's 1 + (1/ø)(1 - e/ø + ...) = 1 + 1/ø - (e/ø^2)(1 - e/ø + ...) = 1 + 1/ø - (e/ø^2)/(1 + ø/e) = 1 + 1/ø - (e/ø)(1/(ø + e)).  Now, ø = (1 + sqrt(5))/2, so 1 + 1/ø = ø.  Therefore, r(k + 1) = ø - (e/ø)(1/(ø + e), or, in other words, r(k + 1) = ø + f, where f = -(e/ø)(1/(ø + e)).  For all e < -ø^3 or e > -ø, |f| < |e|, though we must make sure that -ø < e < ø for the geometric series trickery to work, and for the Fibonacci sequence in particular, the first real value for r is r(2) = 1, with e = ø - 1 falling well in this range and, subsequent differences being smaller, only produces more values in this range.  Therefore, we can say that growth of the Fibonacci series is bounded between r(n) = 1 and r(n) = 2ø - 1.  However, the sequence being bounded by 1^n isn't particularly useful, so we'll use the next two ratios of 2 and 3/2 to note that Cl·(3/2)^n ≤ F(n) ≤ Cr·2^n, where Cl and Cr are positive constants.  In particular, F(n) ≥ Cl·(3/2)^n.
The rest can be proved similarly, in as much detail as you care about.