An example of a polynomial f(X) 
R[X] such that the corresponding
function R -> R satisfies f(1) = 2 and
f(2) = 5,
is f(X) = X2 + 1, but also 3X - 1. One can look for
such a polynomial as follows. Choose a degree, preferably equal to
the number of interpolation points minus 1; but let us now take 2.
Then write f(X) = f0
+ f1 X
+ f2X2 and substitute the given values. This
leads to the following system of linear equations:
f 0 + f 1 · 1 + f 2 · 12 = 2,
f 0 + f 1 · 2 + f 2 · 22 = 5.
Solving it gives f 0 = 2r - 1, f
1 = 3 - 3r and f2 = r,
with r
R, so that there are many polynomials with the required properties.
It also depends on the degree how many polynomials you find. No
polynomials of degree 
0 do the job, exactly one polynomial of
degree 1 works, and
an infinite number of degree 2. This is in accordance with the theorem, applied for n = 2.