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Question

Define a sequence f0(x),f1(x),f2(x), of functions by f0(x)=1,f1(x)=x, (fn(x))21=fn+1(x)fn1(x), for n1.
Prove that each fn(x) is a polynomial with integer coefficients.

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Solution

We have,
f2n(x)fn1(x)fn+1(x)=1=f2n1(x)fn2(x)fn(x).
fn(x)(fn(x)+fn2(x))=fn1(fn1(x)+fn+1(x)).
We write this as
fn1(x)+fn+1(x)fn(x)=fn2(x)+fn(x)fn1(x).
Using induction, we get
fn1(x)+fn+1(x)fn(x)=f0(x)+f2(x)f1(x).
Observe that
f2(x)=f21(x)1f0(x)=x21.
Hence
fn1(x)+fn+1(x)fn(x)=1+(x21)x=x.
Thus we obtain
fn+1(x)=xfn(x)fn1(x).
Since f0(x),f1(x)andf2(x) are polynomials with integer coefficients, induction again shows that fn(x) is a polynomial with integer coefficients.
Note: We can get fn(x) explicitly:
fn(x)=xn(n11)xn2+(n22)xn4(n33)xn6+

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