Using the defining relationship

P_{n + 2} (x) = {xP}_{n + 1} (x) - P_{n} (x)

we get the sequence:

\begin{matrix} P_{0} & = 0 \\ P_{1} & = 1 \\ P_{2} & = x \\ P_{3} & = x^{2} - 1 \\ P_{4} & = x^{3} - 2 x \\ P_{5} & = x^{4} - 3 x^{2} + 1 \\ P_{6} & = x^{5} - 4 x^{3} + 3 x \dots \end{matrix}

Using the defining relationship we can express

P_{6}

in terms of previous polynomials in the sequence.

\begin{matrix} P_{6} & = {xP}_{5} - P_{4} \\ = (x^{2} - 1) P_{4} - {xP}_{3} \\ = P_{3} P_{4} - P_{2} P_{3} \\ = P_{3} (P_{4} - P_{2}) . \end{matrix}

This shows that

P_{3}

is a factor of

P_{6}

so all the roots of

P_{3}

are roots of

P_{6}

.

Similarly we can express

P_{8}

in terms of previous polynomials in the sequence.

\begin{matrix} P_{8} & = {xP}_{7} - P_{6} \\ = (x^{2} - 1) P_{6} - {xP}_{5} \\ = (x^{3} - 2 x) P_{5} - (x^{2} - 1) P_{4} \\ = P_{4} (P_{5} - P_{3}) . \end{matrix}

This shows that

P_{4}

is a factor of

P_{8}

so all the roots of

P_{4}

are roots of

P_{8}

.

Again we can express

P_{10}

in terms of previous polynomials in the sequence.

\begin{matrix} P_{10} & = {xP}_{9} - P_{8} \\ = (x^{2} - 1) P_{8} - {xP}_{7} \\ = (x^{3} - 2 x) P_{7} - (x^{2} - 1) P_{6} \\ = (x^{4} - 3 x^{2} + 1) P_{6} - (x^{3} - 2 x) P_{5} \\ = P_{5} P_{6} - P_{4} P_{5} \\ = P_{5} (P_{6} - P_{4}) . \end{matrix}

This shows that

P_{5}

is a factor of

P_{10}

so all the roots of

P_{5}

are roots of

P_{10}

.

This suggests a conjecture that

P_{2 k} = P_{k} (P_{k + 1} - P_{k - 1})

where

k

is any natural number. This is true but the proof is beyond the scope of school mathematics.