## Pell identities: (P(n))^4+(P(n+1))^4+(P(n+2))^4 = 2((P(n))^2+2P(n)P(n+1)+4(P(n+1))^2)^2

Prove the following identities: $( \,P_{n})^4 + 16 \, (P_{n+1})^4 + (P_{n+2} \,)^4 \; = \; 2 \, ( \,(P_{n})^2 + 2 \, P_{n} \, P_{n+1} + 4 \, (P_{n+1})^2 \,)^2$ $( \,Q_{n})^4 + 16 \, (Q_{n+1})^4 + (Q_{n+2} \,)^4 \; = \; 2 \,( \,(Q_{n})^2 + 2 \, Q_{n} \, Q_{n+1} + 4 \, (Q_{n+1})^2 \,)^2$

Let $a$   and $b$   be any real numbers
Then $a^4 + b^4 + (a+b)^4 = 2 \,(a^2 + a \, b + b^2)^2$
can be confirmed algebraically.

This identity has an interesting application to the Pell family.

Let $\{ \, x_n \, \}$   be an integer sequence satisfying the Pell recurrence.

Suppose we let $a = x_n$   and $b = 2 \, x_{n+1}$

Then $( \,x_n)^4 + 16 \, (x_{n+1})^4 + (x_{n+2} \,)^4 = 2 \,( \,(x_n)^2 + 2 \, x_n \, x_{n+1} + 4 \,(x_{n+1})^2 \,)^2$

This gives the abovementioned Pell identities. math grad - Interest: Number theory
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### 2 Responses to Pell identities: (P(n))^4+(P(n+1))^4+(P(n+2))^4 = 2((P(n))^2+2P(n)P(n+1)+4(P(n+1))^2)^2

1. John McMahon says:

The RHS, 2 [ (Pn)^2 + 2(Pn)(Pn+1) + 4(Pn+1)^2 ]^2, when fully expanded, gives:

2(Pn)^4 + 8(Pn)^3(Pn+1) + 24(Pn)^2(Pn+1)^2 + 32(Pn)(Pn+1)^3 + 32(Pn+1)^4

Each term in the Pell Sequence is given by (Pn) = 2(Pn-1) + (Pn-2) , so therefore (Pn+1) = 2(Pn) + (Pn-1) , and (Pn+2) = 2(Pn+1) + (Pn) .

Substituting 2(Pn+1) + (Pn) for (Pn+2), the LHS, (Pn)^4 + 16(Pn+1)^4 + (Pn+2)^4 , can now be re-written, as

= (Pn)^4 + 16(Pn+1)^4 + (2Pn+1 + Pn)^4

= (Pn)^4 + 16(Pn+1)^4 + (Pn)^4 + 8(Pn)^3(Pn+1) + 24(Pn)^2(Pn+1)^2 + 32(Pn)(Pn+1)^3 + 16(Pn+1)^4

= 2(Pn)^4 + 8(Pn)^3(Pn+1) + 24(Pn)^2(Pn+1)^2 + 32(Pn)(Pn+1)^3 + 32(Pn+1)^4 = LHS

• benvitalis says: