INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A15

ON SUMS OF SQUARES OF PELL-LUCAS NUMBERS

Zvonko Čerin

University of Zagreb, Zagreb, Croatia, Europe

                               cerin@math.hr

Gian Mario Gianella

University of Torino, Torino, Italy, Europe

                           gianella@dm.unito.it

Received: 2/18/06, Revised: 4/10/06, Accepted: 4/24/06, Published: 5/15/06

Abstract

In this paper we prove several formulas for sums of squares of even Pell-Lucas numbers, sums of squares of odd Pell-Lucas numbers, and sums of products of even and odd Pell-Lucas numbers. These sums have nice representations as products of appropriate Pell and Pell-Lucas numbers with terms from certain integer sequences.

1. Introduction

The Pell and Pell-Lucas sequences P_{n} and Q_{n} are defined by the recurrence relations

P_{0} = 0,  P_{1} = 2,  P_{n} = 2 P_{n−1} + P_{n−2} for n ⩾ 2,

and

Q_{0} = 2,  Q_{1} = 2,  Q_{n} = 2 Q_{n−1} + Q_{n−2} for n ⩾ 2.

In the Sections 2–4 we consider sums of squares of odd and even terms of the Pell-Lucas sequence and sums of their products. These sums have nice representations as products of appropriate Pell and Pell-Lucas numbers.

The numbers Q_{k} make the integer sequence A002203 from [8] while the numbers \frac{1}{2} P_{k} make A000129. In this paper we shall also need the sequences A001109, A029546, A029547 and A077420 that we shorten to a_{k}, b_{k}, c_{k} and d_{k} with k ⩾ 0. For the convenience of the reader we shall now explicitly define these sequences.

The first ten terms of the sequence a_{k} are 0, 1, 6, 35, 204, 1189, 6930, 40391, 235416 and 1372105, it satisfies the recurrence relations a_{0} = 0, a_{1} = 1 and a_{n} = 6 a_{n−1} − a_{n−2} for n ≥ 2, and it is given by the formula a_{n} = \frac{(3+2 \sqrt{2})^{n}−(3−2 \sqrt{2})^{n}}{4 \sqrt{2}} .

The first seven terms of the sequence b_{k} are 1, 35, 1190, 40426, 1373295, 46651605 and 1584781276, it satisfies the recurrence relations b_{0} = 1, b_{1} = 35, b_{2} = 1190 and

b_{n} = 35 (b_{n−1} − b_{n−2}) − b_{n−3}

for n ≥ 3, and it is given by the formula b_{n} = \frac{1}{192} (f_{+} F_{+}^{n} + f_{−} F_{−}^{n} − 6), where f_{±} = 99 ± 70 \sqrt{2} and F_{±} = 17 ± 12 \sqrt{2}.

The first seven terms of the sequence c_{k} are 1, 34, 1155, 39236, 1332869, 45278310 and 1538129671, it satisfies the recurrence relations c_{0} = 1, c_{1} = 34 and c_{n} = 34 c_{n−1} − c_{n−2} for n ≥ 2, and it is given by the formula c_{n} = \frac{F_{+}^{n+1}−F_{−}^{n+1}}{F_{+}−F_{−}}.

Finally, the first seven terms of the sequence d_{k} are 1, 33, 1121, 38081, 1293633, 43945441 and 1492851361, it satisfies the recurrence relations d_{0} = 1, d_{1} = 33 and d_{n} = 34 d_{n−1} − d_{n−2} for n ≥ 2, and it is given by the formula d_{n} = \frac{3+2 \sqrt{2}}{6} F_{+}^{n} + \frac{3−2 \sqrt{2}}{6} F_{−}^{n}.

In the last three sections we look into the alternating sums of squares of odd and even terms of the Pell-Lucas sequence and the alternating sums of products of two consecutive Pell-Lucas numbers. These sums also have nice representations as products of appropriate Pell and Pell-Lucas numbers with terms from the above four integer sequences.

These formulas for ordinary sums and for alternating sums have been discovered with the help of a PC computer and all algebraic identities needed for the verification of our theorems can be easily checked in either Derive, Mathematica or Maple V. Running times of all these calculations are in the range of a few seconds.

Similar results for Fibonacci and Lucas numbers have recently been discovered by the first author in papers [1], [2], [3], and [4]. They improved some results in [7].

2. Pell-Lucas even squares

The following lemma is needed to accomplish the inductive step in the proof of our first theorem.

Lemma 1. For every m ⩾ 0 and k ⩾ 0 the following equality holds:

32 a_{m+1}^{2} + Q_{2k+2m+2}^{2} + a_{m+1} · Q_{2k} · Q_{2k+2m} = a_{m+2} · Q_{2k} · Q_{2k+2m+2}. (2.1)

Proof. Let α = 1 + \sqrt{2} and β = 1 − \sqrt{2}. Notice that α + β = 2 and α · β = −1 so that the real numbers α and β are solutions of the equation x^{2} − 2 x − 1 = 0. Since Q_{j} = α^{j} + β^{j} and

a_{j} = \frac{7α + β}{16} α^{2j−2} + \frac{α + 7β}{16} β^{2j−2}

for every j ⩾ 0, the difference of the right hand side and the left hand side of the relation (2.1) (after the substitutions β = −\frac{1}{α} and α = 1 + \sqrt{2} and the replacement of 99 + 70\sqrt{2}, 17 + 12\sqrt{2}, 7 + 5\sqrt{2}, 3 + 2\sqrt{2}, 1 + \sqrt{2}, and −1 + \sqrt{2} with α^{6}, α^{4}, α^{3}, α^{2}, α and \frac{1}{α}) is equal to \frac{M}{16 α^{4}} , where M = p_{0} α^{4k+4m} + p_{−} α^{4k} with p_{0} = p_{+} − 9232 − 6528 \sqrt{2} and

p_{±} = 3 α^{8} + 17 α^{7} + 30 α^{6} ± 14 α^{5} − 5 α^{4} ∓ 3 α^{3}.

Since both polynomials p_{0} and p_{−} contain α − 1 − \sqrt{2} as a factor we conclude that M = 0 and the proof is complete.

Theorem 1. For every m ⩾ 0 and k ⩾ 0 the following equality holds:

α_{m} + ∑_{i=0}^{m} Q_{2k+2i}^{2} = \begin{cases}Q_{4 k}, & if m = 0; \\ a_{m+1} · Q_{2k} · Q_{2k+2m}, & if m ⩾ 1, \end{cases} (2.2)

where the sequence α_{m} is defined as follows: α_{0} = −2, α_{1} = 32, and α_{m} − α_{m−1} = 32 · a_{m}^{2} for m ⩾ 2 (i. e., α_{m} = 32 · (∑_{j=1}^{m} a_{j}^{2})).

Proof. When m = 0 we obtain

Q_{2k}^{2} − Q_{4k} − 2 = (α^{2 k} + β^{2 k})^{2} − α^{4 k} − β^{4 k} − 2 = 2 (α^{k})^{2} (β^{k})^{2} − 2 = 0.

For m ⩾ 1 the proof is by induction on m. When m = 1 the relation (2.2) is

32 + Q_{2k}^{2} + Q_{2k+2}^{2} − 6 · Q_{2k} · Q_{2k+2} = 0

which is true since its left hand side is (α^{2} − 2 α − 1)(α^{2} + 2 α − 1)(α^{4k} + α^{−4k−4} − 6 α^{−2}).

Assume that the relation (2.2) is true for m = r. Then

α_{r+1} + ∑_{i=0}^{r+1} Q_{2k+2i}^{2} = 32 a_{r+1}^{2} + α_{r} + Q_{2k+2r+2}^{2} + ∑_{i=0}^{r} Q_{2k+2i}^{2} =

32 a_{r+1}^{2} + Q_{2k+2r+2}^{2} + a_{r+1} · Q_{2k} · Q_{2k+2r} = a_{r+2} · Q_{2k} · Q_{2k+2r+2},

where the last step uses Lemma 1 for m = r + 1. Hence, (2.2) is true for m = r + 1 and the proof is completed.

Remark 1. The following are three additional versions of Theorem 1:

For every j ⩾ 0 and k ⩾ 1 the following equalities holds:

α_{j+1} + ∑_{i=0}^{j} Q_{2k+2i}^{2} = a_{j+1} · Q_{2k−2} · Q_{2k+2j+2}, (2.3)

∑_{i=0}^{j} Q_{2k+2i}^{2} = γ_{j} + a_{j+1} · Q_{2k−1} · Q_{2k+2j+1}, (2.4)

∑_{i=0}^{j} Q_{2k+2i}^{2} = δ_{j} + a_{j+1} · Q_{2k+1} · Q_{2k+2j−1}, (2.5)

where γ_{0} = δ_{0} = 8 and γ_{j} − γ_{j−1} = 2 · P_{2 j+1}^{2} and δ_{j} − δ_{j−1} = 2 · P_{2 j−1}^{2} for every j ⩾ 1.

3. Pell-Lucas odd squares

The initial step in an inductive proof of our second theorem uses the following lemma. Lemma 2. For every k ⩾ 1 the following identity holds:

Q_{2k+1}^{2} − Q_{2k−1} Q_{2k+3} = 32. (3.1)

Proof. By the Binet formula Q_{n} = α^{n} + β^{n} so that we have

Q_{2k+1}^{2} − Q_{2k−1} Q_{2k+3} =

(α^{2k+1} + β^{2k+1})^{2} − (α^{2k−1} + β^{2k−1})(α^{2k+3} + β^{2k+3}) = 2 (α · β)^{2 k+1} − (α · β)^{2 k−1}(α^{4} + β^{4}) = −2 + Q_{4} = −2 + 34 = 32.

The following lemma is needed to accomplish the inductive step in the proof of our second theorem. Lemma 3. For every j ⩾ 0 and k ⩾ 1 the following equality holds:

Q_{2k+2j+3}^{2} + Q_{2k−1}(a_{j+1} Q_{2k+2j+3} − a_{j+2} Q_{2k+2j+5}) = 32 a_{j+2}^{2}. (3.2)

Proof. The difference of the left hand side and the right hand side of (3.2) (after the substitutions β = −\frac{1}{α} and α = 1 + \sqrt{2} and the replacement of 99 + 70\sqrt{2}, 17 + 12\sqrt{2}, 7 + 5\sqrt{2}, 3 + 2\sqrt{2}, 1 + \sqrt{2}, and −1 + \sqrt{2} with α^{6}, α^{4}, α^{3}, α^{2}, α and \frac{1}{α}) is equal to \frac{M}{16 α^{8}} , where

M = p_{0} α^{4k+4j} + p_{−} α^{4k}

with p_{0} = p_{+} − 1827888 − 1292512 \sqrt{2} and

p_{±} = 3 α^{14} ± 17 α^{13} + 30 α^{12} ± 14 α^{11} − 5 α^{10} ∓ 3 α^{9}.

Since both polynomials p_{0} and p_{−} contain α − 1 − \sqrt{2} as a factor we conclude that M = 0 and the proof is complete. Theorem 2. For every j ⩾ 0 and k ⩾ 1 the following equality holds:

∑_{i=0}^{j} Q_{2k+2i+1}^{2} = α_{j+1} + a_{j+1} · Q_{2k−1} · Q_{2k+2j+3}. (3.3)

Proof. The proof is by induction on j. For j = 0 the relation (3.3) is

Q_{2k+1}^{2} = 32 + Q_{2k−1} Q_{2k+3}

which is true by the relation (3.1) in Lemma 2. Assume that the relation (3.3) is true for j = r. Then

∑_{i=0}^{r+1} Q_{2k+2i+1}^{2} = Q_{2k+2r+3}^{2} + ∑_{i=0}^{r} Q_{2k+2i+1} =

Q_{2k+2r+3}^{2} + α_{r+1} + a_{r+1} · Q_{2k−1} · Q_{2k+2r+3} =

α_{r+1} + 32 a_{r+2}^{2} + a_{r+2} · Q_{2k−1} · Q_{2k+2r+5} =

α_{(r+1)+1} + a_{r+2} · Q_{2k−1} · Q_{2k+2(r+1)+3},

where the third step uses Lemma 3. Hence, (3.3) is true also for j = r + 1 and the proof is complete.

Another version of Theorem 2 is the following statement:

Theorem 3. For every j ⩾ 0 and k ⩾ 1 the following equality holds:

β_{j} + ∑_{i=0}^{j} Q_{2k+2i+1}^{2} = a_{j+1} · Q_{2k−2} · Q_{2k+2j+4}, (3.4)

with β_{0} = 200 and β_{j} − β_{j−1} = 2 · P_{2j+3}^{2} for j ⩾ 1.

Proof. We shall only outline the key steps in an inductive proof leaving the details to the reader because they are analogous to the proof of Theorem 2.

The initial step is the equality 200 + Q_{2k+1}^{2} = Q_{2k−2} Q_{2k+4} which holds for every k ⩾ 1. On the other hand, the inductive step is realized with the following equality:

2 P_{2r+5}^{2} + Q_{2k+2r+3}^{2} + a_{r+1} Q_{2k−2} Q_{2k+2r+4} = a_{r+2} Q_{2k−2} Q_{2k+2r+6},

which holds for every k ⩾ 1 and every r ⩾ 1.

4. Pell-Lucas products

For the first two steps in a proof by induction of our next theorem we require the following lemma.

Lemma 4. For every k ⩾ 1 the following equalities hold:

Q_{2k} Q_{2k+1} = Q_{4k+1} + 2. (4.1)
Q_{2k} Q_{2k+1} + Q_{2k+2} Q_{2k+3} = 6 Q_{2k} Q_{2k+3} − 80. (4.2)

Proof of (4.1). By the Binet formula we have

Q_{2k} Q_{2k+1} = (α^{2k} + β^{2k}) (α^{2k+1} + β^{2k+1}) =

α^{4k+1} + (α · β)^{2k} (α + β) + β^{4k+1} = Q_{4k+1} + 2.

Proof of (4.2). When we apply the Binet formula to the terms in the difference of the left hand side and the right hand side of (4.2) we get

80 + α^{4 k+1} + α (α β)^{2 k} + β (α β)^{2 k} + β^{4 k+1} + α^{4 k+5} + α^{3}β^{2} (α β)^{2 k} +

β^{3}α^{2} (α β)^{2 k} + β^{4 k+5} − 6α^{4 k+3} − 6α^{3} (α β)^{2 k} − 6β^{3} (α β)^{2 k} − 6β^{4 k+3}.

Since α β = −1, this simplifies to p(α) α^{4k} + p(β) β^{4k} + q, where p(x) = x^{5} − 6 x^{3} + x and q = α^{3} β^{2} + α^{2} β^{3} − 6 α^{3} − 6 β^{3} + α + β + 80. Now it is easy to check (by the substitutions α = 1 + \sqrt{2} and β = 1 − \sqrt{2}) that p(α) = 0, p(β) = 0 and q = 0. This implies that the relation (4.2) holds.

With the following lemma we shall make the inductive step in the proof of the third theorem.

Lemma 5. For every r ⩾ 1 and k ⩾ 1 the following equality holds:

Q_{2k}(a_{r+2} Q_{2k+2r+3} − a_{r+1} Q_{2k+2r+1}) − Q_{2k+2r+2} Q_{2k+2r+3} = 2 P_{2r+2} P_{2r+3}.

Proof. Let R denote the difference of the left hand side and the right hand side of the above relation. We need to show that R = 0.

Let A = α^{2k}, B = β^{2k}, U = β^{2r}, V = β^{2r}, u = \frac{7α+β}{16} and v = \frac{α+7β}{16} . Note that a_{r+1} is equal to u U + v V. By the Binet formula we get

R = (A + B) [(u α^{2} U + v β^{2} V)(α^{3} A U + β^{3} B V) − (u U + v V) (α A U + β B V)] − (α^{2} A U + β^{2} B V)(α^{3} A U + β^{3} B V) −

(α^{2} U + β^{2} V)(α^{3} U + β^{3} V) = ℓ_{+} [u A B + \frac{(8 u − m) A^{2} − m}{8}] U^{2}

− 2 (A B − 1) U V + (ℓ_{−} v A B − n) V^{2},

where ℓ_{±} = 4(10 ± 7 \sqrt{2}), m = 4 + 3 \sqrt{2} and n = 41 − 29 \sqrt{2}. Since A B = 1, u = \frac{m}{8} and ℓ_{−} v = n it follows that R = 0.

Theorem 4. For every j ⩾ 0 and k ⩾ 1 the following equality holds:

α_{j} + ∑_{i=0}^{j} Q_{2k+2i} · Q_{2k+2i+1} = \begin{cases}Q_{4 k}, & if j = 0; \\ a_{j+1} · Q_{2k} · Q_{2k+2j+1}, & if j ⩾ 1, \end{cases} (4.3)

where α_{0} = −2, α_{1} = 80, and α_{j+1} − α_{j} = 2 · P_{2j+2} · P_{2j+3} for j ⩾ 1.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A15 7

Proof. The proof is by induction on j. For j = 0 the relation (4.3) is

Q_{2k} Q_{2k+1} − 2 = Q_{4k+1}

which is true by (4.1). For j = 1 the relation (4.3) is

80 + Q_{2k} Q_{2k+1} + Q_{2k+2} Q_{2k+3} = 6 Q_{2k} Q_{2k+3}

which is true by (4.2).

Assume that the relation (4.3) is true for j = r. Then

α_{r+1} + ∑_{i=0}^{r+1} Q_{2k+2i} Q_{2k+2i+1} = α_{r} + 2 P_{2r+2} P_{2r+3}

+ ∑_{i=0}^{r} Q_{2k+2i} Q_{2k+2i+1} + Q_{2k+2r+2} Q_{2k+2r+3} = a_{r+1} Q_{2k} Q_{2k+2r+1}

+ 2 P_{2r+2} P_{2r+3} + Q_{2k+2r+2} Q_{2k+2r+3} = a_{r+2} Q_{2k} Q_{2k+2(r+1)+1},

where the last step uses Lemma 5. Hence, (4.3) is true also for j = r + 1.

5. Alternating Pell-Lucas even squares

In this section we look for formulas that give closed forms for alternating sums of squares of Pell-Lucas numbers with even indices. Lemma 6. For every k ⩾ 0 we have

Q_{2k}^{2} = 8 + Q_{2k+1} · Q_{2k−1}. (5.1)

Proof. By the Binet formula we get

8 + Q_{2k−1} Q_{2k+1} = 8 + (α^{2k−1} + β^{2k−1})(α^{2k+1} + β^{2k+1}) = 8 + α^{4k} + β^{2} (α β)^{2k−1} + α^{2} (α β)^{2k−1} + β^{4k} = α^{4k} + 2 + β^{4k} = (α^{2k} + β^{2k})^{2} = Q_{2k}^{2}.

Lemma 7. For every k ⩾ 0 we have

Q_{2k}^{2} − Q_{2k+2}^{2} = 32 · (1 − a_{k+1} Q_{2k}). (5.2)

Proof. For real numbers a, b and c let 𝔭_{b}^{a}, 𝔮_{a} and 𝔯_{c}^{a;b} denote \frac{α^{a}−β^{a}}{α^{b}−β^{b}}, α^{a} + β^{a} and \frac{a α+b β}{c}. The difference of the left hand side and the right hand side in (5.2) is

(𝔮_{2k})^{2} − (𝔮_{2k+2})^{2} − 32 + 32 (𝔯_{16}^{7;1} α^{2k} + 𝔯_{16}^{1;7} β^{2k}) 𝔮_{2k}.

Let α^{2k} = A and β^{2k} = B. The above expression reduces to p(α) A^{2} + q A B + p(β) B^{2} − 32, where p(α) = 1 + 14 α + 2 β − α^{4} and q = 2 + 16 α + 16 β − 2 α^{2} β^{2}. If we replace α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get p(α) = p(β) = 0 and q = 32. Hence, the difference is 32 A B − 32 = 0 because A B = 1.

Lemma 8. For every k ⩾ 0 and every r ⩾ 0 we have

32 · c_{r} · (1 − a_{k+r+1} Q_{2k+2r}) + Q_{2k+4r+4}^{2} − Q_{2k+4r+6}^{2} (5.3)

= 32 · c_{r+1} · (1 − a_{k+r+2} Q_{2k+2r+2}).

Proof. Notice that c_{r} = 𝔭_{4}^{4r+4} for every r ⩾ 0. The difference of the left hand side and the right hand side in (5.3) is

32 · 𝔭_{4}^{4r+4} [1 − (𝔯_{16}^{7;1} α^{2k+2r} + 𝔯_{16}^{1;7} β^{2k+2r}) 𝔮_{2k+2r}] + (𝔮_{2k+4r+4})^{2} −

(𝔮_{2k+4r+6})^{2} − 32 · 𝔭_{4}^{4r+8} [1 − (𝔯_{16}^{7;1} α^{2k+2r+2} + 𝔯_{16}^{1;7} β^{2k+2r+2}) 𝔮_{2k+2r+2}] .

Let α^{2k} = A, β^{2k} = B, α^{2r} = U and β^{2r} = V. The above expression reduces to \frac{M}{α^{4}−β^{4}} , where M is

p(α, β) A^{2} U^{4} − q(α, β) A^{2} U^{2} V^{2} + r(α, β) A B U^{3} V + 2 s A B U^{2} V^{2} − p(β, α)B^{2} V^{4} + q(β, α)B^{2} U^{2} V^{2} − r(β, α)A B U V^{3} − 32(t(α)U^{2} − t(β)V^{2}),

where

p(α, β) = t(α)(α^{4} β^{4} − α^{8} + 14 α^{5} + 2 α^{4} β + 14 α + 2 β),
q(α, β) = 2 β^{4}(7 α + β)(α^{4} β^{4} − 1),
r(α, β) = 16 α^{4}(α + β)(α^{6} β^{2} − 1),

s = α^{4} β^{4}(α^{4} − β^{4})(α^{2} β^{2} − 1) and t(α) = α^{4} (α^{4} − 1). Replacing α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get

M = u_{+} A B U^{3} V − u_{−} A B U V^{3} − u_{+} U^{2} + u_{−} V^{2},

where u_{±} = 128(140 ± 99 \sqrt{2}). Hence, M = 0 so that the difference is zero because A B = 1 and U V = 1.

Lemma 9. For every k ⩾ 0 and every r ⩾ 0 we have

d_{r} · Q_{2k+2r}^{2} − 64 b_{r−1} − Q_{2k+4r+2}^{2} + Q_{2k+4r+4}^{2} = d_{r+1} · Q_{2k+2r+2}^{2} − 64 b_{r}.

Proof. Notice that

b_{r} = 𝔯_{384}^{169;29} α^{4r} + 𝔯_{384}^{29;169} β^{4r} − \frac{1}{32}

and

d_{r} = 𝔯_{12}^{5;1} α^{4r} + 𝔯_{12}^{1;5} β^{4r},

for every r ⩾ 0. The difference of the left hand side and the right hand side is

(𝔯_{12}^{5;1} α^{4r} + 𝔯_{12}^{1;5} β^{4r}) (𝔮_{2k+2r})^{2} − 𝔯_{6}^{169;29} α^{4r−4} − 𝔯_{6}^{29;169} β^{4r−4} − (𝔮_{2k+4r+2})^{2} +

(𝔮_{2k+4r+4})^{2} − (𝔯_{12}^{5;1} α^{4r+4} + 𝔯_{12}^{1;5} β^{4r+4}) · (𝔮_{2k+2r+2})^{2} + 𝔯_{6}^{169;29} α^{4r} − 𝔯_{6}^{29;169} β^{4r}.

Let α^{2k} = A, β^{2k} = B, α^{2r} = U and β^{2r} = V. The above expression reduces to \frac{−M}{12 α^{4} β^{4}} , where M is

p(α, β)A^{2}U^{4} + q(α, β)(AUV)^{2} + r(α, β)ABU^{3}V − 24 s AB(UV)^{2} + p(β, α)B^{2}V^{4} + q(β, α)(BUV)^{2} + r(β, α)ABUV^{3} − t(α, β)U^{2} − t(β, α)V^{2},

with P = (α β)^{4},

p(α, β) = P (α^{4} − 1)(5 α^{5} + α^{4} β − 12 α^{4} + 5 α + β),
q(α, β) = P (P − 1)(α + 5 β),
r(α, β) = 2 P (α^{6} β^{2} − 1)(5 α + β),

s = P (P − α^{2} β^{2}) and t(α, β) = 2 (P − β^{4})(169 α + 29 β). Replacing α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get

M = u_{+} A B U^{3} V + u_{−} A B U V^{3} − u_{+} U^{2} − u_{−} V^{2},

where u_{±} = 16(24 ± 17 \sqrt{2}). Hence, M = 0 so that the difference is zero because A B = 1 and U V = 1.

Theorem 5. a) For every m ⩾ 0 and k ⩾ 0 the following equality holds:

∑_{i=0}^{2 m+1} (−1)^{i} · Q_{2k+2i}^{2} = 32 · c_{m} · (1 − a_{k+m+1} · Q_{2k+2m}), (5.4)

b) For every m ⩾ 1 and k ⩾ 0 the following equality holds:

∑_{i=0}^{2 m} (−1)^{i} · Q_{2k+2i}^{2} = d_{m} · Q_{2k+2m}^{2} − 64 · b_{m−1}. (5.5)

Proof of a). The proof is by induction on m. For m = 0 the relation (5.4) is

Q_{2k}^{2} − Q_{2k+2}^{2} = 32 · (1 − a_{k+1} Q_{2k})

(i. e., the relation (5.2)) which is true by Lemma 7.

Assume that the relation (5.4) is true for m = r. Then

∑_{i=0}^{2(r+1)+1} (−1)^{i} · Q_{2k+2i}^{2} = ∑_{i=0}^{2r+1} (−1)^{i} · Q_{2k+2i}^{2} + Q_{2k+4r+4}^{2} − Q_{2k+4r+6}^{2}

= 32 · c_{r} · (1 − a_{k+r+1} · Q_{2k+2r}) + Q_{2k+4r+4}^{2} − Q_{2k+4r+6}^{2}

= 32 · c_{r+1} · (1 − a_{k+(r+1)+1} · Q_{2k+2(r+1)}), where the last step uses Lemma 8. Hence, (5.4) is true for m = r + 1 and the proof is completed.

Proof of b). The proof is by induction on m. For m = 0 the relation (5.5) is

Q_{2k}^{2} = d_{0} Q_{2k}^{2} − 64 b_{−1}

which is true since d_{0} = 1 and b_{−1} = 0.

Assume that the relation (5.5) is true for m = r. Then

∑_{i=0}^{2(r+1)} (−1)^{i} · Q_{2k+2i}^{2} = ∑_{i=0}^{2r} (−1)^{i} · Q_{2k+2i}^{2} − Q_{2k+4r+2}^{2} + Q_{2k+4r+4}^{2} =

d_{r} · Q_{2k+2r}^{2} − 64 b_{r−1} − Q_{2k+4r+2}^{2} + Q_{2k+4r+6}^{2} = d_{r+1} · Q_{2k+2r+2}^{2} − 64 b_{r},

where the last step uses Lemma 9. Hence, (5.5) is true for m = r + 1 and the proof is completed.

6. Alternating Pell-Lucas odd squares

Lemma 10. For every k ⩾ 0 we have

Q_{2k+3}^{2} − Q_{2k+1}^{2} = 32 · a_{k+1} · Q_{2k+2}. (6.1)

Proof. By the Binet formulas the difference of the left hand side and the right hand side in (6.1) is

Q_{2k+3}^{2} − Q_{2k+1}^{2} − 32 a_{k+1} Q_{2k+2} = (α^{2k+1} + β^{2k+1})^{2} − (α^{2k+3} + β^{2k+3})^{2}

+ 32 (\frac{7α + β}{16} α^{2k} + \frac{α + 7β}{16} β^{2k}) (α^{2k+2} + β^{2k+2}),

Let α^{2k} = A and β^{2k} = B. The above expression reduces to

p(α)A^{2} − q A B + p(β)B^{2},

where q = 2 α^{3}β^{3} − 2 α β − 2 α^{3} − 14 β^{2}α − 14 α^{2}β − 2 β^{3} and p(α) = α^{2}(14α − α^{4} + 2β + 1). It is easy to check (by substitutions α = 1 + \sqrt{2} and β = 1 − \sqrt{2}) that p(α) = 0, p(β) = 0 and q = 0. This implies that the relation (6.1) is true.

Lemma 11. For every k ⩾ 0 we have

8 + Q_{2k+1}^{2} = Q_{2k} · Q_{2k+2}. (6.2)

Proof. Using the Binet representation of Pell-Lucas numbers we get

8 + Q_{2k+1}^{2} − Q_{2k}Q_{2k+2} = 8 + (α^{2k+1} + β^{2k+1})^{2} − (α^{2k} + β^{2k})(α^{2k+2} + β^{2k+2})

= 8 + 2 (α β)^{2k+1} − (α β)^{2k}(α^{2} + β^{2}) = 8 − 2 − 6 = 0,

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A15 11

because α β = −1 and α^{2} + β^{2} = 6. Hence, 8 + Q_{2k+1}^{2} = Q_{2k}Q_{2k+2}. Lemma 12. For all k ⩾ 0 and r ⩾ 0 we have L = 0, where L is

Q_{2k+4r+5}^{2} − Q_{2k+4r+7}^{2} − 32 c_{r} · a_{k+r+1} · Q_{2k+2r+2} + 32 c_{r+1} · a_{k+r+2} · Q_{2k+2r+4}.

Proof. The expression L is in fact

(𝔮_{2k+4r+5})^{2} − (𝔮_{2k+4r+7})^{2} − 32 𝔭_{4}^{4r+4} (𝔯_{16}^{7;1}α^{2k+2r} + 𝔯_{16}^{1;7}β^{2k+2r}) 𝔮_{2k+2r+2} + 32 𝔭_{4}^{4r+8} (𝔯_{16}^{7;1}α^{2k+2r+2} + 𝔯_{16}^{1;7}β^{2k+2r+2}) 𝔮_{2k+2r+4}.

Let α^{2k} = A, β^{2k} = B, α^{2r} = U and β^{2r} = V. The above expression reduces to \frac{M}{α^{4}−β^{4}} , where M is

p(α, β)A^{2}U^{4} + 2 q(α, β)(AUV)^{2} + 2 r(α, β)ABU^{3}V −

2 s AB(UV)^{2} − p(β, α)B^{2}V^{4} + 2 q(β, α)(BUV)^{2} − 2 r(β, α)ABUV^{3},

with

p(α, β) = α^{6} (α^{4} − 1)(α^{4} β^{4} − α^{8} + 14 α^{5} + 2 α^{4} β + 14 α + 2 β),
q(α, β) = α^{2}β^{4}((αβ)^{4} − 1)(7α + β),
r(α, β) = α^{4}(α + β)(α^{6} β^{2} − 1)(α^{2} + 6αβ + β^{2})

and s = α^{5}β^{5}((αβ)^{2} − 1)(α^{4} − β^{4}). Replacing α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we see easily that all coefficients of the above polynomial vanish so that L = 0. Lemma 13. For every k ⩾ 0 and every r ⩾ 0 we have

2(P_{2r+3}^{2} − P_{2r+1}^{2}) + d_{r} · Q_{2k+2r} · Q_{2k+2r+2} − Q_{2k+4r+3}^{2} + Q_{2k+4r+5}^{2} = d_{r+1} · Q_{2k+2r+2} · Q_{2k+2r+4}.

Proof. The difference of the left hand side and the right hand side is

8 [(𝔭_{1}^{2r+3})^{2} − (𝔭_{1}^{2r+1})^{2}] + (𝔯_{12}^{5;1} α^{4r} + 𝔯_{12}^{1;5} β^{4r}) 𝔮_{2k+2r}𝔮_{2k+2r+2} − (𝔮_{2k+4r+3})^{2} + (𝔮_{2k+4r+5})^{2} − (𝔯_{12}^{5;1} α^{4r+4} + 𝔯_{12}^{1;5} β^{4r+4}) 𝔮_{2k+2r+2}𝔮_{2k+2r+4}.

Let α^{2k} = A, β^{2k} = B, α^{2r} = U and β^{2r} = V. The above expression reduces to \frac{−M}{12 (α−β)^{2}} , where M is

(α − β)^{2} [p(α, β)A^{2}U^{4} + q(α, β)(AUV)^{2} + r(α, β)ABU^{3}V −

24 s AB(UV)^{2} + p(β, α)B^{2}V^{4} + q(β, α)(BUV)^{2} + r(β, α)ABUV^{3}]

− 96 [t(α)U^{2} − 2wUV + t(β)V^{2}] ,

with t(α) = α^{2}(α^{4} − 1),

p(α, β) = t(α)(5 α^{5} + α^{4} β − 12 α^{4} + 5 α + β),

q(α, β) = t(αβ)(α + 5 β)/β^{2},
r(α, β) = (α^{2} + β^{2})(α^{6} β^{2} − 1)(5 α + β),

s = α^{3} β^{3}((αβ)^{2} − 1) and w = αβ((αβ)^{2} − 1). Replacing α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get

M = u_{+} A B U^{3} V + u_{−} A B U V^{3} − u_{+} U^{2} − u_{−} V^{2},

where u_{±} = 384(24 ± 17 \sqrt{2}). Hence, M = 0 so that the difference is zero since A B = 1 and U V = 1.

Theorem 6. a) For every j ⩾ 0 and k ⩾ 0 the following equality holds:

∑_{i=0}^{2 j+1} (−1)^{i} · Q_{2k+2i+1}^{2} = −32 · c_{j} · a_{k+j+1} · Q_{2k+2j+2}, (6.3)

b) For every j ⩾ 1 and k ⩾ 1 the following equality holds:

2 · P_{2j+1}^{2} + ∑_{i=0}^{2 j} (−1)^{i} · Q_{2k+2i+1}^{2} = d_{j} · Q_{2k+2j} · Q_{2k+2j+2}. (6.4)

Proof of a). The proof is by induction on j. For j = 0 the relation (6.3) is

Q_{2k+3}^{2} − Q_{2k+1}^{2} = 32a_{k+1}Q_{2k+2}

(i. e., the relation (6.1)) which is true by Lemma 10.

Assume that the relation (6.3) is true for j = r. Then

∑_{i=0}^{2(r+1)+1} (−1)^{i} Q_{2k+2i+1}^{2} = ∑_{i=0}^{2r+1} (−1)^{i} Q_{2k+2i+1}^{2} + Q_{2k+4r+5}^{2} − Q_{2k+4r+7}^{2}

= −32 c_{r} a_{k+r+1} Q_{2k+2r+2} + Q_{2k+4r+5}^{2} − Q_{2k+4r+7}^{2}

= −32 c_{r+1} a_{k+r+2} Q_{2k+2r+4},

where the last step uses Lemma 12. Hence, (6.3) is true for j = r + 1 and the proof is completed.

Proof of b). The proof is again by induction on j. For j = 0 the relation (6.4) is

2 P_{1}^{2} + Q_{2k+1}^{2} = d_{0} Q_{2k} Q_{2k+2}

which is true by Lemma 11 since d_{0} = 1 and P_{1} = 2.

Assume that the relation (6.4) is true for j = r. Then

2 P_{2r+3}^{2} + ∑_{i=0}^{2(r+1)} (−1)^{i} Q_{2k+2i+1}^{2} = 2(P_{2r+3}^{2} − P_{2r+1}^{2})

+ (2 P_{2r+1}^{2} + ∑_{i=0}^{2r} (−1)^{i} Q_{2k+2i+1}^{2}) − Q_{2k+4r+3}^{2} + Q_{2k+4r+5}^{2} =

2(P_{2r+3}^{2} − P_{2r+1}^{2}) + d_{r} Q_{2k+2r} Q_{2k+2r+2} − Q_{2k+4r+3}^{2} + Q_{2k+4r+5}^{2} = d_{r+1} Q_{2k+2r+2} Q_{2k+2r+4},

where the last step uses Lemma 13. Hence, (6.4) is true for m = r + 1 and the proof is completed.

7. Alternating Pell-Lucas products

Lemma 14. For every k ⩾ 0 we have

Q_{2k} Q_{2k+1} − Q_{2k+2} Q_{2k+3} + 8 P_{2k+3} Q_{2k} = 80. (7.1)

Proof. The difference of the left hand side and the right hand side in (7.1) is

𝔮_{2k} 𝔮_{2k+1} − 𝔮_{2k+2} 𝔮_{2k+3} + 16 𝔭_{1}^{2k+3} 𝔮_{2k} − 80.

Let α^{2k} = A and β^{2k} = B. The above expression reduces to \frac{M}{α−β} where M is

p(α) A^{2} − q A B − p(β) B^{2} − 80(α − β),

with p(α) = α(β α^{4} − α^{5} + 16 α^{2} + α − β) and

q = (α − β)(α^{3} β^{2} + α^{2} β^{3} − 16 α^{2} − 16 α β − 16 β^{2} − α − β).

If we replace α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get p(α) = p(β) = 0 and q = −160 \sqrt{2} and 80(α − β) = 160 \sqrt{2}. Hence, M = 0 because A B = 1.

Lemma 15. For all k ⩾ 0 and r ⩾ 0 we have

8 c_{r+1} · (P_{2k+2r+5} · Q_{2k+2r+2} − 10) + Q_{2k+4r+4} · Q_{2k+4r+5} = 8 c_{r} · (P_{2k+2r+3} · Q_{2k+2r} − 10) + Q_{2k+4r+6} · Q_{2k+4r+7}.

Proof. The difference L of the left hand side and the right hand side is

8 𝔭_{4}^{4r+8} [2 𝔭_{1}^{2k+2r+5} 𝔮_{2k+2r+2} − 10] + 𝔮_{2k+4r+4} 𝔮_{2k+4r+5}

− 𝔮_{2k+4r+6} 𝔮_{2k+4r+7} − 8 𝔭_{4}^{4r+4} [2 𝔭_{1}^{2k+2r+3} 𝔮_{2k+2r} − 10]

Let α^{2k} = A, β^{2k} = B, α^{2r} = U and β^{2r} = V. The above expression reduces to

\frac{M}{(α + β)(α − β)^{2}(α^{2} + β^{2})},

14 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A15

where M is

p(α, β)A^{2}U^{4} − 16 q(α, β)(AUV)^{2} + 16 r(α, β)ABU^{3}V − s AB(UV)^{2} +

p(β, α)B^{2}V^{4} − 16 q(β, α)(BUV)^{2} − 16 r(β, α)ABUV^{3} − t(α)U^{2} + t(β)V^{2},

with

p(α, β) = α^{6} (α^{4} − 1)(α^{6}β − α^{7} + α^{3}β^{4} − α^{2}β^{5} + 16 α^{4} + 16),
q(α, β) = α^{3}β^{4}((αβ)^{4} − 1),
s = α^{4}β^{4}((αβ)^{2} − 1)(α^{2} + β^{2})(α^{2} − β^{2})^{2},
r(α, β) = α^{4}(α − β)(α^{6} β^{2} − 1)(α^{2} + αβ + β^{2})

and t(α) = 80 α^{4}(α − β) (α^{4} − 1). Replacing α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get

M = u_{+} A B U^{3} V + u_{−} A B U V^{3} − u_{+} U^{2} − u_{−} V^{2},

where u_{±} = 1280(99 ± 70 \sqrt{2}). Hence, M = 0 so that the difference is zero since A B = 1 and U V = 1. Lemma 16. For all k ⩾ 0 and r ⩾ 0 we have

64 c_{r} + d_{r} · Q_{2k+2r} · Q_{2k+2r+1} + Q_{2k+4r+4} · Q_{2k+4r+5} = d_{r+1} · Q_{2k+2r+2} · Q_{2k+2r+3} + Q_{2k+4r+2} · Q_{2k+4r+3}.

Proof. The difference L of the left hand side and the right hand side is

64 𝔭_{4}^{4r+4} + (𝔯_{12}^{5;1} α^{4r} + 𝔯_{12}^{1;5} β^{4r}) 𝔮_{2k+2r} 𝔮_{2k+2r+1} − 𝔮_{2k+4r+2} 𝔮_{2k+4r+3} + 𝔮_{2k+4r+4} 𝔮_{2k+4r+5} − (𝔯_{12}^{5;1} α^{4r+4} + 𝔯_{12}^{1;5} β^{4r+4}) 𝔮_{2k+2r+2} 𝔮_{2k+2r+3}.

Let α^{2k} = A, β^{2k} = B, α^{2r} = U and β^{2r} = V. The above expression reduces to \frac{M}{12(α^{4}−β^{4})}, where M is

p(α, β)A^{2}U^{4} − q(α, β)(AUV)^{2} + r(α, β)ABU^{3}V + 12 s AB(UV)^{2} −

p(β, α)B^{2}V^{4} − q(β, α)(BUV)^{2} − r(β, α)ABUV^{3} + t(α)U^{2} − t(β)V^{2},

with

p(α, β) = α (α^{4} − 1)(β^{4} − α^{4})(5α^{5} + α^{4}β − 12α^{4} + 5 α + β),
s = α^{2}β^{2}((αβ)^{2} − 1)(β^{4} − α^{4})(α + β),
r(α, β) = (β^{4} − α^{4})(α + β)(5α + β)(α^{6} β^{2} − 1),
q(α, β) = α((αβ)^{4} − 1)(β^{4} − α^{4})(α + 5β)

and t(α) = 768 α^{4}. Replacing α and β with 1 + \sqrt{2} and 1 − \sqrt{2} we get M = −u_{+} A B U^{3} V + u_{−} A B U V^{3} + u_{+} U^{2} − u_{−} V^{2}, where u_{±} = 768(17 ± 12 \sqrt{2}). Hence, M = 0 so that the difference is zero since A B = 1 and U V = 1.

Theorem 7. a) For every j ⩾ 0 and k ⩾ 0 the following equality holds:

∑_{i=0}^{2 j+1} (−1)^{i}·Q_{2k+2i}·Q_{2k+2i+1} = −8·c_{j} ·(P_{2k+2j+3}·Q_{2k+2j} −10), (7.2)

b) For every j ⩾ 0 and k ⩾ 0 the following equality holds:

β_{j} + ∑_{i=0}^{2 j} (−1)^{i} · Q_{2k+2i} · Q_{2k+2i+1} = d_{j} · Q_{2k+2j} · Q_{2k+2j+1}, (7.3)

where the sequence β_{j} is determined by the conditions: β_{0} = 0 and β_{j+1} = β_{j} + 64 · c_{j} for every j ⩾ 0.

Proof of a). The proof is by induction on j. For j = 0 the relation (7.2) is

Q_{2k} Q_{2k+1} − Q_{2k+2} Q_{2k+3} = −8 P_{2k+3} Q_{2k} + 80

(i. e., the relation (7.1)) which is true by Lemma 14.

Assume that the relation (7.2) is true for j = r. Then

∑_{i=0}^{2(r+1)+1} (−1)^{i} Q_{2k+2i} Q_{2k+2i+1} = ∑_{i=0}^{2r+1} (−1)^{i} Q_{2k+2i} Q_{2k+2i+1} +

Q_{2k+4r+4} Q_{2k+4r+5} − Q_{2k+4r+6} Q_{2k+4r+7} = −8c_{r}(P_{2k+2r+3} Q_{2k+2r} − 10) +

Q_{2k+4r+4} Q_{2k+4r+5} − Q_{2k+4r+6} Q_{2k+4r+7} = −8c_{r+1}(P_{2k+2r+5} Q_{2k+2r+2} − 10),

where the last step uses Lemma 15. Hence, (7.2) is true for j = r + 1 and the proof is completed.

Proof of b). The proof is once again by induction on j. For j = 0 the relation (7.3) is β_{0} + Q_{2k} Q_{2k+1} = d_{0} Q_{2k} Q_{2k+1} which is true since d_{0} = 1 and β_{0} = 0.

Assume that the relation (7.3) is true for j = r. Then

β_{r+1} + ∑_{i=0}^{2(r+1)} (−1)^{i} Q_{2k+2i} Q_{2k+2i+1} = β_{r+1} − β_{r} + β_{r} +

∑_{i=0}^{2r} (−1)^{i} Q_{2k+2i} Q_{2k+2i+1} − Q_{2k+4r+2} Q_{2k+4r+3} + Q_{2k+4r+4} Q_{2k+4r+5} =

64 c_{r} + d_{r} Q_{2k+2r} Q_{2k+2r+1} − Q_{2k+4r+2} Q_{2k+4r+3} + Q_{2k+4r+4} Q_{2k+4r+5} = d_{r+1} Q_{2k+2r+2} Q_{2k+2r+3},

where the last step uses Lemma 16. Hence, (7.3) is true for j = r + 1 and the proof is completed.

We mention also the following equalities that have been discovered while attempting to prove Theorem 7.

Theorem 8. For every k ⩾ 0 we have

Q_{2k} · Q_{2k+1} = Q_{2k+3} · Q_{2k−2} − 80 = 2 · (P_{2k+2} · P_{2k−1} − 6).

References

[1] Z. Čerin, Properties of odd and even terms of the Fibonacci sequence, Demonstratio Mathematica, 39 (1) (2006), 55–60.

[2] Z. Čerin, On sums of squares of odd and even terms of the Lucas sequence, Proccedings of the 11th Fibonacci Conference, (to appear).

[3] Z. Čerin, Some alternating sums of Lucas numbers, Central European Journal of Mathematics 3 (1) (2005), 1 – 13.

[4] Z. Čerin, Alternating Sums of Fibonacci Products, (preprint).

[5] V. E. Hoggatt, Jr., Fibonacci and Lucas numbers, The Fibonacci Association, Santa Clara, 1969.

[6] R. Knott, Fibonacci numbers and the Golden Section,

   http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html.

[7] V. Rajesh and G. Leversha, Some properties of odd terms of the Fibonacci sequence, Mathematical Gazette, 88 (2004), 85–86.

[8] N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences,

   http://www.research.att.com/~njas/sequences/.

[9] S. Vajda, Fibonacci and Lucas numbers, and the Golden Section: Theory and Applications, Halsted Press, Chichester (1989).