SOLUTIONS

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41. [1992, 27] Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Define the sequence {L_{n}}_{n=1}^{∞} by L_{1} = a, L_{2} = b and L_{n+2} = L_{n+1} + L_{n} for n ≥ 1, where a and b are arbitrary integers. If a = 1 and b = 2, then L_{i} = i for three consecutive integers i.

(i) Are there other values of a and b with this property? (ii) Are there values of a and b such that L_{i} = i for four consecutive values of i? (iii)∗ What happens if the ‘consecutive’ restriction is removed in (i) and (ii)?

Solution by N. J. Kuenzi and Kandasamy Muthuvel, University of Wisconsin-Oshkosh, Oshkosh, Wisconsin (jointly).

(i) We will answer the more general question: Are there any other values of a and b such that three consecutive terms L_{n}, L_{n+1}, and L_{n+2} yield three consecutive integers i, i + 1, and i + 2?

Let F_{1} = 1, F_{2} = 1, and F_{n+2} = F_{n+1} + F_{n} for n ≥ 1 denote the Fibonacci sequence. Then the sequence L_{1} = a, L_{2} = b, and L_{n+2} = L_{n+1} + L_{n} for n ≥ 1 can be expressed in terms of the Fibonacci sequence by

L_{n+2} = aF_{n} + bF_{n+1} for n ≥ 1.

If L_{n} = i, L_{n+1} = i + 1, and L_{n+2} = i + 2 then 2i + 1 = i + 2 and i = 1. So a = 1 and b = 2 are the only values for a and b with the property that L_{i} = i for three consecutive integers.

Next, suppose that

\begin{aligned}L_{n} & = aF_{n−2} + bF_{n−1} = 1 \\ L_{n+1} & = aF_{n−1} + bF_{n} = 2. \end{aligned}

Eliminating the variable b from these two equations yields

a(F_{n−2}F_{n} − F_{n−1}^{2}) = F_{n} − 2F_{n−1} = −F_{n−3}.

Since

F_{n−2}F_{n} − F_{n−1}^{2} = (−1)^{n−1},
a = (−1)^{n}F_{n−3}.

Similarly,

b = (−1)^{n+1}F_{n−4}.

Example: If a = F_{7} = 13 and b = −F_{6} = −8, then L_{10} = 1, L_{11} = 2, and L_{12} = 3. (ii) In part (i), we saw that if any three consecutive terms of the sequence yield three

consecutive integers then the integers are 1, 2, and 3. Hence, it is not possible to have four consecutive integers as four consecutive terms of the sequence.

(iii) We will address the following question: Is it possible to have three distinct values i < j < k, other than 1, 2, 3 with L_{i} = i, L_{j} = j, and L_{k} = k?

Let i be the smallest subscript such that L_{i} = i and let c = L_{i+1} − L_{i}.

(a) Suppose c ≥ 0. Then

\begin{aligned}L_{i+2} & = 2i + c ≥ i + 1 ≥ 2, \\ L_{i+3} & = 3i + 2c ≥ i + 2, \\ L_{i+4} & = 5i + 3c ≥ i + 4, \end{aligned}

and

L_{i+n} > i + n for n ≥ 5.

So the only way we could have L_{j} = j and L_{k} = k would be to find values for i and c which would satisfy two of the four equations:

\begin{aligned}i + c & = i + 1, \\ 2i + c & = i + 2, \\ 3i + 2c & = i + 3, \\ 5i + 3c & = i + 4. \end{aligned}

Exploring each of these possible cases yields only the solution i = 1, c = 1, L_{2} = 2, and L_{3} = 3. So if L_{i} = i and c ≥ 0 it is not possible to have L_{i} = i, L_{j} = j, and L_{k} = k, for i < j < k, unless i = 1, j = 2, and k = 3.

(b) Suppose c < 0. Let j be the next smallest subscript such that L_{j} = j. Note that j ≥ i + 2 ≥ 3.

If the sequence of terms

L_{i}, L_{i+1}, …, L_{i+n} = L_{j}

contained two consecutive negative terms, then all successive terms in the sequence would be negative and L_{j} < 0. So the sequence cannot contain two consecutive negative terms.

If the sequence of terms

L_{i}, L_{i+1}, …, L_{i+n} = L_{j}

alternate in sign between positive and negative terms, then

i = L_{i} > L_{i+2} > L_{i+4} > ⋯ > L_{i+n} = L_{j} = j.

But i < j and so the sequence cannot just alternate in sign between positive and negative terms.

Since the sequence

L_{i}, L_{i+1}, …, L_{i+n} = L_{j}

cannot alternate in sign and cannot contain two consecutive negative terms, it follows that L_{j−1} ≥ 0.

If L_{j−1} ≠ 1 then L_{j+1} ≠ j + 1,

\begin{aligned}L_{j+1} & ≥ j, \\ L_{j+2} & ≥ 2j > j + 2, \end{aligned}

and

L_{j+n} > j + n for successive terms,

(i.e., L_{k} ≠ k for all k > j).

This leaves only the case L_{i} = i, L_{j−1} = 1, L_{j} = j to be considered. Working backward from L_{j} yields L_{j} = j, L_{j−1} = 1, L_{j−2} = j − 1, L_{j−3} = 2 − j, L_{j−4} = 2j − 3, and in general

L_{j−n} = (−1)^{n}(jF_{n−1} − F_{n}).

But then

L_{i} = i = jF_{n−1} − F_{n}

for some even integer n ≥ 2. So

i = jF_{n−1} − F_{n} ≥ (i + 2)F_{n−1} − F_{n}

≥ iF_{n−1} + 2F_{n−1} − F_{n} ≥ i + 1.

But this is a contradiction and so L_{j−1} ≠ 1.

Hence it is not possible to have three distinct values i < j < k, other than 1, 2, 3, with L_{i} = i, L_{j} = j, and L_{k} = k.

(iv) In this section we address the following question: Is it possible to have three terms of the sequence L_{k}, L_{m}, and L_{n} such that L_{k} = i,

L_{m} = i + 1, and L_{n} = i + 2?

(a) Let L_{k} = i and L_{k+1} = 1. Then L_{k+2} = i + 1 and L_{k+3} = i + 2. To find initial values for L_{1} and L_{2} simply work backward from L_{k} = i and L_{k+1} = 1.

Example: Suppose L_{6} = 18 and L_{7} = 1, then L_{8} = 19 and L_{9} = 20. Working backward yields

L_{5} = −17, L_{4} = 35, L_{3} = −52, L_{2} = 87, and L_{1} = −139.

(b) Let L_{n} = i + 2, and L_{n+1} = −1. Then L_{n+2} = i + 1 and L_{n+3} = i. To find initial values for L_{1} and L_{2}, simply work backward from L_{n} = i + 1 and L_{n+1} = −1.

Example: Suppose L_{6} = 18 and L_{7} = −1, then L_{8} = 17 and L_{9} = 16. Working backward yields

L_{5} = −19, L_{4} = 37, L_{3} = −56, L_{2} = 93, and L_{1} = −149.

(v) Finally we close with a question for the reader: Is it possible to have three terms of the sequence L_{k}, L_{m}, and L_{n} with no two consecutive

such that L_{k} = i, L_{m} = i + 1, and L_{n} = i + 2 for some i?

(i) and (ii) were also solved by the proposer.

42. [1992, 27] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Find the general solution to the differential equation

∑_{k=0}^{1990}(1990)^{k+1}y^{(k)} = 0,

where y^{(k)} represents the kth derivative of y.

Solution by the proposer and Jayanthi Ganapathy, University of Wisconsin – Oshkosh, Oshkosh, Wisconsin (independently).

The characteristic equation for this differential equation is

(1990)^{1991}t^{1990} + (1990)^{1990}t^{1989} + ⋯ + (1990)^{2}t + 1990 = 0.

Next, dividing this equality by 1990 first, and then letting u = 1990t, we get

(1)  u^{1990} + u^{1989} + ⋯ + u + 1 = 0.

Now, using the fact that

x^{n} − 1 = (x − 1)(x^{n−1} + x^{n−2} + ⋯ + x + 1),

the roots of (1) are 1990 distinct complex roots of unity

u_{k} = cos \frac{2kπ}{1991} + i sin \frac{2kπ}{1991}, k = 1, 2, …, 1990.

Thus,

t_{k} = \frac{u_{k}}{1990}.

But, since t_{1991−k} is the complex conjugate of t_{k}, the general solution of the given differential equation is

y = ∑_{k=1}^{995} e^{α_{k}x} (A_{k} cos β_{k}x + B_{k} sin β_{k}x) ,

where

α_{k} = \frac{1}{1990} cos \frac{2kπ}{1991} and β_{k} = \frac{1}{1990} sin \frac{2kπ}{1991}.

Also solved by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

43. [1992, 28] Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, Missouri.

Let n ≥ 3 be a positive integer and m = \frac{n(n+1)}{2}. Evaluate

∑_{1≤a,b,c≤n a,b,c all distinct} \frac{abc}{m(m − a)(m − a − b)}.

Solution by Mangho Ahuja, Southeast Missouri State University, Cape Girardeau, Missouri, N. J. Kuenzi, University of Wisconsin-Oshkosh, Oshkosh, Wisconsin, and Russell Euler, Northwest Missouri State University, Maryville, Missouri (independently).

∑_{1≤a,b,c≤n a,b,c all distinct} \frac{abc}{m(m − a)(m − a − b)}

= ∑_{1≤a,b≤n ; a,b distinct} \frac{ab}{m(m − a)(m − a − b)} ∑_{{1≤c≤n \atop c≠a,b}} c

= ∑_{1≤a,b≤n ; a,b distinct} \frac{ab}{m(m − a)(m − a − b)}(m − a − b)

= ∑_{1≤a,b≤n ; a,b distinct} \frac{ab}{m(m − a)}

= ∑_{1≤a≤n} \frac{a}{m(m − a)} ∑_{{1≤b≤n \atop b≠a}} b

= ∑_{1≤a≤n} \frac{a}{m(m − a)}(m − a)

= ∑_{1≤a≤n} \frac{a}{m} = \frac{1}{m} ∑_{1≤a≤n} a = \frac{1}{m}m = 1.

Generalized Solution by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

∑_{1≤a_{i}≤n; i=1,··· ,r a_{i}’s all distinct} \frac{a_{1}a_{2} ⋯ a_{r}}{m(m − a_{1})(m − a_{1} − a_{2}) ⋯ (m − a_{1} − a_{2} − ⋯ − a_{r−1})}

= \frac{1}{m} ∑_{a_{1}=1}^{n} \frac{a_{1}}{m − a_{1}} ∑_{{a_{2}=1 \atop a_{2}≠a_{1}}}^{n} \frac{a_{2}}{m − a_{1} − a_{2}} ∑_{{a_{3}=1 \atop a_{3}≠a_{2},a_{3}≠a_{1}}}^{n} \frac{a_{3}}{m − a_{1} − a_{2} − a_{3}}

⋯ ∑_{{a_{r−1}=1 \atop a_{r−1}≠a_{i}; i=1,··· ,r−2}}^{n} \frac{a_{r−1}}{m − a_{1} − a_{2} − ⋯ − a_{r−1}} ∑_{{a_{r}=1 \atop a_{r}≠a_{i}; i=1,··· ,r−1}}^{n} a_{r}

= 1.

Also solved by the proposers.

44. [1992, 28] Proposed by Kandasamy Muthuvel, University of Wisconsin-Oshkosh, Oshkosh, Wisconsin.

It is well known that the set of all real numbers ℝ is a field under ordinary addition and multiplication, the set of all positive real numbers ℝ^{+} is a subgroup of the multiplicative group ℝ^{∗} (the set of all nonzero real numbers), r^{2} ∈ ℝ^{+} for all r ∈ ℝ^{∗}, the characteristic of ℝ is 0, and ℝ^{+} − ℝ^{+} = ℝ.

Prove the generalized result, “Let G be a subgroup of the multiplicative group F^{∗} of a field F such that f^{2} ∈ G for all f in F^{∗}. If the characteristic of F is not equal to 2, 3, and 5, then G − G = F.”

Remark. The above result is not true if the characteristic of F is equal to 2, 3, or 5.

Solution by the proposer.

It is immediate that G − G ⊆ F. To show F ⊆ G − G, we first let f ∈ F such that f ≠ ±1. Then 1 + f, 1 − f ∈ F^{∗}. Next, 2 ≠ 0 since the characteristic of F is not 2. Thus 4 = 2^{2} ∈ G. So 4^{−1} ∈ G, since G is a group. Therefore,

f = 4^{−1} · 4f
= 4^{−1} ((1 + f)^{2} − (1 − f)^{2})
= 4^{−1}(1 + f)^{2} − 4^{−1}(1 − f)^{2}

so f ∈ G − G. Finally, we need to show ±1 ∈ G − G. First of all, 4^{−1} ≠ ±1 since the characteristic of F is not 3 and 5, respectively. Thus

1 = 4 · 4^{−1} = (1 + 4^{−1})^{2} − (1 − 4^{−1})^{2} ,

so 1 ∈ G − G. Similarly, −1 ∈ G − G. Thus F ⊆ G − G. Therefore, G − G = F and this completes the proof.