SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the problem editor.

53. [1993, 39] Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Prove analytically that

\sqrt[3]{19 + 9\sqrt{6}} + \sqrt[3]{19 − 9\sqrt{6}}

is an integer.

Solution I by Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin; Gregory Bruton, Cape Girardeau, Missouri; and Sherri Palmer, Ste. Genevieve, Missouri.

Since (1 + \sqrt{6})^{3} = 19 + 9\sqrt{6} and (1 − \sqrt{6})^{3} = 19 − 9\sqrt{6}, the desired sum equals 2.

Solution II by Robert L. Doucette, McNeese State University, Lake Charles, Louisiana; Donald P. Skow, University of Texas–Pan American, Edinburg, Texas; Kanadasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin; Joseph E. Chance, University of Texas–Pan American, Edinburg, Texas; Seung–Jin Bang, Albany, California; Joe Howard, New Mexico Highlands University, Las Vegas, New Mexico; J. Sriskandarajah, University of Wisconsin Center–Richland, Richland Center, Wisconsin; and the proposer.

Let s = \sqrt[3]{19 + 9\sqrt{6}} and t = \sqrt[3]{19 − 9\sqrt{6}}. Note that st = \sqrt[3]{19^{2} − 9^{2}6} = −5 and s^{3} + t^{3} = (19 + 9\sqrt{6}) + (19 − 9\sqrt{6}) = 38. Since

(s + t)^{3} = s^{3} + 3s^{2}t + 3st^{2} + t^{3} = (s^{3} + t^{3}) + 3st(s + t) = 38 − 15(s + t),

we see that s + t is a root of the equation x^{3} + 15x − 38 = 0. Noting that x = 2 is a root of this equation, we have x^{3} + 15x − 38 = (x − 2)(x^{2} + 2x + 19). Since x^{2} + 2x + 19 has no real roots, it follows that s + t = 2.

Generalized Solution I by Donald P. Skow, University of Texas–Pan American, Edinburg, Texas.

Let x ≥ 0 and define

f(x) = \sqrt[3]{(3x + 1) + (x + 3)\sqrt{x}} + \sqrt[3]{(3x + 1) − (x + 3)\sqrt{x}}.

Then f(6) = \sqrt[3]{19 + 9\sqrt{6}} + \sqrt[3]{19 − 9\sqrt{6}}. But f(6) = 2 since

f(x) = \sqrt[3]{(1 + \sqrt{x})^{3}} + \sqrt[3]{(1 − \sqrt{x})^{3}} = 2

for all x.

Generalized Solution II by Joseph E. Chance, University of Texas–Pan American, Edinburg, Texas.

If

S = \sqrt[3]{a + c\sqrt{b}} + \sqrt[3]{a − c\sqrt{b}} = s + t,

then

(∗)  S^{3} − 3\sqrt[3]{a^{2} − c^{2}b} · S − 2a = 0.

If we let a^{2} − c^{2}b = w^{3}, then other representations of this form for 2 can be found if 2 is a root of (∗), or

w = \frac{4 − a}{3}.

Some experimenting yields at least two other choices,

\begin{matrix}\sqrt[3]{31 + 13\sqrt{10}} + \sqrt[3]{31 − 13\sqrt{10}} = 2, and \\ \sqrt[3]{37 + 30\sqrt{3}} + \sqrt[3]{37 − 30\sqrt{3}} = 2. \end{matrix}

54. [1993, 39] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Let x_{1} = 1 and

x_{j} − x_{j−1} = ∑_{k=1}^{j}(−1)^{k+1}\binom{j}{k} ∑_{m=1}^{k} \frac{1}{m}, for j ≥ 2.

Prove that for any positive integer n > 1,

∑_{k=1}^{n} k (kx_{k}^{−1})^{k} < (n + 1)! .

Solution by the proposer.

From

\binom{j}{k} = \binom{j − 1}{k} + \binom{j − 1}{k − 1}

and some simple calculations we get

∑_{k=1}^{j}(−1)^{k+1}\binom{j}{k} ∑_{m=1}^{k} \frac{1}{m} = \frac{1}{j}.

(A good reference on this topic is John Riordan, Combinatorial Identities, John Wiley & Sons, Inc., New York, 1968, p. 5.) Hence, x_{j} − x_{j−1} = 1/j, and thus

x_{n} = ∑_{j=1}^{n} \frac{1}{j}.

Now, from the relationship between the arithmetic and the geometric means of the numbers 1, \frac{1}{2}, \frac{1}{3}, …, \frac{1}{n}, we deduce that

\frac{n^{n}}{n!} < (∑_{j=1}^{n} \frac{1}{j})^{n} = x_{n}^{n}.

Therefore, n! > (nx_{n}^{−1})^{n}. This implies that n(n!) > n(nx_{n}^{−1})^{n}. Consequently,

∑_{k=1}^{n} k(kx_{k}^{−1})^{k} < ∑_{k=1}^{n} k(k!) = (n + 1)! − 1 < (n + 1)! .

55. [1993, 40] Proposed by Stanley Rabinowitz, Westford, Massachusetts.

Let F_{n} and L_{n} denote the nth Fibonacci and Lucas numbers, respectively. Find a polynomial f(x, y) with constant coefficients such that f(F_{n}, L_{n}) is identically zero for all positive integers n or prove that no such polynomial exists.

Solution by the proposer.

From problem 35 in the Missouri Journal of Mathematical Sciences [1992, 96–97],

4(−1)^{n} = L_{n}^{2} − 5F_{n}^{2}.

Square both sides to get terms with constant coefficients. The desired polynomial is therefore

f(x, y) = 25x^{4} − 10x^{2}y^{2} + y^{4} − 16.

Generalized Solution by Joseph E. Chance, University of Texas–Pan American.

Let A_{n} and B_{n} be any two solutions of the difference equation

X_{n+1} = aX_{n} + bX_{n−1},

written as

\begin{aligned}A_{n} & = C_{1}α^{n} + C_{2}β^{n} \\ B_{n} & = D_{1}α^{n} + D_{2}β^{n} \end{aligned}

where α, β solve x^{2} − ax − b = 0, α ≠ β. Using Cramer’s rule, it follows that

α^{n} = \frac{D_{2}A_{n} − C_{2}B_{n}}{D}

and

β^{n} = \frac{C_{1}B_{n} − D_{1}A_{n}}{D} ,

where we assume that

D = det ( \begin{matrix}C_{1} & C_{2} \\ D_{1} & D_{2} \end{matrix} ) ≠ 0.

Thus

(C_{1}B_{n} − D_{1}A_{n})(C_{2}B_{n} − D_{2}A_{n}) = −D^{2}(αβ)^{n} = −D^{2}(−b)^{n}.

This product is dependent upon n unless b = −1. If b = 1, the product can be squared to remove the dependency on n, as in the case of Fibonacci and Lucas numbers. For other values of b, the product is dependent on n and this technique fails to suggest an appropriate polynomial.

56. [1993, 40] Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, Missouri.

Let

A_{1} = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

and

A_{n+1} = ( \begin{matrix}0 & 1 & 1 & 0 & 0 & ⋯ & 0 \\ 1 & 0 & 0 & 1 & 0 & ⋯ & 0 \\ 1 & 0 \\ 0 & 1 \\ 0 & 0 & & & A_{n} \\ ⋮ & ⋮ \\ 0 & 0 \end{matrix} )

for n ≥ 1. Find det A_{1993}.

Solution by Seung–Jin Bang, Albany, California.

We will prove that det A_{n} = − det A_{n−3}. Replacing the 3rd row by the 3rd row minus the 2nd row and the 4th row by the 4th row minus the 1st row, we have

det ( \begin{matrix}0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & & & A_{n−3} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} )

Next, we expand the determinant by the 3rd and 4th rows. Then

det ( \begin{matrix}0 & 1 & 1 & 0 & 0 & 0 & 0 & ⋯ & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & & & A_{n−3} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 \end{matrix} )

and then expand about the 1st and 2nd columns. Then we have

− det ( \begin{matrix}1 & 0 & 1 & 0 & 0 & ⋯ & 0 \\ 0 & 1 & 0 & 1 & 0 & ⋯ & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 & & & A_{n−3} \\ ⋮ & ⋮ \\ 0 & 0 \end{matrix} ) = − det A_{n−3}.

Since 1993 = 3 · 664 + 1, we have det A_{1993} = det A_{1} = −1. Also solved by the proposers.