4 Miscellaneous
4.1 Proposals
4.1.1
Proposed by B. C. Rennie, James Cook University of North Queensland. Crux Math. [3: 297].
If \(0 < b \leq a\), prove that \[ a + b - 2\sqrt{ab} \geq \frac{1}{2} \frac{(a - b)^2}{a + b}. \]
4.1.2
Proposed by W. J. Blundon. Crux. Math. [7: 240].
Find interesting sets of three distinct real numbers such that their product is equal to their sum.
4.1.3
Proposed by Albert Wilansky, Lehigh University. Amer. Math. Monthly [67: 290].
Is there a field whose additive and multiplicative groups are isomorphic?
4.1.4
Proposed by J. B. Love, Eastern Baptist College. Amer. Math. Monthly [69: 668].
Let \(P(x)\) be a polynomial of degree \(n\) with only real roots and real coefficients. Show that \[ (n - 1)[P^\prime(x)]^2 - nP(x)P^{\prime\prime}(x) \geq 0. \]
4.1.5
Proposed by V. N. Murty, Pennsylvania State University. Crux Math. [6: 213].
If \(x, y, z > 0\), show that \[ \sum_{\textrm{cyclic}} \frac{2x^2(y + z)}{(x + y)(x + z)} \leq x + y + z, \] with equality if and only if \(x = y = z\).
4.1.6
Proposed by Alexander Oppenheim, University of Malaya. Amer. Math. Monthly [71: 317].
The three positive numbers \(x\), \(y\), \(z\) lie between the least and the greatest of the three positive numbers \(a\), \(b\), \(c\). If \(x + y + z = a + b + c\) and \(xyz = abc\), show that, in some order, \(x\), \(y\), \(z\) are equal to \(a\), \(b\), \(c\).
4.1.7
Proposed by Ludwig Bruch. Amer. Math. Monthly [74: 447].
For any positive integer \(n\), prove \[ \sum_{k=0}^{n - 1} \csc^2 \left\{\frac{\pi}{2n}(2k + 1)\right\} = n^2. \]
4.1.8
Proposed by Bruce McColl, St. Lawrence College. Crux Math. [3: 65].
Prove that \[ \tan\frac{\pi}{11} \tan\frac{2\pi}{11} \tan\frac{3\pi}{11} \tan\frac{4\pi}{11} \tan\frac{5\pi}{11} = \sqrt{11}. \]
4.1.9
Proposed by A. A. Mullin, University of California, Livermore. Amer. Math. Monthly [73: 547].
Does there exist an integer for which the \(n \times n\) matrix whose elements are \(a_{ij} = (i,j)\) is singular? \((i,j)\) is the greatest common divisor of the natural numbers \(i\) and \(j\).
4.1.10
Proposed by W. J. Blundon. Crux Math. [9: 142].
Find (a) necessary and (b) sufficient conditions on \(a\), \(b\), \(c\) for the system \[ x + \frac{1}{x} = a, \quad y + \frac{1}{y} = b, \quad xy + \frac{1}{xy} = c \] to be consistent, that is, to have at least one common solution \((x, y)\). (These necessary and sufficient conditions constitute the eliminant of the system.)
4.1.11
Proposed by Hippolyte Charles. Crux Math. [4: 250].
Solve the following system of equations for \(x\) and \(y\):
\[\begin{align*} \frac{(ab + 1)(x^2 + 1)}{x + 1} &= \frac{(a^2 + 1)(xy + 1)}{y + 1} \\ \frac{(ab + 1)(y^2 + 1)}{y + 1} &= \frac{(b^2 + 1)(xy + 1)}{x + 1}. \end{align*}\]
4.1.12
Proposed by R. S. Underwood, Texas Technological College. Amer. Math. Monthly [63: 729].
Find real solutions of the equations \[ x^2 + y^2 + z^2 + u + v = D, \quad x^2 + \frac{y^2}{2} + \frac{z^2}{3} + \frac{u^2}{4} + \frac{v^2}{5} = 1, \] in case (a) \(D = -3\), and (b) \(D = 15/4\). What can be said about the solutions for other values of \(D\)?
4.1.13
Proposed by S. C. Chan. Crux Math. [7: 117].
For which constants \(\lambda\) and \(m\) does the infinite series \[ \frac{1^3\lambda}{1!}e^{-m} + \frac{2^3\lambda^2}{2!}e^{-2m} \frac{3^3\lambda^3}{3!}e^{-3m} + \frac{4^3\lambda^4}{4!}e^{-4m} + \cdots \] converge, and to what sum?
4.1.14
Proposed by E. L. Whitney, University of Alberta. Canad. Math. Bull, [4: 187].
Show that \[ \pi^2 = 10 - \sum_{n=1}^{\infty} \frac{1}{n^3(n + 1)^3}. \]
4.1.15
Proposed by L. Moser and J. R. Pounder, University of Alberta. Canad. Math. Bull. [5: 70].
For real \(\alpha\), \(\beta\), \(\gamma\), \(\delta\) and \((\alpha x + \beta y)(\gamma x + \delta y) = ax^2 + bxy + cy^2\), prove that \(\max(a, b, c) \geq \frac{4}{9} (\alpha + \beta)(\gamma + \delta)\).
4.1.16
Proposed by Kenneth M. Wilke. Crux Math. [3: 190].
Let \(\langle\sqrt{10}\rangle\) denote the fractional part of \(\sqrt{10}\). Prove that for any positive integer \(n\) there exists an integer \(I_n\) such that \[ (\langle\sqrt{10}\rangle)^n = \sqrt{I_n + 1} - \sqrt{I_n}. \]
4.2 Solutions
4.2.1
Solution by W. J. Blundon. Crux Math. [4: 162].
If the two sides of \(2(a+b)(a+b-2\sqrt{ab}) = (\sqrt{a}+\sqrt{b})^2+(\sqrt{a}-\sqrt{b})^2\) are multiplied by \((\sqrt{a}-\sqrt{b})^2\) we obtain \[ 2(a+b)(a+b-2\sqrt{ab}) = (a-b)^2 + (\sqrt{a}-\sqrt{b})^4 \geq (a-b)^2, \] with equality if and only if \(a = b\). Division by \(2(a+b)\) gives the required result.
4.2.2
Solution by W. J. Blundon. Crux Math. [8: 284-285].
I offer two satisfactory sets consisting of an arithmetic progression, \[ \left\{a + \sqrt{a^2 - 3}, a, a - \sqrt{a^2 - 3}\right\} \] and a geometric progression, \[ \left\{\frac{a(a^2-1)+a\sqrt{(a^2+1)(a^2-3)}}{2}, a, \frac{a(a^2-1)-a\sqrt{(a^2+1)(a^2-3)}}{2}\right\}. \]
4.2.3
Solution by W. J. Blundon. Amer. Math. Monthly [67: 925].
In a field \(F\), the equation \(x^2 = 1\) has two solutions if \(\textrm{char}(F) \neq 2\) and one solution if \(\textrm{char}(F) = 2\). On the other hand, the equation \(x + x = 0\) has one solution if \(\textrm{char}(F) \neq 2\) and \(|F|\) solutions if \(\textrm{char}(F) = 2\). Since \(|F| > 1\), there is no case where these two equations have the same number of solutions, and so there is no field whose additive and multiplicative groups are isomorphic.
4.2.4
Solution by W. J. Blundon. Amer. Math. Monthly [70: 443].
It is easily seen that \((n-1){P^\prime}^2 - nPP^{\prime\prime} = 0\) for \(n = 1\). Define \(Q(x) = (ax+b)P(x)\), where \(a\) and \(b\) are real, so that \(Q(x)\) is of degree \(n+1\) with real roots and real coefficients, and suppose \((n-1){P^\prime}^2 - nPP^{\prime\prime} \geq 0\). Then \(n{Q^\prime}^2 - (n+1)Q^{\prime\prime}Q\) equals \[ [(n+1)/n] [(n-1){P^\prime}^2 - n P^{\prime\prime}P] (ax+b)^2 + (1/n)[naP - (ax+b)P^\prime]^2 \geq 0. \]
The result now follows by mathematical induction.
4.2.5
Solution by V. N. Murty, Penssylvania State University. Crux Math. [7: 217].
The identity \[ \sum 2x^2(y+z)^2 + \sum yz(y-z)^2 \equiv (y+z)(z+x)(x+y)(x+y+z) \] is easily established by direct calculation or by noting that its left member is a symmetric and homogeneous polynomial of degree \(4\) in \(x\), \(y\), \(z\) which vanishes when \(y = -z\). It follows that
\[\begin{align*} \sum \frac{2x^2(y+z)}{(x+y)(x+z)} &\equiv x + y + z - \frac{\sum yz(y-z)^2}{(y+z)(z+x)(x+y)} \\ &\leq x + y + z, \end{align*}\] with equality if and only if \(x = y = z\).
Comment by W. J. Blundon.
If we set \(y + z = a\), \(z + x = b\), \(x + y = c\), it is easy to show that \(x,y,z > 0\) if and only if \(a\), \(b\), \(c\) are sides of a triangle, and then the proposed inequality is equivalent to \[ \sum 2a^2(s-a)^2 \leq abcs, \] where \(s\) is the semiperimeter, with equality if and only if the triangle is equilateral.
4.2.6
Solution by W. J. Blundon. Amer. Math. Monthly [72: 185].
It is easily verified that
\[\begin{align*} bc(x-a)(y-a)(z-a) &= ca(x-b)(y-b)(z-b) \\ &= ab(x-c)(y-c)(z-c), \end{align*}\] each expression being equal to \(abc(ab + ac + bc - xy - xz- yz)\).
If no factor in any of the three expressions is zero, then one expression has all positive factors and another expression has exactly three negative factors which is a contradiction. Thus each expression has one zero factors, and the required result follows.
4.2.7
Solution by W. J. Blundon. Amer. Math. Monthly [75: 421].
Let \(\tan (2k+1)\pi/4n = \sqrt{t_k}\). Putting \(z_k = (1+i\sqrt{t_k})/(1-i\sqrt{t_k})\), we have \[ z_k = \cos \frac{(2k+1)\pi}{2n} + i \sin \frac{(2k+1)\pi}{2n}, \] so that \(z_k^{2n} = -1\), which reduces to \((i+\sqrt{t_k})^{2n} + (i-\sqrt{t_k})^{2n} = 0\). Thus the \(t_k\) are the \(n\) roots of the polynomial equation \[ t^n - n(2n-1)t^{n-1} + \cdots + (-1)^n = 0. \]
Consequently, \[ \sum_{k=0}^{n-1}t_k = n(2n-1) = \sum_{k=0}^{n-1}t_k^{-1}, \] the right sum following from the identity \(\cot (\frac{1}{2}\pi-\theta) = \tan \theta\).
Finally, since \(\csc \theta = \frac{1}{2}(\tan \frac{1}{2}\theta + \cot \frac{1}{2}\theta)\), the required sum is equal to \[ \frac{1}{4}\sum_{k=0}^{n-1} (t_k + t_k^{-1} + 2) = n^2. \]
4.2.8
Solution by W. J. Blundon. Crux Math. [3: 200-201].
The stated equation is equivalent to \[ \tan\frac{\pi}{11} \tan\frac{2\pi}{11} \tan\frac{3\pi}{11} \dots \tan\frac{10\pi}{11} = -11. \]
Let \(x = \tan \frac{k\pi}{11}\), where \(k\) is any of the integers \(1, 2, \dots, 10\). Then \[ \cos\frac{2k\pi}{11} + i\sin \frac{2k\pi}{11} = \frac{1-x^2+2xi}{1+x^2} = \frac{(1+xi)^2}{1+x^2}, \] and de Moivre’s Theorem gives \[ (1+xi)^{22} = (1+x^2)^{11}. \]
Equating real parts, we have \[ x^{20}- 110x^{18} + \dotsm + 121 = 0. \]
This is the same as \[ (x^{10} - 55x^8 + 330x^6 - 462x^4 + 165x^2 - 11)^2 = 0. \]
The result follows.
4.2.9
Solution by W. J. Blundon. Amer. Math. Monthly [74: 602].
Let \(X_n\) be the given matrix and let \(\phi\) be Euler’s totient function. Define two \(n \times n\) matrices \(B\) and \(C\) as follows. \[ B = [b_{ij}] \textrm{ where } b_{ij} = \begin{cases} \phi(i) \quad j = i, \\ 0 \quad \textrm{otherwise}; \end{cases} \]
\[ C = [c_{ij}] \textrm{ where } c_{ij} = \begin{cases} 1 \quad j \mid i, \\ 0 \quad \textrm{otherwise} \end{cases} \]
and let \(C^{\prime\prime}\) denote the transpose of \(C\). It follows that \(X_n = CBC^\prime\) since the \(ij\)th term of this matrix is equal to \[ \sum_{k=1}^n c_{ik}b_{kk}c_{kj}^\prime = \sum_{k \mid (i,j)} \phi(k) = (i,j) \] from the well-known summation of the \(\phi\)-function. Note that \(\det C = \det C^\prime = 1\). Thus \(D_n = \det X_n = \det B = \phi(1)\phi(2)\dotsm\phi(n)\). Since \(\det X_N\) is always positive, \(X_n\) is never singular.
(Note the recurrence relation: \(D_n/D_{n-1} = \phi(n)\).)
4.2.10
Solution by Gali Salvatore. Crux Math. [10: 238-239].
We assume that the equations are over the field of complex numbers. Suppose the given system is consistent, and let \((x,y)\) be a common solution. If we multiply together corresponding members of the three equations, we obtain \[ (a^2-2) + (b^2-2) + (c^2-2) + 2 = abc, \] and thus a necessary condition for consistency is \[ a^2 + b^2 + c^2 - 4 = abc, \tag{4.1}\]
Conversely, suppose (Equation 4.1) holds, and let \((x,y)\) be any solution of the first two equations of the system. Then \((x,1/y)\) is also a solution of the same first two equations. Now, from (Equation 4.1), \[ c^2 - abc + (a^2+b^2-4) = 0, \] and so \[\begin{align*} c &= \frac{ab \pm \sqrt{(a^2-4)(b^2-4)}}{2} \\ &= \frac{(x+\frac{1}{x})(y+\frac{1}{y}) \pm (x-\frac{1}{x})(y-\frac{1}{y})}{2} \\ &= xy + \frac{1}{xy} \quad \textrm{or} \quad \frac{x}{y} + \frac{y}{x}. \end{align*}\]
Now \(c = xy + 1/xy\) implies that \((x,y)\) is a common solution of the system, and \(c = x/y + y/x\) implies that \((x,1/y)\) is a common solution. In either case the system is consistent. We conclude that (Equation 4.1) is the required eliminant of the system.
It is clear from the above solution that (Equation 4.1) is also the eliminant of the system \[ x + \frac{1}{x} = a, \quad y + \frac{1}{y} = b, \quad \frac{x}{y} + \frac{y}{x} = c. \]
If we are restricted to the real field, it is easy to see that the eliminant consists of (Equation 4.1) and the additional two conditions \(|a| \geq 2\) and \(|b| \geq 2\). (The condition \(|c| \geq 2\) is then automatically satisfied, so it need not appear in the eliminant.)
4.2.11
Solution by W. J. Blundon. Crux Math. [5: 176-178].
The given system is equivalent to the following: \[ (ab+1)(x^2+1)(y+1) = (a^2+1)(xy+1)(x+1), \tag{4.2}\] and \[ (ab+1)(y^2+1)(x+1) = (b^2+1)(xy+1)(y+1), \tag{4.3}\] where \(x,y \neq -1\).
We give an exhaustive discussion of the nature of the solution sets over the complex field according to the values of the parameters \(a\) and \(b\).
If \(a = b = \pm i\) (where \(i = \sqrt{-1}\)), every pair \((x,y)\) with \(x \neq -1\) and \(y \neq -1\) is a solution.
If \(a = b \neq \pm i\), we subtract (Equation 4.2) and (Equation 4.3) and obtain an equation equivalent to \[ (x-y)(x+y-2) = 0. \] It is easy to verify that if \(x + y - 2 = 0\), the pair \((x,y)\) is not a solution unless \(x = y = 1\). So the solutions consist of all the pairs \((x,y)\) with \(x = y \neq -1\).
If \(a \neq b\) and \(ab = -1\), then \(a^2 + 1 \neq 0\), \(b^2 + 1 \neq 0\), and we conclude from (Equation 4.2) or (Equation 4.3) that \(xy + 1 = 0\). Here the solutions are all the pairs \((x,y)\) with \(y = -1/x\) and \(x \neq \pm 1\).
We assume from now on that \(a \neq b\) and \(ab \neq - 1\). It follows from (Equation 4.2) and (Equation 4.3) that if \(xy + 1 = 0\) then \(x^2 = -1\) and \(y^2 = -1\). This yields the two solutions \((i,i)\) and \((-i,-i)\). The remaining solutions, which we will now find, are those for which \(xy + 1 \neq 0\). For each such solution, there is a unique nonzero number \(t\) such that \[ (ab+1)(x+1)(y+1) = t(xy+1). \tag{4.4}\]
If we multiply (Equation 4.2) and (4.3) by the nonzero numbers \(x+1\) and \(y+1\), respectively, we obtain equations equivalent to \[ t(x^2+1) = (a^2+1)(x+1)^2 \text{ and } t(y^2+1) = (b^2+1)(y+1)^2. \tag{4.5}\]
Equations (Equation 4.4) and (Equation 4.5) are equivalent to \[ ab(x+1)(y+1) = t(xy+1) - (x+1)(y+1), \tag{4.6}\]
\[ a^2(x+1)^2 = t(x^2+1) - (x+1)^2, \tag{4.7}\]
\[ b^2(y+1)^2 = t(y^2+1) - (y+1)^2. \tag{4.8}\]
We now subtract the square of (4.6) from the product of (4.7) and (4.8) and obtain \[ t(t-2)(x-y)^2 = 0. \]
It can be verified from (Equation 4.2) and (Equation 4.3) that \(x = y\) would lead to \(a = b\) or \(xy + 1 = 0\), both of which are excluded. Since \(t \neq 0\), we must have \(t = 2\). Substituting this value in Equations (4.5), we have no trouble in getting \[ \left(\frac{x-1}{x+1}\right) = a^2 \quad \textrm{and}\quad \left(\frac{y-1}{y+1}\right)^2 = b^2, \] from which \[ x = \frac{1+a}{1-a} \quad \textrm{or} \quad \frac{1-a}{1+a} \] and \[ y = \frac{1+b}{1-b} \quad \textrm{or} \quad \frac{1-b}{1+b}. \tag{4.9}\]
It can be verified from (Equation 4.2) and (Equation 4.3) that of the four pairs \((x,y)\) defined by (4.9), only \[ (x_1, y_1) = \left(\frac{1+a}{1-a},\frac{1+b}{1-b}\right) \quad \textrm{and} \quad (x_2, y_2) = \left(\frac{1-a}{1+a},\frac{1-b}{1+b}\right) \] can be solutions. Note that these are solutions provided no denominator vanishes. If \(a = 1\), for example, then \(b \neq - 1\) (since \(ab \neq - 1\)) and only \((x_2,y_2)\) is a solution.
4.2.12
Solution by W. J. Blundon. Amer. Math. Monthly [64: 680-681].
Subtracting the first term from three times the second gives \[ 2x^2 + \frac{1}{2}y^2 + \frac{3}{4}\left(u-\frac{2}{3}\right)^2 + \frac{3}{5}\left(v-\frac{5}{6}\right)^2 = \frac{15}{4} - D, \] so that \(D\) is not more than \(15/4\). If \(D = 15/4\), every term on the left vanishes, giving the solutions \((x,y,z,u,v) = (0,0,\pm3/2,2/3,5/6)\).
Adding the first to \(3/2\) times the second gives \[ \frac{5}{2}x^2 + \frac{7}{4}y^2 + \frac{3}{2}z^2 + \frac{3}{8}\left(u + \frac{4}{3}\right)^2 + \frac{3}{10}\left(v + \frac{5}{3}\right)^2 = D + 3. \]
Thus \(D\) cannot be less than \(-3\). For \(D = -3\), the unique solution is \((0,0,0,-4/3,-5/3)\). There are no real solutions for values of \(D\) outside the range \(-3 \leq D \leq 15/4\).
4.2.13
Solution by W. J. Blundon. Crux Math. [8: 117].
Let \(z = \lambda e^{-m}\). For any nonnegative integer \(k\), the series \[ S_k(z) = \sum_{n=1}^\infty \frac{n^kz^n}{n!} \] converges absolutely for all complex \(z\) by the ratio test (hence for all \(\lambda\) and \(m\)), and we have to evaluate \(S_3(z)\). Since \[ n^3 = n + 3n(n-1) + n(n-1)(n-2), \] so that, for \(n \geq 3\), \[ \frac{n^3}{n!} = \frac{1}{(n-1)!} + \frac{3}{(n-2)!} + \frac{1}{(n-3)!}, \] we have \[\begin{align*} S_3(z) &= z + 4z^2 + \sum_{n=3}^\infty \left\{\frac{1}{(n-1)!} + \frac{3}{(n-2)!} + \frac{1}{(n-3)!}\right\} z^n \\ &= z + 4z^2 + z \sum_{n=2}^\infty \frac{z^n}{n!} +3z^2 \sum_{n=1}^\infty \frac{z^n}{n!} + z^3 \sum_{n=0}^\infty \frac{z^n}{n!} \\ &= z + 4z^2 + z(e^z - 1 - z) + 3z^2(e^z - 1) + z^3e^z \\ &= z(1+3z+z^2)e^z. \end{align*}\]
In terms of \(\lambda\) and \(m\), we therefore have, for all complex \(\lambda\) and \(m\), \[ S_3(\lambda,n) = \lambda e^{-m}(1+3\lambda e^{-m} + \lambda^2 e^{-2m}) \exp(\lambda e^{-m}). \]
The sum \(S_k(\lambda,m)\) can be evaluated just as easily for any nonnegative integer \(k\).
4.2.14
Solution by W. J. Blundon. Canad. Math. Bull. [5: 197-198].
Using partial fractions and the well-known sum \[ \frac{\lambda^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2}, \] we have \[\begin{align*} \sum_{n=1}^\infty \frac{1}{n^3(n+1)^3} &= 6 \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) + \sum_{n=1}^\infty \left(\frac{1}{n^3} - \frac{1}{(n+1)^3}\right) -3\sum_{n=1}^\infty \frac{1}{n^2} - 3\sum_{n=1}^\infty \frac{1}{(n+1)^2} \\ &= 6 + 1 - 3\left(\frac{\pi^2}{6}\right) - 3\left(\frac{\pi^2}{6} - 1\right) \\ &= 10 - \pi^2. \end{align*}\]
4.2.15
Solution by W. J. Blundon. Canad. Math. Bull. [6: 120-121].
The problem is equivalent to providing for real \(a\), \(b\), \(c\) the inconsistency of the inequalities \[\begin{align*} 5a &< 4b + 4c \\ 5b &< 4c + 5a \\ 5c &< 4a + 4b \\ 4ac &\leq b^2. \end{align*}\]
Case I: \(b = 0\). Addition of the first and third inequalities gives \(a + c < 0\), whereas the second gives \(a + c > 0\). Note that the fourth inequality is not used.
Case II: \(b > 0\). Substitution of \(a = \frac{1}{2}bX\), \(c = \frac{1}{2}bY\) gives \[\begin{align*} -5X + 4Y + 8 &> 0 \\ 2X + 2Y -5 &> 0 \\ 4x - 5Y + 8 &> 0 \\ XY &\leq 1. \end{align*}\]
The first three inequalities of this set define the interior of the triangle with vertices at \((1/2,2)\), \((2,1/2)\), \((8,8)\). It is clear that this triangle lies outside the region defined by \(XY \leq 1\). The inequalities are, therefore, inconsistent. Further, if \(>\) is replaced by \(\geq\), equality will occur only for \((X,Y) = (1/2, 2)\) or \((2,1/2)\). Thus equality in the original relations occurs only when \(b = c = 4a\) or \(b = a = 4c\).
Case III: \(b < 0\). The inequalities are now those of Case II with the sense reversed in the first three inequalities. It follows from the properties of complex polynomial sets that the set of points defined by the transformed inequalities is now empty. Thus the inequalities are again inconsistent. Note that, as in Case I, the fourth inequality is not used.
4.2.16
Solution by Lindsay Reynolds, student in the class of H. L. Ridge, University of Toronto. Crux Math. [4: 79-80].
We will show that, for every nonnegative integer \(n\), there exist nonnegative integers \(a_n\), \(b_n\) verifying \[ a_n^2 - 10b_n^2 = (-1)^n \tag{4.10}\]
such that \[ (\langle\sqrt{10}\rangle)^n = (-1)^n (a_n - b_n \sqrt{10}). \tag{4.11}\]
The required result will then follow by taking \[ I_n = \begin{cases} a_n^2, &\textrm{if }n \textrm{ is odd}; \\ 10b_n^2, &\textrm{if } n \textrm{ is even}. \end{cases} \]
For \(n = 0\), we can choose \(a_0 = 1\), \(b_0 = 0\); these satisfy both (Equation 4.10) and (Equation 4.11). For \(n \geq 0\), assume the existence of \(a_n\), \(b_n\) satisfying (Equation 4.10) and (Equation 4.11). Then \[\begin{align*} (\langle\sqrt{10}\rangle)^{n+1} &= (\langle\sqrt{10}\rangle)(\langle\sqrt{10}\rangle)^n \\ &= (-3 + \sqrt{10}) \cdot (-1)^n (a_n - b_n \sqrt{10}) \\ &= (-1)^{n+1} \left\{(3a_n + 10b_n) - (a_n + 3b_n)\sqrt{10}\right\}. \end{align*}\]
Thus (Equation 4.11) will be satisfied if we choose \[ a_{n+1} = 3a_n + 10b_n, \quad b_{n+1} = a_n + 3b_n; \] and since \[ a_{n+1}^2 - 10b_{n+1}^2 = (3a_n + 10b_n)^2 - 10(a_n + 3b_n)^2 = -(a_n^2 - 10b_n^2) = (-1)^{n+1}, \] our choice satisfies (Equation 4.10) as well as (Equation 4.11), and the induction is complete.
Comment by W. J. Blundon.
Assuming that, for \(n \geq 0\), the existence of an \(I_n\) such that \[ (\langle\sqrt{10}\rangle)^n = (\sqrt{10}-3)^n = \sqrt{I_n + 1} - \sqrt{I_n} \tag{4.12}\]
has been demonstrated (as in the above solution or otherwise), we will find a simple recurrence relation that will generate all \(I_n\) beyond \(I_0 = 0\) and \(I_1 = 9\), which are obvious from (Equation 4.12).
Taking reciprocals in (Equation 4.12) gives \[ (\sqrt{10} + 3)^n = \sqrt{I_n + 1} + \sqrt{I_n} \tag{4.13}\] and adding (Equation 4.12) and (Equation 4.13) and squaring the result yields \[ 4I_n + 2 = \theta^n + \theta^{-n}, \] where \(\theta = (\sqrt{10} - 3)^2 = 19 - 6\sqrt{10}\) (hence \(\theta^{-1} = 19 + 6\sqrt{10}\)). If \(n \geq 1\), we now have \[\begin{align*} 38(4I_n + 2) &= (\theta + \theta^{-1})(\theta^n + \theta^{-n}) \\ &= (\theta^{n+1} + \theta{-n - 1}) + (\theta^{n-1} + \theta^{-n + 1}) \\ &= (4I_{n+1} + 2)(4I_{n-1} + 2), \end{align*}\] which reduces to \[ I_{n+1} = 38I_n - I_{n-1} + 18. \]
Values for \(I_n\) for small \(n\) are set forth in the following table.
| \(n\) | \(I_n\) |
|---|---|
| 0 | 0 |
| 1 | 9 |
| 2 | 360 |
| 3 | 13689 |
| 4 | 519840 |
| 5 | 19740249 |
| 6 | 749609640 |