Chapter 19

Of three workmen, \(A\) can finish a given job once in three weeks, \(B\) can finish it three times in eight weeks, while \(C\) can finish it five times in twelve weeks. How long will it take for the three workmen to complete the job together? (This exercise and the next two are from Newton's Universal Arithmetic.)

Suppose that the distance between London and Edinburgh is 360 miles and that a courier for London sets out from the Scottish city running at \(10 \mathrm{mph}\) at the same time that one sets out from the English capital for Edinburgh at \(8 \mathrm{mph}\). Where will the couriers meet? (This exercise and the next two are from Maclaurin's Treatise of Algebra.)

Derive Cramer's rule for three equations in three unknowns from the rule for two equations in two unknowns: Given the system $$ \begin{aligned} &a x+b y+c z=m \\ &d x+e y+f z=n \\ &g x+h y+k z=p \end{aligned} $$ solve each equation for \(x\) in terms of \(y\) and \(z\), then form two equations in those variables and solve for \(z\). Finally, determine \(y\) and \(x\) by substitution.

A company dining together find that the bill amounts to \$175. Two were not allowed to pay. The rest found that their shares amounted to \(\$ 10\) per person more than if all had paid. How many were in the company?

Twenty persons, men and women, dine at a tavern. The share of the bill for one man is \(\$ 8\), for one woman \(\$ 7\), and the entire bill amounts to \(\$ 145\). Required, the number of men and women separately. (This exercise and the next two are from Euler's Introduction to Algebra.)

A horse dealer bought a horse for a certain number of crowns and sold it again for 119 crowns, by which means his profit was as much per cent as the horse cost him. What was the purchase price?

Three brothers bought a vineyard for \(\$ 100\). The youngest says that he could pay for it alone if the second gave him half the money which he had; the second says that if the eldest would give him only a third of his money, he could pay for the vineyard singly; lastly, the eldest asks only a fourth part of the money of the youngest, to pay for the vineyard himself. How much money did each have?

Factor Leibniz's polynomial \(x^{4}+a^{4}\) into two real quadratic polynomials. ( Hint: Add and subtract \(2 a^{2} x^{2}\).)

Factor \(x^{5}-1\) into linear and real quadratic factors.

Nicolaus Bernoulli claimed that the polynoma1 \(x^{2}-4 x=\) \(2 x^{2}+4 x+4\) could not be factored into a product of two quadratic polynomials. Show that in fact the factors are $$ \left(x^{2}-(2+\sqrt{4+2 \sqrt{7}}) x+(1+\sqrt{4+2 \sqrt{7}}+\sqrt{7})\right) $$ and $$ \left(x^{2}-(2-\sqrt{4+2 \sqrt{7}}) x+(1-\sqrt{4+2 \sqrt{7}}+\sqrt{7})\right) $$.

Use Euler's procedure from his proof that all real quartics factor to determine the factorization of \(x^{4}-2 x^{2}+8 x-3\) as a product of two quadratic polynomials.

Consider the following system of linear equations given by Euler: $$ \begin{array}{r} 5 x+7 y-4 z+3 v-24=0 \\ 2 x-3 y+5 z-6 v-20=0 \\ x+13 y-14 z+15 v+16=0 \\ 3 x+10 y-9 z+9 v-4=0 \end{array} $$ Show that these four equations are "worth only two," so that they do not determine a unique 4 -tuple as a solution.

Show that if \(n\) is prime, then the roots of \(x^{n}-1=0\) can all be expressed as powers of any such root \(\alpha \neq 1\).

Let \(x_{1}, x_{2}\) be the two roots of the quadratic equation \(x^{2}+\) \(b x+c=0\). Since \(t=x_{1}+x_{2}\) is invariant under the two permutations of the two roots, while \(v=x_{1}-x_{2}\) takes on two distinct values, \(v\) must satisfy an equation of degree 2 in \(t\). Find the equation. Similarly, \(x_{1}\) is invariant under the same permutations as \(x_{1}-x_{2}\). Thus, \(x_{1}\) can be expressed rationally in terms of \(x_{1}-x_{2}\). Find such a rational expression. Use the rational expression and equation you found to "solve" the original quadratic equation.

Determine the three roots \(x_{1}, x_{2}, x_{3}\) of \(x^{3}-6 x-9=0\) Use Lagrange's procedure to find the sixth-degree equation satisfied by \(y\), where \(x=y+2 / y\). Determine all six solutions of this equation and express each explicitly as \(\frac{1}{3}\left(x^{\prime}+\omega x^{\prime \prime}+\omega^{2} x^{\prime \prime \prime}\right)\), where \(\left(x^{\prime}, x^{\prime \prime}, x^{\prime \prime \prime}\right)\) is a permutation of \(\left(x_{1}, x_{2}, x_{3}\right)\) and \(\omega\) is a complex root of \(x^{3}-1=0\).

Show that the expression \(x_{1} x_{2}+x_{3} x_{4}\) takes on only three distinct values under the 24 permutations of four elements.

Let \(x_{1}, x_{2}, x_{3}, x_{4}\) denote the four roots of the quartic equation \(x^{4}+a x^{3}+b x^{2}+c x+d=0\). Set \(\alpha=x_{1} x_{2}+x_{3} x_{4}\), \(\beta=x_{1} x_{3}+x_{2} x_{4}, \gamma=x_{1} x_{4}+x_{2} x_{3} .\) Show that \(\alpha+\beta+\) \(\gamma=b\), that \(\alpha \beta+\alpha \gamma+\beta \gamma=a c-4 d\), and that \(\alpha \beta \gamma=\) \(a^{2} d+c^{2}-4 b d\). Show that this implies that \(\alpha, \beta\), and \(\gamma\) are the roots of the cubic equation \(y^{3}-b y^{2}+(a c-4 d) y-\) \(\left(a^{2} d+c^{2}-4 b d\right)=0\).

In Euler's proof of the case \(n=3\) of Fermat's Last Theorem, show that \(\frac{1}{4} p\) and \(p^{2}+3 q^{2}\) are relatively prime if \(p\) is not divisible by 3 . (Recall that \(x=p+q, y=p-q\) are odd and relatively prime.)

In Euler's proof of the case \(n=3\) of Fermat's Last Theorem, we now consider the situation where \(p=3 r\). We then know that \(\frac{3}{4} r\left(9 r^{2}+3 q^{2}\right)=\frac{9}{4} r\left(3 r^{2}+q^{2}\right)\) must be a cube. Show that the two factors in this expression are relatively prime. It follows that each must be a cube. In particular, \(q^{2}+3 r^{2}\) must be a cube. Factor this expression as in the text, using complex numbers of the form \(a+b \sqrt{-3}\), and conclude that \(q=t\left(t^{2}-9 u^{2}\right), r=3 u\left(t^{2}-u^{2}\right)\), where \(t\) is odd and \(u\) is even. Also, since \(\frac{9}{4} r\) is a cube, show that \(\frac{2}{3} r=2 u(t+\) u) \((t-u)\) is a cube where the factors are relatively prime. Conclude as in the case detailed in the text that we can now find three integers smaller than the original set for which the sum of their cubes is a cube.

Calculate the distinct residues \(1, \alpha, \beta, \ldots\) of \(1,5,5^{2}, \ldots\) modulo 13 . Then pick a nonresidue \(x\) of the sequence of powers and determine the coset \(x, x \alpha, x \beta, \ldots\) Continue to pick nonresidues and determine the cosets until you have divided the group of all 12 nonzero residues modulo 13 into nonoverlapping subsets, the cosets of the group of powers of \(5 .\)

Determine the quadratic residues modulo 13 .

Prove that \(-1\) is a quadratic residue with respect to a prime \(q\) if and only if \(q \equiv 1(\bmod 4)\).

Benjamin Banneker was fond of solving mathematical puzzles and recorded many in his notebook, including his own version of the old hundred fowls problem: A gentleman sent his servant with \(£ 100\) to buy 100 cattle, with orders to give \(£ 5\) for each bullock, 20 shillings for each cow, and 1 shilling for each sheep. (Recall that 20 shillings equals \(£ 1\).) What number of each sort of animal did he bring back to his master? \(?^{23}\)

Divide 60 into four parts such that the first increased by 4, the second decreased by 4, the third multiplied by 4, and the fourth divided by 4 shall each equal the same number (Banneker).

Suppose a ladder 60 feet long is placed in a street so as to reach a window on one side 37 feet high, and without moving it at the bottom, to reach a window on the other side 23 feet high. How wide is the street? (Banneker))