4.3: Factors, roots, polynomials and surds

Problem 113

(a) (i) Find a prime number which is one less than a square.

(ii) Find another such prime.

(b) (i) Find a prime number which is one more than a square.

(ii) Find another such prime.

(c) (i) Find a prime number which is one less than a cube.

(ii) Find another such prime.

(d) (i) Find a prime number which is one more than a cube.

(ii) Find another such prime.

Problem 114 Factorise x⁴ + 1 as a product of two quadratic polynomials with real coefficients.

4.3.1 Standard factorisations

The challenge to factorise unfamiliar expressions, may at first leave us floundering. But if we assume that each such problem is solvable with the tools at our disposal, we then have no choice but to fall back on the standard tools we have available (in particular, the standard factorisation of a difference of two squares, in which “cross terms” cancel out). The next problem extends this basic repertoire of standard factorisations.

Problem 115

(a)(i) Factorise a³ − b³.

(ii) Factorise a⁴ − b⁴ as a product of one linear factor and one factor of degree 3, and as a product of two linear factors and one quadratic factor.

(iii) Factorise aⁿ − bⁿ as a product of one linear factor and one factor of degree n − 1.

(b)(i) Factorise a³ + b³.

(ii) Factorise a⁵ + b⁵ as a product of one linear factor and one factor of degree 4.

(iii) Factorise a²ⁿ⁺¹ + b²ⁿ⁺¹ as a product of one linear factor and one factor of degree 2n.

Problem 115 develops the ideas that were implicit in Problem 113. The clue lies in Problem 113(a), and in the comment made in the main text in Chapter 1 (after Problem 4 in Chapter 1), which we repeat here:

“The last part [of Problem 113(a)] is included to emphasise a frequently neglected message:

Words and images are part of the way we communicate.
But most of us cannot calculate with words and images.

To make use of mathematics, we must routinely translate words into symbols. So “numbers” need to be represented by symbols, and points in a geometric diagram need to be properly labelled before we can begin to calculate, and to reason, effectively.”

As soon as one reads the words “one less than a square”, one should instinctively translate this into the form “x² − 1”. Bells will then begin to ring; for it is impossible to forget the factorisation

$$x^2-1=(x-1)(x+1).$$

And it follows that:

for a number that factorises in this way to be prime, the smaller factor x − 1 must be equal to 1;
$\therefore x=2$ , so there is only one such prime.

The integer factorisations in Problem 113(c) − namely

$$3^3-1=2\times 13,4^3-1=3\times 21,5^3-1=4\times 31,6^3-1=5\times 43,\dots$$

may help one to remember (or to discover) the related factorisation

$$x^3-1=(x-1)\left(x^2+x+1\right).$$

∴ For a number that factorises in this way to be prime, the smaller factor “x − 1” must be equal to 1;
∴ x = 2, so there is only one such prime.

Problem 113 parts (a) and (c) highlight the completely general factorisation (Problem 115(a)(iii)):

$$x^n-1=(x-1)\left(x^{n-1}+x^{n-2}+\cdots +x^2+x+1\right).$$

This family of factorisations also shows that we should think about the factorisation of x² − 1 as (x − 1)(x + 1), with the uniform factor (x − 1) first (rather than as (x + 1)(x − 1)). Similarly, the results of Problem 115 show that we should think of the familiar factorisation of a² − b² as ^{(a − b)(a} + b), (not as (a ' b)(a − b), but always with the factor (a − b) first).

The integer factorisations in Problem 113(d) − namely

$$3^3+1=4\times 7,4^3+1=5\times 13,5^3+1=6\times 21,6^3+1=7\times 31,7^3+1=8\times 43,\dots$$

may help one to remember (or to discover) the related factorisation

$$x^3+1=(x+1)\left(x^2-x+1\right)$$

∴ For such a number to be prime, one of the factors must be equal to 1.

This time one has to be more careful, because the first bracket may not be the “smaller factor” – so there are two cases to consider:

(i) if x + 1 = 1 , then x = 0, and x³ + 1 = 1 is not prime;

(ii) if x² − x = 1 then x = 0 or x = 1, so x = 1 and we obtain the prime 2 as the only solution.

The factorisation for x³ + 1 works because “3 is odd”, which allows the alternating +/− signs to end in a “+” as required. Hence Problem 113(d)(iii) highlights the completely general factorisation for odd powers:

$$x^{2n+1}+1=(x+1)\left(x^{2n}-x^{2n-1}+x^{2n-2}-\cdots +x^2-x+1\right)$$

You probably know that there is no standard factorisation of x² + 1, or of x⁴ + 1 (but see Problem 114 above).

Problem 116

(a) Derive a closed formula for the sum of the geometric series

$$1+r+r^2+r^3+\cdots +r^n.$$

(The meaning of closed formula was discussed in the Note to the solution to Problem 54(b) in Chapter 2.)

(b) Derive a closed formula for the sum of the geometric series

$$a+ar+a{r}^2+a{r}^3+\cdots +a{r}^n$$

We started this subsection by looking for prime numbers of the form x² − 1. A simple-minded approach to the distribution of prime numbers might look for formulae that generate primes - all the time, or infinitely often, or at least much of the time. In Chapter 1 (Problem 25) you showed that no prime of the form 4k + 3 can be “represented” as a sum of two squares (i.e. in the form “x² + y²”), and we remarked that every other prime can be so represented in exactly one way. It is true (but not obvious) that roughly half the primes fall into the second category; so it follows that substituting integers for the two variables in the polynomial x² + y² produces a prime number infinitely often.

Problem 117 Experiment suggests, and Goldbach (1690–1764) showed in 1752 that no polynomial in one variable, and with integer coefficients, can give prime values for all integer values of the variable. But Euler (1707–1783) was delighted when he discovered the quadratic

$$f(x)=x^2+x+41.$$

Clearly f (0) = 41 is prime. And f (1) = 43 is also prime. What is the first positive integer n for which f (n) is not prime?

Problem 117 should be seen as a particular instance of the question as to whether prime numbers can be captured by a polynomial with integer coefficients, and in particular by a quadratic. The next two problems consider the simplest instances of representing prime numbers by expressions involving exponentials (that is, where the variable is in the exponent).

Problem 118

(a)(i) Suppose aⁿ − 1 = p is a prime. Prove that a = 2 and that n must itself be prime.

(ii) How many primes are there among the first five such numbers

$$2^2-1,2^3-1,2^5-1,2^7-1,2^{11}-1?$$

(b) (i) Suppose aⁿ + 1 = p is a prime. Prove that either a = 1, or a must be even and that n must then be a power of 2.

(ii) In the simplest case, where a = 2, how many primes are there among the first five such numbers

$$2^1+1,2^2+1,2^4+1,2^8+1,2^{16}+1?$$

Primes of the form 2^p − 1 are called Mersenne primes (after Marin Mersenne (1588–1648)). We now know at least fifty such primes (with the exponent p ranging up to around 80 million). Finding new primes is not in itself important, but the search for Mersenne primes has been used as a focus for many new developments in programming, and in number theory.

Primes of the form 2ⁿ + 1 are called Fermat primes (after Pierre de Fermat (1601–1665)). The story here is very different. We now refer to the number 2ⁿ + 1 with n = 2^k as the k^th Fermat number f_k. You showed in Problem 118 (as Fermat did himself) that f₀, f₁, f₂, f₃, f₄ are all prime. Fermat then rather rashly claimed that f_n is always prime. However, Euler showed (100 years later) that the very next Fermat number f₅ fails to be prime. And despite all the power of modern computers, we have still not found another Fermat number that is prime!

4.3.2 Quadratic equations

The general solution of quadratic equations dates back to the ancient Babylonians ($\approx 1700\text{BC}$ ). Our modern understanding depends on two facts:

• an equation of the form x² = a where a > 0, has exactly two solutions: $x=\pm \sqrt{a}$ ;
• any product X . Y is equal to 0 precisely when one of the two factors X, Y is equal to 0.

Problem 119 Solve the following quadratic equations:

(a) $x^2-3x+2=0$

(b) $x^2-1=0$

(c) $x^2-2x+1=0$

(d) $x^2+\sqrt{2}x-1=0$

(e) $x^2+x-\sqrt{2}=0$

(f) $x^2+1=0$

(g) $x^2+\sqrt{2}x+1=0$

Problem 120 Let

$$p(x)=x^2+\sqrt{2}x+1.$$

Find a polynomial q(x) such that the product p(x)q(x) coefficients.

Problem 121

(a) I am thinking of two numbers, and am willing to tell you their sum s and their product p. Express the following procedure algebraically and explain why it will always determine my two unknown numbers.

Halve the sum s, and square the answer.

Then subtract the product p and take the square root of the result, to get the answer.

Add “the answer” to half the sum and you have one unknown number; subtract “the answer” from half the sum and you have the other unknown number.

(b) I am thinking of the length of one side of a square. All I am willing to tell you are two numbers b and c, where when I add b times the side length to the area I get the answer c. Express the following procedure algebraically and explain why it will always determine the side length of my square.

Take one half of b, square it and add the result to c.

Then take the square root.

Finally subtract half of b from the result.

(c) A regular pentagon ABCDE has sides of length 1.

(i) Prove that the diagonal AC is parallel to the side ED.

(ii) If AC and BD meet at X, explain why AXDE is a rhombus.

(iii) Prove that triangles ADX and CBX are similar.

(iv) If AC has length x, set up an equation and find the exact value of x.

Problem 121(a), (b) link to Problem 111(a) (and to Problem 129 below), in relating the roots and the coefficients of a quadratic. If we forget for the moment that the coefficients are usually known, while the roots are unknown, then we see that if α and β are the roots of the quadratic

$$x^2+bx+c,$$

then

$$(x-\alpha )(x-\beta )=x^2+bx+c,$$

so

$$\alpha +\beta =-b\mathrm{and}\alpha \beta =c.$$

In other words, the two coefficients b, c are equal to the two simplest symmetric expressions in the two roots α and β. Part (a) of the next problem is meant to suggest that all other symmetric expressions in α and β (that is, any expression that is unchanged if we swap α and β) can then be written in terms of b and c. The full result proving this fact is generally attributed to Isaac Newton (1642–1727). Part (b) suggests that, provided one is willing to allow case distinctions, something similar may be true of anti-symmetric expressions (where the effect of swapping a and ft is to multiply the expression by “−1”).

Problem 122 Let α and β be the roots of the quadratic equation

$$x^2+bx+c=0.$$

(a) (i) Write α² + β² in terms of b and c only.

(ii) Write α²β + β²α in terms of b and c only.

(iii) Write α³ + β³ − 3αβ in terms of b and c only.

(b) (i) Write α − β in terms of b and c only.

(ii) Write α²β − β²α in terms of b and c only.

(iii) Write α³ − β³ in terms of b and c only.

Problem 123 (Nested surds, simplification of surds)

(a)(i) For any positive real numbers a, b, prove that $$\sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4ab}}$$

(ii) Simplify $\sqrt{5+\sqrt{24}}$ .

(b) (i) Find a similar formula for $\sqrt{a}-\sqrt{b}$ .

(ii) Simplify $\sqrt{5-\sqrt{16}}$ and $\sqrt{6-\sqrt{20}}$.

Problem 124 (Integer polynomials with a given root) We know that α = 1 is a root of the polynomial equation x² − 1 = 0; that $\alpha =\sqrt{2}$ is a root of x² − 2 = 0; and that $\alpha =\sqrt{3}$ is a root of x² − 3 = 0.

(a) Find a quadratic polynomial with integer coefficients which has

$$\alpha =1+\sqrt{2}$$

as a root.

(b) Find a quadratic polynomial with integer coefficients which has

$$\alpha =1+\sqrt{3}$$

as a root.

(c) Find a polynomial with integer coefficients which has

$$\alpha =\sqrt{2}+\sqrt{3}$$

as a root. What are the other roots of this polynomial?

(d) Find a polynomial with integer coefficients which has

$$\alpha =\sqrt{2}+\frac{1}{\sqrt{3}}$$

as a root. What are the other roots of this polynomial?

Problem 125

(a) Prove that the number $\sqrt{2}+\sqrt{3}$ is irrational.

(b) Prove that the number $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational.

Problem 126 (Polynomial long division) Find

(i) the quotient and the remainder when we divide x¹⁰ + 1 by x³ − 1

(ii) the remainder when we divide x²⁰¹³ + 1 by x² − 1

(iii) the quotient and the remainder when we divide x^m + 1 by xⁿ − 1, for $m>n⩾1$ .

Problem 127 Find the remainder when we divide x²⁰¹³ + 1 by x² + x + 1.