Here is an interesting mathematical puzzle alike problem involving the use of Egyptian fractions, whose solution sufficiently uses the basic algebra.

Problem

Let a, b, c, d and e be five non-zero complex numbers, and;

$ a + b + c + d + e = -1$ … (i)

$ a^2+b^2+c^2+d^2+e^2=15$ …(ii)

$ \dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} +\dfrac{1}{d} +\dfrac{1}{e}= -1$ …(iii)

$ \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{1}{e^2}=15$ …(iv)

$ abcde = -1 $ …(v).

Solve and find the values of a, b, c, d and e.

Remark

According to the problems of solution of algebraic equations, we know that, if $ \alpha, \beta, \ldots , \delta, \ldots$ are the roots of an equation, then $ (x-\alpha)(x-\beta) \ldots (x-\delta) \ldots =0$ .

Solution

The algebraic equation with roots (a, b, c, d, e ) is $ (x – a)(x – b)(x – c)(x – d)(x – e)=0$

After multiplying terms to each other, we get the following polynomial:
$ x^5-(a + b + c + d + e) \cdot x^4+(ab+ac+ad+ae+bc+bd+be+cd+ce+de)\cdot x^3-(abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde) \cdot x^2+(abcd+abce+abde+acde+bcde) \cdot x -abcde=0$
… (vi)

Squaring both sides of equation (i), we get

$ a^2+b^2+c^2+d^2+e^2+ 2ab + 2ac + 2ad + 2ae + 2bc + 2bd + 2be + 2cd + 2ce + 2de = 1$
Or,
$ a^2+b^2+c^2+d^2+e^2+ 2(ab + ac + ad + ae + bc + bd + be + cd + ce + de) = 1$ …(vii)

Subtracting equation (ii) from the equation (vii)

$ 2(ab + ac + ad + ae + bc + bd + be + cd + ce + de) = -14$
Or, $ ab+ac+ad+ae+bc+bd+be+cd+ce+de=-7$ …(viii)

Now, multiplying equation (iii) by (v) :: multiplying left side by abcde and right side by -1:: we have

$ bcde + acde + abde + abce + abcd =1$
or, $ abcd+abce+abde+acde+bcde=1$ … (ix)

Again, multiplying equation (iv) by the square of equation (v), we get

$ (bcde)^2 + (acde)^2+ (abde)^2+ (abce)^2+ (abcd)^2= 15$

Or, $ (abcd)^2+ (bcde)^2+ (cdea)^2+ (deab)^2+ (eabc)^2= 15$ … (x)

Squaring $ abcd + bcde +cdea +deab + eabc = 1$ we get

$ (abcd)^2+(bcde)^2+(cdea^2)+(deab)^2+(eabc)^2+2abcde \cdot (abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde)=1$ … (xi)

Substitute known values in (xi):

$ 15 – 2(abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde) = 1$

Or, $ abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde =7$ … (xii)

Putting values from equation, (i), (viii), (xii), (ix) and (v) respectively, to the equation (vi), we obtain the algebraic equation
$ x^5+x^4-7x^3-7x^2+x+1=0$ … (xiii).

Now, Equation (xiii) has the solutions a, b, c, d and e.

Factorizing (xiii) we get:
$ (x + 1)(x^4- 7x^2 + 1) = 0$
$ x = -1 $ or $ x^4- 7x^2+ 1 = 0$
As, $ x^4- 7x^2+ 1 = 0$ has roots:

  1. $ \dfrac{1}{2} (3+\sqrt{5})$
  2. $ \dfrac{1}{2} (3-\sqrt{5})$
  3. $ \dfrac{1}{2} (-3+\sqrt{5})$
  4. $ \dfrac{1}{2} (-3-\sqrt{5})$

Thus, a, b, c, d and e are

  • $ -1$
  • $ \dfrac{1}{2} (3+\sqrt{5})$
  • $ \dfrac{1}{2} (3-\sqrt{5})$
  • $ \dfrac{1}{2} (-3+\sqrt{5})$
  • $ \dfrac{1}{2} (-3-\sqrt{5})$

irrespective of any distinct order due to their symmetry.

You May Also Like

On Ramanujan’s Nested Radicals

Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory as well as in Numerical Methods . The simple binomial theorem of degree 2 can be written as: $ {(x+a)}^2=x^2+2xa+a^2 \…

Three Children, Two Friends and One Mathematical Puzzle

Two close friends, Robert and Thomas, met again after a gap of several years. Robert Said: I am now married and have three children. Thomas Said: That’s great! How old they are? Robert: Thomas! Guess it yourself with some clues provided by me. The product of the ages of my children is 36. Thomas: Hmm… Not so helpful clue. Can…

Some good, OK, and useless revision techniques

Exam have been haunting student since forever, and although you’re willing to do whatever you can to retain essential information, sometimes you end up spending weeks studying with useless revision techniques. We’re accustomed to employ our own techniques when it comes to studying such as making sticky notes, highlighting, or drawing charts. However, recent studies conducted in the US have…

The Cattle Problem

This is a famous problem of intermediate analysis, also known as ‘Archimedes’ Cattle Problem Puzzle’, sent by Archimedes to Eratosthenes as a challenge to Alexandrian scholars. In it one is required to find the number of bulls and cows of each of four colors, the eight unknown quantities being connected by nine conditions. These conditions ultimately form a Pell equation…

Dedekind’s Theory of Real Numbers

Intro Let $ \mathbf{Q}$ be the set of rational numbers. It is well known that $ \mathbf{Q}$ is an ordered field and also the set $ \mathbf{Q}$ is equipped with a relation called “less than” which is an order relation. Between two rational numbers there exists an infinite number of elements of $ \mathbf{Q}$. Thus, the system of rational numbers seems…

Everywhere Continuous Non-differentiable Function

Weierstrass had drawn attention to the fact that there exist functions which are continuous for every value of $ x$ but do not possess a derivative for any value. We now consider the celebrated function given by Weierstrass to show this fact. It will be shown that if $ f(x)= \displaystyle{\sum_{n=0}^{\infty} } b^n \cos (a^n \pi x) \ \ldots (1)…