Interesting Egyptian Fraction Problem

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.

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Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. gauravtiwari.org is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

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