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.