Diophantine Equation, make a choice

Before tackling the question above, consider the following equation for $a,b,n \in \mathbb{N}$ .

$\begin{aligned} \dfrac{1}{a} + \dfrac{1}{b} &= \dfrac{1}{n} \\ \\ an + bn &= ab \\ \\ ab - an - bn &= 0 \qquad \small \text{[Note the similarity to the expansion of } (x - z)(y - z)] \\ \\ ab - an - bn + n^2 &= n^2 \\ \\ (a - n)(b - n) &= n^2 \end{aligned}$

Thus $a - n$ and $b - n$ must be factor pairs of $n^2$ ; if we additionally require that $a \not= b$ , then we don't consider the pair $(n,n)$ .

For example, when $n = 15$ and excluding the pair $(15,15)$ , the possible factor pairs of $n^2$ are $(1,225), (3,75), (5,45)$ and $(9,25)$ . Then the possible solutions for $(a,b)$ , assuming $a < b$ , are respectively $(16,240), (18,90), (20,60)$ and $(24,40)$ , i.e.

$\dfrac{1}{15} = \dfrac{1}{16} + \dfrac{1}{240} = \dfrac{1}{18} + \dfrac{1}{90} = \dfrac{1}{20} + \dfrac{1}{60} = \dfrac{1}{24} + \dfrac{1}{40}$

$\text{}$

To tackle the actual question, we start with a couple of observations: $\text{} \\ \text{}$

• $\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} < 1$ so there can be no solutions with $a \ge 3 \\$

• $\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} > 1 \$ and $\ \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{8} < 1$ so there can be no solutions with $b \ge 5 \\$

So any solutions would have $a = 2$ and either $b = 3$ or $b = 4$ . We consider the two cases separately. $$

Case 1: $a = 2, b = 3 \qquad$ We need $\dfrac{1}{c} + \dfrac{1}{d} = \dfrac{1}{6}$ .

Excluding $(6,6)$ , the factor pairs of $6^2$ are $(1,36), (2,18), (3,12)$ and $(4,9)$ so the possible solutions for $(c,d)$ are $(7,42), (8,24), (9,18)$ and $(10,15) \\$ .

Case 2: $a = 2, b = 4 \qquad$ We need $\dfrac{1}{c} + \dfrac{1}{d} = \dfrac{1}{4}$ .

Excluding $(4,4)$ , the only factor pairs of $4^2$ are $(1,16)$ and $(2,8)$ so the only possible solutions for $(c,d)$ are $(5,20)$ and $(6,12)$ .

$$

Finally, the possible solutions for the given equation are

$\begin{aligned} 1 &= \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{42} \\ \\ &= \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{8} + \dfrac{1}{24} \\ \\ &= \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{18} \\ \\ &= \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{10} + \dfrac{1}{15} \\ \\ &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{20} \\ \\ &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} \end{aligned}$

Because a<b<c<d,

1= 1/a + 1/b + 1/c + 1/d < 4/a

Therefore, a<4 : a = 2 or 3

i) a = 3

2/3 = 1/b + 1/c + 1/d < 3/b

b<4.5

so, b = 4

then, 1/c + 1/d = 5/12 : 4<c<d : 1/c + 1/d is less than 1/5+1/6 =11/30 : no solution

ii) a = 2

1/2 = 1/b + 1/c + 1/d < 3/b

b<6

so, b = 3, 4, 5

1)) b=3

1/c + 1/d = 1/6 : cd-6c-6d=0 : (c-6)(d-6)=36 : c-6 : 1,2,3,4 : 4 solution

2)) b=4

1/c + 1/d = 1/4 : cd-4c-4d=0 : (c-4)(d-4)=16 : c-4 : 1.2 : 2 solution

3)) b=5

1/c + 1/d = 3/10 : 3cd-10c-10d = 0 : (3c-10)(3d-10)=100 : no solution

so, by i), ii) : we have 6 solutions