Given $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ , which $a,b,c ,d$ are an positive integer , all solution of $a,b,c,d$ are members of Set "S" ,then find sum all of members in set "S"
I don't know how to start to solve this problem but I've got 2 solution that is $\frac{2}{52} + \frac{3}{598} + \frac{4}{437} + \frac{5}{266} = \frac{1}{14}$ and $\frac{2}{224} + \frac{3}{304} + \frac{4}{437} + \frac{5}{115} = \frac{1}{14}$
question 2 If problem say that $a,b,c,d$ are an all integer , they will have negative integer to this solution ? please give me hint or theorem relevant. Thank you in advance .
Hint: For first question you can use following algorithm; it is a bit long operation.
$2/a+3/b+4/c+5/d=1/14$
we rewrite the equation as:
$$\frac{1}{\frac{a}{28}}+\frac{1}{\frac{b}{42}}+\frac{1}{\frac{c}{56}}+\frac{1}{\frac{d}{70}}=1$$
Now we solve equation:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1$$
Number of terms is 4 ,so:
$\frac{1}{x}≥\frac{1}{4}$ ⇒ $x≤4$, let $x=2$ then we have:
$\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1-\frac{1}{2}=\frac{1}{2}$
$\frac{1}{y}<\frac{1}{2}$⇒ $y>2$, meanwhile $\frac{3}{y}≥\frac{1}{2}$ so $y≤6$ so $3≤y≤6$; let $y=3$ then we have:
$\frac{1}{z}+\frac{1}{t}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$
$\frac{1}{z}<\frac{1}{6}$⇒ $z>6$. Meanwhile, $\frac{3}{z}≥\frac{1}{6}$⇒$z<18$
, so $6<z<18$ , let $z=7$ then we must have:
$\frac{1}{t}=\frac{1}{6}-\frac{1}{7}=\frac{1}{42}$⇒ $t=42$ .
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1$
So following numbers can be a set of solutions to your equation:
$a=2\times 28=56$
$b=3\times 42=126$
$c=7\times 56=392$
$d=42\times 70=2940$
x can be $2, 3, 4$, each of these numbers leads to one set of solutions. y can be $3, 4, 5, 6 $ , z can be $ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18$ among which certain values give integer solutions for t.
Note that equation of parameters x, y, x and t is symmetric for initial equation i.e each term of this equation can be equal to any term of initial equation.