Sierpinski's Conjecture The Sierpinski's conjecture states that for all integer $n>1$, we have $\frac{5}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ where $(a,b,c) \in \mathbb{N}_*^3$.
But is it easier to prove that $\frac{5}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ ?
Thanks,
B.L. 
 A: Well, first of all, this conjecture is implied by the Erdos-Straus Conjecture $(\forall n>1, \frac{4}{n} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ for some $(a,b,c) \in \mathbb{N}_*^3)$.  So any counterexamples must also be counterexamples of Erdos-Straus -- and hence very unlikely.
But specifically, any counterexample must have $n \equiv 1 \mod 24$.
Furthermore, $n \neq 0 \mod 5$ for obvious reasons.
Next, if $n = 5k + 4$, take $a = k+1, b = n*a$.  (We can ignore c and d, because we can always replace b by b+1 and set c = b*(b+1) and similarly for d.)
If $n = 5k + 3$, take $a = b = c = 3k+2$ and $d = n*a$.
If $n = 5k + 2$, take $a = b = 2k+1$ and $c = n*a$.
So we're left with $n = 1 \mod 120$ as potential counterexamples.
A: One might suspect this to be the case, especially since it's known that $\frac{5}{n}$ can be written as the sum of five distinct unit fractions. (See, for example, page 7 of http://kevingong.com/Math/EgyptianFractions.pdf ,)  However, Sierpinski's conjecture is still open and no one seems to have attempted B.L.'s four-term variation, so it's not known which statement is easier to prove or even if either statement is true.  
