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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.

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    +1 for @Doug for providing an interesting document, albeit at the irrelevant place.^^ At least from the point of view to solve the problem, we need not this document to back it up.2011-08-21

2 Answers 2

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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.

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    Very nice connection to Erdos-Strauss.2011-08-31
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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.