- 05 Jun '04 00:22All the numbers to follow are positive integers .

For what n is it possible to express n/M as the sum of two fractions with numerator 1 for all M? For what M, for each n, is this decoposition unique?

For example, it is possible for n = 1, and if M is prime the decomposition is unique (are there nonprimes for which it is)? - 05 Jun '04 00:25

Does k/1 count as a fraction?*Originally posted by royalchicken***All the numbers to follow are positive integers .**

For what n is it possible to express n/M as the sum of two fractions with numerator 1 for all M? For what M, for each n, is this decoposition unique?

For example, it is possible for n = 1, and if M is prime the decomposition is unique (are there nonprimes for which it is)? - 05 Jun '04 00:36

In that case, it doesn't work for n at least 3.*Originally posted by royalchicken***Yes, but it will only help you sometimes. Since you took STEP in 2002, you've done a special case of this problem unless you're one of those silly people who even attempts the mechanics ones .**

For n=2, there's the obvious decomposition 2/M = 1/M + 1/M. Similarly 1/M = 1/2M + 1/2M.

As for uniqueness, I'll have to think about that. (I don't remember much about STEP) - 05 Jun '04 00:41 / 1 editYou're right; I made a mistake by saying 'two'. Take that away and you can at least get it big enough since, eg, the harmonic series diverges.

I should also say that they go in lowest terms.

If you leave the 'two' bit, you can still have the parts be distinct though, rather than the 'obvious' 1/m + 1/m... - 06 Jun '04 01:26

Without the "two" bit, n=1 is a trivial solution*Originally posted by royalchicken***You're right; I made a mistake by saying 'two'. Take that away and you can at least get it big enough since, eg, the harmonic series diverges.**

I should also say that they go in lowest terms.

If you leave the 'two' bit, you can still have the parts be distinct though, rather than the 'obvious' 1/m + 1/m...

1/1 = 1/1

1/2 = 1/2

etc - 08 Jun '04 08:54OK, I think I can do this indirectly. I'm assuming 'two or more (but finitely many) distinct' below.

Let f(1) = 1/2, 1/3, 1/6

f(1/(n-1)) = 1/n, 1/n(n-1) for n > 2.

Then f decomposes 1/k into distinct fractions with numerator 1 and larger denominator.

Let S1 = f(1).

By applying f repeatedly to the terms in Sn and the resulting terms, we can obtain another decomposition S(n+1) of 1 that does not involve the terms of S1, ..., Sn. Hence we can decompose any natural number n by induction, by combining S1, ..., Sn. Now multiply the denominator of each term in the resulting decomposition by M, and we have a decomposition of n/M. This decomposition is not unique, as we can certainly apply f to the term with the highest denominator to get a different decomposition satisfying the conditions. - 08 Jun '04 10:30 / 1 editNow for the 'exactly two distinct' case:

Suppose n/M = 1/ck + 1/dk, where c and d are coprime and aren't both 1

<=> cdkn = M(c+d)

Now cd, c+d are coprime; n, M are coprime

<=> cd|M, (c+d)|kn, n|(c+d), M|cdk

<=> cdp = M, nq = (c+d), k = pq

So a decomposition is possible iff we can find coprime factors of M whose sum is a multiple of n (then we can choose k,p,q to fit).

I can't think of a general answer right now, but two things are immediate:

n = 1: decomposition is always possible

M prime: decompositions, if they exist, are unique

Come to think of it,

n = 1: since we can choose any pair of factors, if M is composite then the decomposition is not unique.

n = 2: decomposition is possible for M > 2, since M must have an odd factor f greater than 1, and f+1 is even. The decomposition is unique iff these are the only two odd factors of M => M is a power of 2 times a prime => M is prime.