Representing every positive rational number in the form of (an+bn)/(cn+dn)(a^n+b^n)/(c^n+d^n)

About a month ago, I got the following :

For every positive rational number r, there exists a set of four positive integers (a,b,c,d) such that
r=a3+b3c3+d3.

For r=p/q where p,q are positive integers, we can take
(a,b,c,d)=(3ps3t+9qt4, 3ps3t9qt4, 9qst3+ps4, 9qst3ps4)
where s,t are positive integers such that 3<r(s/t)3<9.

For r=2014/89, for example, since we have (2014/89)(2/3)36.7, taking (p,q,s,t)=(2014,89,2,3) gives us 201489=2098893+801273754783+110303.

Then, I began to try to find every positive integer n such that the following proposition is true :

Proposition : For every positive rational number r, there exists a set of four positive integers (a,b,c,d) such that r=an+bncn+dn.

The followings are what I’ve got. Let r=p/q where p,q are positive integers.

  • For n=1, the proposition is true. We can take (a,b,c,d)=(p,p,q,q).

  • For n=2, the proposition is false. For example, no such sets exist for r=7/3.

  • For even n, the proposition is false because the proposition is false for n=2.

However, I’ve been facing difficulty in the case of odd n5. I’ve tried to get a similar set of four positive integers (a,b,c,d) as the set for n=3, but I have not been able to get any such set. So, here is my question.

Question : How can we find every odd number n5 such that the following proposition is true?

Proposition : For every positive rational number r, there exists a set of four positive integers (a,b,c,d) such that r=an+bncn+dn.

Update : I posted this question on MO.

Added : Problem N2 of IMO 1999 Shortlist asks the case n=3.

Answer

For the above problem there are four sets of solutions (this is intuitive: for a, b, c, & d). In the case of positive rational r and any odd number n we can eliminate all but one of the solutions:

dn=5cn=1an+bn=30r=5anZ

In the case of any odd number n≥3 we refer to the generating function:

a2n+1+b2n+1=30c2n+1=1d2n+1=5r=5a2n+1Z

As well as the case of every odd number n≥5 (et. al):

a2n+3+b2n+3=30c2n+3=1d2n+3=5r=5a2n+3Z

Quickly we discover that it doesn’t matter the value of n, as long as it’s odd and positive, leading to the generalization:

r=c51a2n+1+b2n+1=(c1+c4+1)(c5+1)c2n+1+c3=c1+c2+1c2+d2n+1=c3+c4(c5|c4|c3|c2|c1|a2n+1)Z

For all n:

r=c51

an+bn=(c1+c4+1)(c5+1)

cn+c3=c1+c2+1

c2+dn=c3+c4

(c5|c4|c3|c2|c1|an)Z

Note: this isn’t a complete answer so it might be more appropriate as a comment, but pending reputation I may as well take a naive crack at it. Excuse any abuse of notation or lack of comprehension–it’s been over a decade since I’ve had any formal mathematics. Lastly, I welcome criticism, especially if it’s informative and friendly!

Attribution
Source : Link , Question Author : mathlove , Answer Author : David Scott Kirby

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