RMO 2012 Solution to Question No. 2

2. Let a, b, c be positive integers such that a divides \(b^5\) , b divides \(c^5\) and c divides \(a^5\). Prove that abc divides \((a+b+c)^{31}\).

Solution:

A general term of the expansion of \((a+b+c)^{31}\) is \(\frac {31!}{p!q!r!} a^p b^q c^r\) where p+q+r = 31 (by multinomial theorem; this may reasoned as following: from 31 factors (a+b+c), choose p factors and from those chosen p factors take out 'a'. From remaining 31-p factors choose q factors and from these chosen q factors take out 'b'. From the remaining r factors take out 'c'.)

Now the terms of the expansion can be of three types:

Case 1

p, q and r are all non-zero. These terms are straight away divisible by abc as all of a, b, and c are present in them.

Case 2

Exactly one of p, q, r is zero and rest two are non-zero. Let us examine the subcase where r=0 and p,q are non zero. Other two subcases will be similar.

Then p+q+0 = 31 or p+q=31

Term of the expansion will have :

\(a^p b^q = ab(a^{p-1} b^{q-1})\)

We will show that \(a^{p-1} b^{q-1}\) is divisible by c where p+q=31

Suppose \((p-1)\ge 5 , (q-1)\ge 5\) then c divides \(a^{p-1}\) as it contains \(a^5\)  and by problem c divides \(a^5\)

Again if p-1 < 5 then \((q-1) \ge 25 \) as p-1 + q-1 = 29 (as p+q = 31)

Now a divides \(b^5 \) or \(a^5\) divides \(b^{25}\). As c divides \(a^5\) and \(a^5\) divides \(b^{25}\) hence c divides \(b^{25}\) implying c divides \(b^{q-1}\) as \((q-1)\ge 25 \)

Case 3.

Exactly two of p, q and r are zero. Let us again examine of the three subcases where q=0, r=0 and p nonzero. Other two subcases will be similar.

Then p = 31.

\(a^{31} = a\times a^5 \times a^{25}\). c divides \(a^5\) and b divides \(c^5\) which divides \(a^{25}\).

Hence we have checked all possible terms and have shown than abc divides each of them.