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.
