Determine the total number of ways in which 111 can be written as a sum of three distinct nonzero integers which are in geometric sequence.

Prove that there are no others.

let the 3 terms be: a,a*r,a*r^2
then their sum is:
a+a*r+a*r^2=a*(1+r+r^2)=111
r^2+r+1=111/a=v
r^2+r+1-v=0
r=(-1+-sqrt(1-4(1-v)))/2
r=(-1+-sqrt(4v-3))/2

now for each of a,a*r,a*r^2 to be integers, we need r to be rational.
thus for r to be real
4v-3>=0
v>=3/4
111/a>=3/4
0<a<=148

for each of these a's, we get 2 possible values for r.  And for each of
these pairs of (a,r) we get a possible sequence.  Below are all the
pairs of (a,r) with their corresponding sequences.  I have already
eliminated the (a,r)'s where r is not rational.
(1,10) : {1,10,100}
(1,-11) : {1,-11,121}
(27,4/3) : {27,36,48}
(27,-7/3) : {27,-63,147}
(37,1) : {37,37,37}
(37,-2) : {37,-74,148}
(48,3/4) : {48,36,27}
(48,-7/4) : {48,-84,147}
(100,1/10) : {100,10,1}
(100,-11/10) : {100,-110,121}
(111,0) : {111,0,0}
(111,-1) : {111,-111,111}
(121,-1/11) : {121,-11,1}
(121,-10/11) : {121,-110,100}
(147,-3/7) : {147,-63,27}
(147,-4/7) : {147,-84,48}
(148,-1/2) : {148,-74,37}

now we can further eliminate sequences which either contain zero or
a duplicate value.  Which gives us:
(1,10) : {1,10,100}
(1,-11) : {1,-11,121}
(27,4/3) : {27,36,48}
(27,-7/3) : {27,-63,147}
(37,-2) : {37,-74,148}
(48,3/4) : {48,36,27}
(48,-7/4) : {48,-84,147}
(100,1/10) : {100,10,1}
(100,-11/10) : {100,-110,121}
(121,-1/11) : {121,-11,1}
(121,-10/11) : {121,-110,100}
(147,-3/7) : {147,-63,27}
(147,-4/7) : {147,-84,48}
(148,-1/2) : {148,-74,37}

this list can be further simplified using the fact that if you have
(a,r) (a,b,c) as a sequence, then
(c,1/r) (c,b,a) is also a sequence
so by requiring a<c we can narrow it down to the following solutions:
(1,10) : {1,10,100}
(1,-11) : {1,-11,121}
(27,4/3) : {27,36,48}
(27,-7/3) : {27,-63,147}
(37,-2) : {37,-74,148}
(48,-7/4) : {48,-84,147}
(100,-11/10) : {100,-110,121}

Posted by Daniel on 2013-08-08 10:55:54