Fractions, Floors, Arithmetic and Geometric Series!

If $b < c$ , the middle term of the arithmetic series is zero (requiring one of the other terms to be negative), and the middle term of the geometric series is one (requiring one of the other terms to be fractional). Therefore we conclude $c > b$ .

CASE 1 : $a < b$ . The arithmetic series now takes the form $0, s, 2s$ , and the geometric series is $1, t, t^2$ for positive integers $s$ and $t$ ; we see that $s = \left\lfloor \frac bc \right\rfloor,\ \ t = \left\lceil \frac bc \right\rceil,$ $2s = \left\lfloor \frac ca \right\rfloor,\ \ t^2 = \left\lceil \frac ca \right\rceil.$ Moreover, $t = s + \delta$ , where $\delta = 0$ if $b$ divides $c$ and $\delta = 1$ otherwise; likewise, $t^2 = 2s + \epsilon$ depending on whether $c$ divides $a$ . Thus we get the equation $2s + \epsilon = (s+\delta)^2.$ The combinations $(\delta,\epsilon) = (0, 1); (1, 0); (1,1)$ yield no positive integer solutions.

If $(\delta,\epsilon) = (0,0)$ , we solve $2s = s^2$ to find $s = t = 2$ , so that $c = 4a, b = 2c = 8a$ . This results in 12 solutions: $(a,b,c) = (n,8n,4n)\ \ \ n = 1\dots 12.$

CASE 2 : $a > b$ . The arithmetic series now takes the form $2s, s, 0$ , and the geometric series is $t^2, t, 1$ . This case is nearly identical to Case 1. We get $a = 4b$ , $b = 2c$ , and 12 solutions of the form $(a,b,c) = (8n,2n,n)\ \ \ n = 1\dots 12.$

In total we have therefore $\boxed{24}$ solutions.

Lazy solution

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int N = 0;

for (int a = 1; a < 100; a++) for (int b = 1; b < 100; b++)
    if (a != b) for (int c = 1; c < 100; c++) if (a != c && b != c) {

        int s1 = a/b, s2 = b/c, s3 = c/a;
        int t1 = (a+b-1)/b, t2 = (b+c-1)/c, t3 = (c+a-1)/a;

        if (s1 + s3 == s2 + s2 && t1 * t3 == t2 * t2) {

            N++;
            System.out.println(N + ": " + a + ", " + b + ", " + c);
        }
    }

Output:

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24

1: 1, 8, 4
2: 2, 16, 8
3: 3, 24, 12
4: 4, 32, 16
5: 5, 40, 20
6: 6, 48, 24
7: 7, 56, 28
8: 8, 2, 1
9: 8, 64, 32
10: 9, 72, 36
11: 10, 80, 40
12: 11, 88, 44
13: 12, 96, 48
14: 16, 4, 2
15: 24, 6, 3
16: 32, 8, 4
17: 40, 10, 5
18: 48, 12, 6
19: 56, 14, 7
20: 64, 16, 8
21: 72, 18, 9
22: 80, 20, 10
23: 88, 22, 11
24: 96, 24, 12

It is impossible that $a>b$ , $b>c$ and $c>a$ , as from $a>b$ and $b>c$ we get $a>c$ , which is a contradiction. Therefore, either $a<b$ , $b<c$ or $c<a$ is true (but not all true).

$b<c$ is not possible as then $\left \lceil \dfrac{b}{c} \right \rceil=1$ , and because $\left \lceil \dfrac{a}{b} \right \rceil$ and $\left \lceil \dfrac{c}{a} \right \rceil$ cannot both be 1, then condition b) cannot be satisfied as there would have to be a positive integer less than 1. Therefore, $a<b$ or $c<a$ .

Say $a<b$ , then $\left \lceil \dfrac{a}{b} \right \rceil=1$ , so we can say $\left \lceil \dfrac{b}{c} \right \rceil=k$ and $\left \lceil \dfrac{c}{a} \right \rceil=k^{2}$ for k is a positive integer greater than 1.

Also, then $\left \lfloor \dfrac{a}{b} \right \rfloor=0$ , $\left \lfloor \dfrac{b}{c} \right \rfloor=k,k-1$ and  $\left \lfloor \dfrac{c}{a} \right \rfloor=k^{2},k^{2}-1$ .

Now we can find the values of $\left \lfloor \dfrac{b}{c} \right \rfloor$ and $\left \lfloor \dfrac{c}{a} \right \rfloor$ . As condition a) is an arithmetic series, there are 4 cases:

1) $k-1=k^{2}-1-(k-1)$ ,

$k^{2}-2k+1=0$ ,

$(k-1)^{2}=0$ ,

$k=1$ .

However, then $\left \lfloor \dfrac{b}{c} \right \rfloor=0$ and $\left \lfloor \dfrac{c}{a} \right \rfloor=0$ , which is impossible, as then $a<b$ , $b<c$ and $c<a$ which is untrue (above).

2) $k-1=k^{2}-(k-1)$ ,

$k^{2}-2k+2=0$ and this has no integral solutions.

3) $k=k^{2}-1-k$ ,

$k^{2}-2k-1=0$ and this has no integral solutions.

4) $k=k^{2}-k$ ,

$k^{2}-2k=0$ .

As k is not 0, $k=2$ .

Therefore, $\left \lceil \dfrac{b}{c} \right \rceil=2=\left \lfloor \dfrac{b}{c} \right \rfloor$ , and $\left \lceil \dfrac{c}{a} \right \rceil=4=\left \lfloor \dfrac{c}{a} \right \rfloor$ . As the floor functions are the same as the roof functions, this means $b=2c$ and $c=4a$ . Therefore, $b=8a$ and $c=2a$ .

As b is not more than 100, the maximum of a is 12 so there are 12 possibilities for a. Similarly, there are 12 possibilities for when $c<a$ , so in total there are 12+12=24 ordered triples.

Apologies if the method looks messy!