Let us write  

x = (a x 10²) + (b x 10) + c; y = (b x 10²) + (c x 10) + a; z = (c x 10²) + (a x 10) + b.  

We are given that y² = xz. This means  

(b x 10²) + (c x 10) + a)²  =  ((a x 10²) + (b x 10) + c))((c x 10²) + (a x 10) + b)).  

We can solve for c and get  

c = (10b² - a²)/(10a - b).  

If a, b, c are digits leading to a solution, and if d = gcd(a; b) then d|c. Consequently, we may  assume that gcd(a; b) = 1. Now  

c = (999a²)/(10a - b) - (10b + 100a),  

showing that 10a - b divides 999a². Since a; b are relatively prime, this is possible only if  10a - b is a factor of 999. It follows that 10a - b takes the values 1,3,9,27,37. These values  lead to the pairs  

(a, b) = (1; 9), (1; 7), (1; 1), (4; 3).  

We can discard the first two pairs as they lead to a value of c > 10. The third gives the  trivial solution (111, 111, 111). Taking d = 2, 3, 4, 5, 6, 7, 8, 9, we get 9 solution.  (abc)10 = 111, 222, 333 444, 555, 666, 777, 888, 999.  

The last pair gives c = 2 and hence the solution (432,324, 243). Another solution is obtained  on multiplying by 2: (864, 648, 486).  

Thus we have  

(abc)₁₀ = 111, 222, 333, 444, 555, 666, 777, 888, 999, 432,864.