Maybe using this knowledge we have c. Then we can get our c^2 record which could be interesting. Then, for records c^2 through c^2 + 2c we know that the d for these is c. Also, the continued fractions would be terminated with the value 2c. Also, if the length of the repeated segment of the repeated fraction 2, then we would know that the value e for that would be a factor of 2c. If e odd, then e | c, if e even, then e/2 | c. So theoretically we would only need to do 2c iterations of this and then we would also only need to compute the first 3 terms of the continued fraction for the square root (ie for [12; 1, 2] where 12 is the original integer value) of that value.
METHOD:
for each value c^2+e:
do the first 3 terms of the continued fraction
if the third term (or the second after the start) is == 2c:
e is a factor of 2c
else:
continue