Consider the following polynomial equation [source: Berkeley Problems in Mathematics, problem 6.13.10]:
Let’s try to figure out the rational values of for which is an integer. Clearly, if then is an integer. So let’s consider the case when where and . Substituting this value of we get:
Since, we conclude that . Also it’s clear that . Hence, and we just need to find the possible values of .
For we get:
Hence we have . Since , we have , that is, .
Similarly, for we get . Hence we conclude that the non-integer values of which lead to integer output are: