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:

for all

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