How many positive integers $n$ satisfy $ floor sqrt n floor = 5$?
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Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.
Because we know 3 and -1 (multiplicity 2) are both roots, the last root is such that we can write
There are a few ways we can go about finding , but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be
.
Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have
so the missing root is 3.
Other things we could have tried that spring to mind:
- three rounds of division, dividing the quartic polynomial by , then by
twice, and noting that the remainder upon each division should be 0
- rational root theorem