Rational input, integer output

Standard

Consider the following polynomial equation [source: Berkeley Problems in Mathematics, problem 6.13.10]:

f(t) = 3t^3 + 10t^2 - 3t

Let’s try to figure out the rational values of t for which f(t) is an integer. Clearly, if t\in\mathbb{Z} then f(t) is an integer. So let’s consider the case when t=m/n where \gcd(m,n)=1 and m\neq \pm 1. Substituting this value of t we get:

\displaystyle{f\left(\frac{m}{n}\right) = \frac{3m^3}{n^3} + \frac{10m^2}{n^2} - \frac{3m}{n}= \frac{m(3m^2+10mn-3n^2)}{n^3}=k \in \mathbb{Z}}

Since, n^3\mid (3m^2+10mn-3n^2) we conclude that n\mid 3. Also it’s clear that m\mid k. Hence, n=\pm 3 and we just need to find the possible values of m.

For n=3 we get:

\displaystyle{f\left(\frac{m}{3}\right) =  \frac{m(m^2+10m-9)}{9}=k \in \mathbb{Z}}

Hence we have 9\mid (m^2+10m). Since \gcd(m,n)=\gcd(m,3)=1, we have 9\mid (m+10), that is, m\equiv 8\pmod 9.

Similarly, for m=-3 we get n\equiv 1 \pmod 9. Hence we conclude that the non-integer values of t which lead to integer output are:

\displaystyle{t = 3\ell+ \frac{8}{3}, -3\ell-\frac{1}{3}} for all \ell\in\mathbb{Z}

 

2 responses

  1. I hope you don’t mind…but you seem to be progressing in your chosen passion/profession v well…and sharing these pearls of eternal mathematics with other interested people…this is good “seva”…please keep it up…..

    Liked by 1 person