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Is there any way in mathematics to test if a function evaluates to an integer? Or in a better case, retrieve only those function members $i$ and $k$ with whom the function evaluates to integer?

$$j = \dfrac{1}{6}\left(\dfrac{6k-1}{6i-1}-1\right). \qquad \text{for} \qquad i,k\in\mathbb{N}$$

Mike Miller
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T.Kalnius
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1 Answers1

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You have: $${6k-1 \over 6i-1} = 6j+1 \\ 6k-1 = (6i-1)(6j+1) = 36 ij + 6i - 6j - 1 \\ k = 6ij+i-j, i \in \mathbb{N}, j \in \mathbb{Z}$$ For $i=0$ any $k \in \mathbb{N}$ is a solution (if you use an agreement $0 \in \mathbb{N}$). Otherwise, the last expression gives you all possible $k$ for all integer values of $j \ge -{i \over 6i-1} \implies j \ge 0$.

Abstraction
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  • Thank you for your answer, I know that the function I presented evaluates to all possible values, but my question is if there exist some magic tricks or formula to put in any function and do some math with it and get some dependencies or something that would lead me towards my function ajustments into new function witch could result with only integers. – T.Kalnius Jun 16 '16 at 10:49
  • I think it's impossible or at best impractical in general case (take ${1 \over 1 + |(x+1)^{n+3}+(y+1)^{n+3}-(z+1)^{n+3}|}$ for example: proving there are no integral values is the same as proving the Fermat theorem). – Abstraction Jun 16 '16 at 11:48
  • In Excel I would test it using "=int()" what rounds down to the nearest integer and if $f_x=int(f_x)$ then its members are acceptable. Is there any way to express such a round down in Mathematics? – T.Kalnius Jun 16 '16 at 12:04
  • There is "floor" notation ($\lfloor{x}\rfloor$), but the problem (one of the problems) is, there is no simple way to compute something like $\lfloor{\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}}\rfloor$. MS Excel performs approximate calculations and may well get a mistake here too (the expression under $\lfloor{}\rfloor$ operation actually equals $1$). – Abstraction Jun 16 '16 at 12:53